\subsubsection*{Acknowledgment.}}
\defF_{\sigma\delta}{F_{\sigma\delta}}
\def\mathbb{N}_0{\mathbb{N}_0}

\renewcommand{\geq}{\geqslant}
\renewcommand{\leq}{\leqslant}

\newcommand{transcendental entire function}{transcendental entire function}
\newcommand\qfor{\text{for }}

\DeclarePairedDelimiter{\floor}{\langle}{\rangle}

\makeatletter
\def\xdef\@thefnmark{}\@footnotetext{\xdef\@thefnmark{}\@footnotetext}

\makeatother

\begin{document}

\title[Escaping sets of continuous functions]{Escaping sets of continuous functions}
\author{Ian Short, \, David J. Sixsmith}
\address{Department of Mathematics and Statistics \\ The Open University \\   Milton Keynes MK7 6AA\\   UK}
\email{Ian.Short@open.ac.uk}
\address{School of Mathematical Sciences \\ University of Nottingham \\ Nottingham
NG7 2RD \\ UK}
\email{David.Sixsmith@nottingham.ac.uk}

\subjclass[2000]{Primary: 37F10; Secondary: 30C65, 30D05}
\thanks{The second author was supported by Engineering and Physical Sciences Research Council grant EP/L019841/1.}
\begin{abstract}
Our objective is to determine which subsets of $\mathbb{R}^d$ arise as escaping sets of continuous functions from $\mathbb{R}^d$ to itself. We obtain partial answers to this problem, particularly in one dimension, and in the case of open sets. We give a number of examples to show that the situation in one dimension is quite different from the situation in higher dimensions. Our results demonstrate that this problem is both interesting and perhaps surprisingly complicated.
\end{abstract}
\maketitle

\section{Introduction}
\subsection{Background}
Given a function $f : \mathbb{R}^d \to \mathbb{R}^d$, we denote the $k$th iterate of $f$ by $f^k$, for any positive integer $k$, and we define the \emph{escaping set} $I(f)$ of $f$ to be
\[
I(f) = \{ x \in \mathbb{R}^d : f^k(x) \rightarrow\infty \text{ as } k\rightarrow\infty \}.
\]
The escaping set of an entire function is a familiar object in complex dynamics. Such sets were first studied in generality by Eremenko \cite{MR1102727}, who showed that if $f$ is a transcendental entire function, then $I(f)$ is never empty and  the closure of $I(f)$ has no bounded components. Eremenko conjectured that, in fact, $I(f)$ has no bounded components; this conjecture remains open and has stimulated much further investigation.

In \cite{MR2448586}, escaping sets of quasiregular functions were studied; we refer to, for example, \cite{MR950174} for a definition of a quasiregular function. It was shown in \cite{MR2448586} that if $f : \mathbb{R}^d \to \mathbb{R}^d$ is a quasiregular function, then, providing $f$ satisfies a certain growth condition, the escaping set $I(f)$ necessarily contains an unbounded component. Also, an example was given of a quasiregular function with an essential singularity at infinity for which the closure of the escaping set has a bounded component. This shows that Eremenko's conjecture fails for quasiregular functions.

The escaping set can be defined for \emph{any} function from $\mathbb{R}^d$ to itself -- the function need not be complex differentiable or quasiregular -- so it is reasonable to study escaping sets in a more general context. In this paper, we investigate the structure of escaping sets of \emph{continuous} functions from $\mathbb{R}^d$ to itself (our functions need not even be surjective). More explicitly, we address the following question: \emph{given a subset $S$ of $\mathbb{R}^d$, does there exist a continuous function $f : \mathbb{R}^d \to \mathbb{R}^d$ such that $S = I(f)$?} We give examples to show that this question, in its full generality, is difficult. We make headway with some special cases, particularly in one dimension, and highlight some interesting differences between the theory of escaping sets of continuous functions and the theory of escaping sets of transcendental entire functions. We are not aware of any other study of escaping sets in this generality.

To facilitate our discussion, we make the following definition.

\begin{definition}
Suppose that $S$ is a subset of $\mathbb{R}^d$ . Then we say that $S$ is an \emph{$I_d$-set} if there is a continuous function $f : \mathbb{R}^d \to \mathbb{R}^d$ such that $S = I(f)$.
\end{definition}

Notice that the collection of $I_d$-sets is invariant under homeomorphisms, in the following sense. Let us denote the set $\mathbb{R}^d\cup\{\infty\}$ by $\overline{\mathbb{R}^d}$, and endow it with the usual topology of a one-point compactification. Suppose that $S$ and $S'$ are subsets of $\mathbb{R}^d$, and there is a homeomorphism $\phi : \overline{\mathbb{R}^d} \to \overline{\mathbb{R}^d}$ such that $\phi(S) = S'$ and $\phi(\infty) = \infty$. Then $S$ is an $I_d$-set if and only if $S'$ is an $I_d$-set. This follows from the fact that if $S = I(f)$, then $S' = I(\phi \circ f \circ \phi^{-1})$.

If $f$ is continuous, then $I(f)$ is either empty or unbounded. Moreover, we can write
\[
I(f) = \bigcap_{p=1}^\infty \bigcup_{q=1}^\infty \bigcap_{m \geq q} \{ x : |f^m(x)| \geq p \}.
\]
 It follows that a necessary condition for a non-empty set $S$ to be an $I_d$-set is that $S$ is an unbounded $F_{\sigma\delta}$ set. However, there are many examples in this paper of unbounded $F_{\sigma\delta}$ sets that are not $I_d$-sets. In order to obtain sufficient conditions for sets to be $I_d$-sets, we must impose further restrictions on them; first we look at open sets, and then sets in one dimension. The reader may care to peruse Figure~\ref{fig:sets} and speculate as to which of the six open planar sets, shown in black, are $I_2$-sets; all will be revealed later.

\begin{figure}[ht]
\subfloat[]{\includegraphics[width=.28\textwidth]{figA}}\quad
\subfloat[]{\includegraphics[width=.28\textwidth]{figB}}\quad
\subfloat[]{\includegraphics[width=.28\textwidth]{figC}}\hfill
\subfloat[]{\includegraphics[width=.28\textwidth]{figD}}\quad
\subfloat[]{\includegraphics[width=.28\textwidth]{figE}}\quad
\subfloat[]{\includegraphics[width=.28\textwidth]{figF}}\hfill
\caption{(a) $\mathbb{R}^2\setminus (\mathbb{Z}\times\{0\})$; (b) $\bigcup_{n\in\mathbb{Z}} D(n,\tfrac14)$; (c) $\bigcup_{n=1}^\infty S(0,n,\pi/2^{n-1})$;\\ (d) $D(0,1)\cup \bigcup_{n=1}^\infty S(1,n,\pi/2^{n-1})$; (e) an infinite snake-like region that accumulates on $[0,+\infty)$; (f) the same snake-like region together with the set $D(5,1)\cap \{(x,y)\,:\,y<0\}$}
\label{fig:sets}
\end{figure}

In the caption of Figure~\ref{fig:sets}, we denote the open disc with radius $r$ and centre $c$ by $D(c,r)$. Also, for positive numbers $r$, $s$ and $\phi$, we write
\[
S(r,s,\phi) = \left\{(r\cos\phi+t\cos\theta,r\sin\phi+t\sin\theta)\,:\, |\theta-\phi|<\phi/4, \ 0<t<s\right\},
\]
which is an open sector of the disc $D((r\cos\phi,r\sin\phi),s)$ of angle $\phi/2$. We remark that none of the open sets displayed in the figure are the escaping sets of transcendental entire functions, as the escaping sets of such functions are never open. This follows from the observation that if, on the contrary, $f$ is a transcendental entire function and $I(f)$ is open, then the \emph{Julia set} of $f$, which is equal to $\partial I(f)$, does not meet $I(f)$, which contradicts a result of \cite{MR1102727}; we refer to \cite{MR1216719} for further background and definitions.

It is useful to observe now that, if $f:\mathbb{R}^d\to\mathbb{R}^d$, then $I(f)$ is \emph{completely invariant}, in the sense that $f(I(f))\subset I(f)$ and $f^{-1}(I(f)) \subset I(f)$.

\subsection{Open sets}

The results in this subsection are about classifying \emph{open} $I_d$-sets. We follow the convention here (and later on) that if $d$ is not specified, then it can take any positive integer value. We define an \emph{unbounded curve} to be a continuous injective function $\gamma:[0,+\infty)\to \mathbb{R}^d$ such that the image set $\gamma([0,+\infty))$ is unbounded. We say that infinity is \emph{accessible} from an unbounded open subset $U$ of $\mathbb{R}^d$ if there is an unbounded curve $\gamma:[0,+\infty)\to U$ that does not  accumulate at any point of $\mathbb{R}^d$, in the sense that $\gamma(t)\to \infty$ as $t\to\infty$. Such a curve is a homeomorphism onto its image.

\begin{theorem}\label{theo:hasanunboundedcomponent}
Any open, unbounded subset of $\mathbb{R}^d$ from which infinity is accessible is an $I_d$-set.
\end{theorem}

We remark that Baker \cite[Theorem 2]{MR980793} showed that infinity is always accessible in a so-called \emph{Baker domain} of a {transcendental entire function}; a Baker domain is an open set in which the iterates tend to infinity locally uniformly, and so always lies in the escaping set. Later on we will see two examples, Examples~\ref{example:snake} and \ref{example:failsnake}, of unbounded subsets of $\mathbb{R}^2$ from which infinity is \emph{not} accessible. In the first example, the unbounded set \emph{is} an $I_2$-set, and in the second, the unbounded set is \emph{not} an $I_2$-set. The set in the second example, however, has two components; it is not a domain. The question of whether every unbounded domain in $\mathbb{R}^2$ is an $I_2$-set remains open.

Next we consider open sets without unbounded components. There certainly are open sets without unbounded components that are $I_2$-sets: Example~\ref{example.open} (to come) provides an example of one. The following theorem gives necessary conditions for an open set without unbounded components to be an $I_d$-set.

\begin{theorem}\label{theo:hasnounboundedcomponent}
Suppose that $S$ is a non-empty open subset of $\mathbb{R}^d$ that has no unbounded components. If $S$ is an $I_d$-set, then there exist a sequence $(V_k)_{k\in\mathbb{N}}$ of components of $S$ and $r>0$, such that
\begin{enumerate}[label=\emph{(\roman*)}]
\item $\sup\{|x|\,:\, x \in V_k\} \rightarrow +\infty$ as $k\rightarrow\infty$;\label{t21}
\item $\inf\{|x|\,:\,x \in V_k\} \leq r$, for $k\in\mathbb{N}$.\label{t22}
\end{enumerate}
\end{theorem}

We show later (see Example~\ref{example.nastyone}) that the conditions of Theorem~\ref{theo:hasnounboundedcomponent}  are not sufficient  for an open set with no unbounded components to be an $I_d$-set.

\subsection{One dimension} \label{d1case}

Let us now consider escaping sets in one dimension. We can immediately classify the open $I_1$-sets using Theorems~\ref{theo:hasanunboundedcomponent} and \ref{theo:hasnounboundedcomponent}, because it is impossible for an open set in $\mathbb{R}$ to satisfy the conditions of Theorem~\ref{theo:hasnounboundedcomponent}.

\begin{theorem} \label{theo:nounb}
A non-empty open subset of $\mathbb{R}$ is an $I_1$-set if and only if it has an unbounded component.
\end{theorem}

Our next result shows that even in one dimension, escaping sets can have a complicated structure.

\begin{theorem}\label{theo:R1irrational}
The set of irrationals is an $I_1$-set.
\end{theorem}

It is well known (see, for example, \cite{Sierpinski1920}) that if $X$ is a countable, dense subset of $\mathbb{R}$, then there is a homeomorphism $\phi:\overline{\mathbb{R}}\to\overline{\mathbb{R}}$ with $\phi(\infty)=\infty$ and $\phi(X)=\mathbb{Q}$. Therefore the complement of $X$ in $\mathbb{R}$ is an $I_1$-set.

In view of Theorem~\ref{theo:R1irrational}, it is natural to ask whether the set of integers, or the set of rational numbers, is an $I_1$-set. It follows from the next result that this is not the case.

\begin{theorem}\label{theo:R1uncountable}
Suppose that $S$ is an $I_1$-set, and that $x \in S$. Then every neighbourhood of $x$ contains uncountably many elements of $S$.
\end{theorem}

Our final result in the one-dimensional setting can be seen as step towards classifying $I_1$-sets.

\begin{theorem}\label{theo:R1open}
If $S$ is an $I_1$-set, then all components of the complement of $S$ are closed.
\end{theorem}

\subsection{Higher dimensions}

Classifying $I_d$-sets when $d\geq 2$ is far more difficult than classifying $I_1$-sets. We give several examples that illustrate the differences between the one- and higher-dimensional cases. Here are brief descriptions of three such examples.

\begin{enumerate}[label=(\roman*),leftmargin=20pt,topsep=0pt,itemsep=0pt]
\item In contrast to Theorem~\ref{theo:R1open}, we give an $I_2$-set, the complement of which is a domain (Example~\ref{example.halfplane}).
\item In contrast to Theorem~\ref{theo:R1uncountable}, we give a countable  nowhere dense set, which is an $I_2$-set (Example~\ref{example.Z}).
\item In contrast to the previous example, we give a countable nowhere dense set, the complement of which is an $I_2$-set (Example~\ref{example.Zcomp}).
\end{enumerate}

Although, for simplicity, all these examples are in two dimensions, it is clear that they can be extended to any dimension greater than one.

\subsection{Structure of the paper}

In Section~\ref{sec:Rmore} we prove Theorems~\ref{theo:hasanunboundedcomponent} and~\ref{theo:hasnounboundedcomponent}. Then, in Section~\ref{sec:R1}, we prove Theorems~\ref{theo:R1irrational}, \ref{theo:R1uncountable} and~\ref{theo:R1open}. Finally, in Section~\ref{sec:examples} we give all the examples mentioned above.

\section{Open sets}\label{sec:Rmore}

We require the following lemma, which is a variation, in Euclidean space, of the Tietze extension theorem; see, for example, \cite[Theorem 15.8]{MR0264581}. In the proof of the lemma (and elsewhere in the paper) we use the notation $\text{dist}(x, K)$ to represent the Euclidean distance from a point $x$ to a set $K$ in $\mathbb{R}^d$; that is, $\text{dist}(x, K)= \inf_{y\in K} |x-y|$.

\begin{lemma}\label{lemma:open}
Suppose that $K$ is a closed subset of $\mathbb{R}^d$ and $\phi : K \to [0,+\infty)$ is a continuous function. Then there exists a continuous function $\psi : \mathbb{R}^d \to [0,+\infty)$ such that $\psi|_K = \phi$ and $\phi^{-1}(0) = \psi^{-1}(0)$.
\end{lemma}
\begin{proof}
Let
\begin{equation*}
\psi(x) =
\begin{cases}
\phi(x), &\qfor x \in K, \\
\inf_{y \in K} \left( \phi(y) + \dfrac{|x-y|}{\operatorname{dist}(x, K)} - 1 \right) + \operatorname{dist}(x, K), &\qfor x \notin K.
\end{cases}
\end{equation*}
It is easy to check that $\psi$ has the required properties.
\end{proof}
\begin{proof}[Proof of Theorem~\ref{theo:hasanunboundedcomponent}]
Suppose that $U$ is an unbounded component of an open set $S$ in $\mathbb{R}^d$, and that $\gamma:[0,+\infty)\to U$ is an unbounded curve that does not accumulate at any point of $\mathbb{R}^d$. Clearly $\mathbb{R}^d$ is an $I_d$-set, so we can assume that $U \ne \mathbb{R}^d$. Choose a point $x_1$ in the image of $\gamma$. Consider the set of straight lines emanating from $x_1$. At least one of these lines must meet $\partial U$. Let the first point where one such line meets $\partial U$ be the point $x_0$. Without loss of generality -- replacing $x_1$ with another point of $\gamma$ if necessary -- we can assume that $\gamma$ does not intersect the open line segment from $x_0$ to $x_1$. After reparameterizing we can assume that $\gamma(1)=x_1$.

Now let $\Gamma : [0,+\infty) \to U\cup\{x_0\}$ be an unbounded curve comprised of the line segment from $x_0$ to $x_1$ followed by the part of $\gamma$ from $x_1$ to $\infty$; explicitly,
\[
\Gamma(t) =
\begin{cases}
tx_1+(1-t)x_0, &\qfor 0\leq t\leq 1,\\
\gamma(t), &\qfor t\geq 1.
\end{cases}
\]
We denote the image of $\Gamma$ by $|\Gamma|$; that is, $|\Gamma| = \Gamma([0,+\infty))$.

We now define a continuous function $f : \mathbb{R}^d \to \mathbb{R}^d$ such that $I(f) = S$. First set $f(x) = x_0$, for $x \notin S$. Thus we certainly have that $I(f) \subset S$.

Next we define $f$ on $|\Gamma|$. Define $h : [0,+\infty)\to[0,+\infty)$ by $h(t) = 2t$. Then set
\[
f(x) = (\Gamma \circ h \circ \Gamma^{-1})(x),\quad \text{for}\quad x \in |\Gamma|.
\]
This definition of $f$ agrees with the previous definition at the point $x_0$. As $f$ is conjugate to $h$ on $|\Gamma|$, it follows from the fact that $\Gamma(t)\rightarrow\infty$ as $t\rightarrow\infty$ that $|\Gamma|\setminus\{x_0\} \subset I(f)$.

Next, if $V$ is a component of $S$ other than $U$, then we define
\[
f(x) = \Gamma\left(\operatorname{dist}(x,\partial V)\right),\quad \text{for}\quad  x \in V.
\]
Since $f(x)\rightarrow\Gamma(0) = x_0$ as $x\rightarrow v$, for $v\in\partial V$, this definition ensures that $f$ is continuous in $\mathbb{R}^d\setminus U$. Moreover, $f$ maps all points of $S\setminus U$ to $|\Gamma|\setminus\{x_0\}$. Hence $S\setminus U \subset I(f)$.

It remains to define $f$ in $U \setminus |\Gamma|$. To do this we first use Lemma~\ref{lemma:open} to construct an auxiliary function. Let $K = (\mathbb{R}^d \setminus U) \cup |\Gamma|$. Note that, since $\Gamma$ does not accumulate in $\mathbb{R}^d$, $K$ is closed. Define a continuous function $\phi : K \to [0,+\infty)$ by
\[
\phi(x) =
\begin{cases}
\Gamma^{-1}(x), &\qfor x \in |\Gamma|, \\
0, &\text{otherwise}.
\end{cases}
\]
It follows, by Lemma~\ref{lemma:open}, that there exists a continuous function $\psi~:~\mathbb{R}^d~\to~[0,+\infty)$ such that $\psi|_K = \phi$ and $\psi^{-1}(0) = \phi^{-1}(0) = \mathbb{R}^d \setminus U$. We can now see that
\begin{enumerate}[label=(\alph*)]
\item $\psi(x) = 0$, for $x \notin U$;
\item $\psi(x) > 0$, for $x \in U$;
\item $\psi(x) = \Gamma^{-1}(x)$, for $x \in |\Gamma|$.
\end{enumerate}
To finish, we define $f(x) = (\Gamma \circ h \circ \psi)(x)$, for $x \in U \setminus |\Gamma|$. Conditions (a) and (c)  ensure that this definition of $f$ is consistent with previous definitions, and that $f : \mathbb{R}^d \to \mathbb{R}^d$ is continuous. Condition (b)  ensures that if $x \in U$, then $f(x) \in |\Gamma| \setminus \{ x_0 \}$. It follows that $U = I(f)$, as required.
\end{proof}

In the remainder of the paper we write $\mathbb{N}_0$ for the set $\mathbb{N}\cup\{0\}$.

\begin{proof}[Proof of Theorem~\ref{theo:hasnounboundedcomponent}]
Suppose that $S=I(f)$, where $f : \mathbb{R}^d \to \mathbb{R}^d$ is a continuous function, and, as stated in the hypotheses of the theorem, suppose that $S$ is open with no unbounded components. Let $U$ be a component of $S$. Let $U'$ be the component of  $S$ containing $f(U)$. Clearly $U \ne U'$. Notice also that, by continuity, $f(\partial U)\subset \partial U'$.

Using these observations we can construct a sequence $(U_i)_{i\in\mathbb{N}_0}$ of components of $S$ such that $f(U_{i-1})\subset U_i$ and $f(\partial U_{i-1})\subset \partial U_i$, for $i\in\mathbb{N}$. Clearly the sets $U_i$ are pairwise disjoint. Now choose any point $x_0$  in $\partial U_0$; because $x_0\notin I(f)$, there is a sequence $(k_i)_{i\in\mathbb{N}}$ of positive integers and a positive constant $r$ such that $|f^{k_i}(x_0)|\leq r$, for $i\in\mathbb{N}$. Define $V_i=U_{k_i}$, for $i\in\mathbb{N}$. Since $f^{k_i}(x_0)\in \partial V_i$, we see that $\inf\{|x|\,:\,x \in V_i\} \leq r$, for $i\in\mathbb{N}$. Also, $\sup\{|x|\,:\, x \in V_i\} \rightarrow +\infty$ as $i\rightarrow\infty$ because $f^{k_i}(U_0)\subset V_i$ and $ U_0$ is contained in the escaping set of $f$. This completes the proof of the theorem.
\end{proof}
\section{One dimension}\label{sec:R1}

In this section we prove various results about $I_1$-sets; later we will see that there are no comparable results for $I_d$-sets, where $d>1$. Before we begin we have some remarks about notation. One way in which the real line $\mathbb{R}$ is distinguished among the spaces $\mathbb{R}^d$, for $d\in\mathbb{N}$, is that it has two topological ends (two `infinities'), whereas all the other spaces have one end. A consequence of this is that  statements about convergence to infinity in one dimension can be ambiguous. So far we have followed the convention that if $(x_n)_{n\in\mathbb{N}}$ is a sequence in $\mathbb{R}^d$, then the statement $x_n\to\infty$ as $n\to\infty$ means that the sequence eventually leaves any compact subset of $\mathbb{R}^d$. We will continue to follow this convention, even when $d=1$. In the one-dimensional case, we write $x_n\to +\infty$ as $n\to\infty$ to mean that the sequence $(x_n)_{n\in\mathbb{N}}$ eventually leaves any interval of the form $(-\infty,K)$, and we write $x_n\to -\infty$ as $n\to\infty$ if $-x_n\to+\infty$ as $n\to\infty$. (In fact, we have used this notation for convergence to $+\infty$ a couple of times already.)

\begin{proof}[Proof of Theorem~\ref{theo:R1irrational}]

We define a continuous function $f : \mathbb{R} \to \mathbb{R}$ as follows (see Figure~\ref{fig1}):
\[
f(x) = \begin{cases}
   x + 1,    &\qfor x < \frac{3}{4},\\
   4n - 3x, &\qfor n - \frac{1}{4} \leq x < n,  n\in\mathbb{N},\\
   5x - 4n, &\qfor n \leq x < n + \frac{1}{4}, n \in\mathbb{N}, \\
   4n + 2 - 3x, &\qfor n + \frac{1}{4} \leq x < n + \frac{1}{2}, n\in\mathbb{N},\\
   5x - 4n - 2, &\qfor n + \frac{1}{2} \leq x < n + \frac{3}{4}, n \in\mathbb{N}.
\end{cases}
\]

\begin{figure}[ht]
\includegraphics[width=.6\textwidth]{figG}
\caption{Graph of the function $f$}
\label{fig1}
\end{figure}

Notice that $f(x)\geq x$, for $x\in\mathbb{R}$, and if $x>\tfrac34$, then for all but countably many values of $x$, the derivative $f'(x)$ exists and satisfies  $|f'(x)| \geq 3$.

Suppose that $x = n + a$, where $n\in\mathbb{N}$ and $a \in \left\{-\frac{1}{4}, \frac{1}{4}\right\}$. Then $f(x) = x+1$, and it follows that $x \in I(f)$. On the other hand, suppose that $x = n + b$, where $n\in\mathbb{N}$ and $b \in \left\{0, \frac{1}{2}\right\}$. Then $f(x) = x$, so $x \notin I(f)$.

Next, suppose that $\Delta$ is a bounded open interval in $\mathbb{R}$. It is easy to see that there exists a positive integer $N$ such that the interval $f^N(\Delta)$ lies in $\left(\frac{3}{4}, +\infty\right)$ and has diameter greater than $\frac{1}{2}$. It follows that $f^N(\Delta)$ meets both $I(f)$ and $\mathbb{R}\setminus I(f)$. Hence $\Delta$ meets both $I(f)$ and $\mathbb{R}\setminus I(f)$, so $I(f)$ and $\mathbb{R}\setminus I(f)$ are dense in $\mathbb{R}$.

Let $y$ be a point of $\mathbb{R}\setminus I(f)$. As $y\leq f(y)\leq f^2(y)\leq \dotsb$, the iterates $f^k(y)$ must be bounded, so there exists a fixed point $p$ of $f$ such that $f^k(y)\rightarrow p$ as $k\rightarrow\infty$. If $f^k(y)$ lies within a distance of $\tfrac14$ from $p$, but is not equal to $p$, then $f^{k+1}(y)$ is greater than $p$, which is impossible. Therefore $f^m(y)= p$, for some positive integer $m$. We deduce that $\mathbb{R}\setminus I(f)$ is countable.

To summarise, the set $\mathbb{R}\setminus I(f)$ is a countable, dense subset of the real line. It  follows that there exists a homeomorphism $\phi : \overline{\mathbb{R}} \to \overline{\mathbb{R}}$ with $\phi(\infty)=\infty$ and $\phi(\mathbb{R}\setminus I(f)) =  \mathbb{Q}$. We deduce that
\[
I(\phi \circ f \circ \phi^{-1})=\mathbb{R}\setminus \mathbb{Q},
\]
and this completes the proof.
\end{proof}

We use the following two lemmas in the proof of Theorem~\ref{theo:R1uncountable}. Recall that $\mathbb{N}_0=\mathbb{N}\cup\{0\}$.

\begin{lemma}\label{lemm:sequence}
Suppose that $f : \mathbb{R}^d \to \mathbb{R}^d$ is a continuous function, $(n_k)_{k\in\mathbb{N}_0}$ is a sequence of positive integers, and $(E_k)_{k\in\mathbb{N}_0}$ is a sequence of non-empty compact subsets of $\mathbb{R}^d$ such that $E_{k+1} \subset f^{n_k}(E_k)$, for $k\in\mathbb{N}_0$. Then there exists $y\in E_0$ such that $f^{s_k}(y) \in E_{k+1}$, where $s_k=\sum_{i=0}^{k}n_i$, for $k\in\mathbb{N}_0$.
\end{lemma}
\begin{proof}
Let
\[
F_k = \{x\in E_0\,:\, f^{s_i}(x)\in E_{i+1}, i=0,1,\dots,k\},\quad\text{for}\quad k\in\mathbb{N}_0.
\]
Then $(F_k)_{k\in\mathbb{N}_0}$ is a decreasing sequence of non-empty compact sets, so $F = \bigcap_{k=0}^\infty F_k$ is non-empty. We can choose any point $y$ in $F$ to complete the proof.
\end{proof}

In the next lemma, we use the special case of Lemma~\ref{lemm:sequence} when each $n_k=1$ to construct points in an escaping set. Later on, we apply the more general version of Lemma~\ref{lemm:sequence} to construct a point $x$ for which the iterates $(f^k(x))_{k\in\mathbb{N}_0}$ have a bounded subsequence.

\begin{lemma}\label{jdf}
Let $f:\mathbb{R}\to\mathbb{R}$ be a continuous function. Suppose that there is an element $x$ of $I(f)$ such that the sequence $(f^n(x))_{n\in\mathbb{N}}$ is unbounded above. Then given any positive constant $R$ there exists a compact interval $J\subset (R,+\infty)$ such that $J\cap I(f)$ is uncountable.
\end{lemma}
\begin{proof}
Observe that no two of the iterates $(f^k(x))_{k\in\mathbb{N}_0}$ are equal. For each $k\in\mathbb{N}_0$, let $B_k^+=[f^{k}(x), f^{p_k}(x)]$, where $p_k$ is such that
\[
f^{p_k}(x) = \min_{m\in\mathbb{N}_0} \{f^m(x)\, :\, f^m(x) > f^{k}(x) \}.
\]
Similarly, let $B_k^-=[f^{q_k}(x), f^{k}(x)]$, where $q_k$ is such that
\[
f^{q_k}(x) = \max_{m\in\mathbb{N}_0} \{f^m(x)\,:\, f^m(x) < f^{k}(x) \}.
\]
(If the set $\{m\in\mathbb{N}_0\,:\, f^m(x) < f^{k}(x) \}$ is empty, which can happen for at most one value of $k$, then we define $B_k^-=(-\infty,f^k(x)]$.) Let $\mathfrak{B}$ denote the collection of all the intervals $B_k^+$ and $B_k^-$. These intervals cover $\mathbb{R}$. All of them are bounded on the right, and at most one is unbounded on the left. Any two intervals from the collection $\mathfrak{B}$ are either disjoint or else intersect at one of their end points.

Let $C = (C_k)_{k\in\mathbb{N}_0}$ be a sequence of intervals from $\mathfrak{B}$. We say that $C$ is a \emph{covering sequence starting at $C_0$} if $C_{k+1} \subset f(C_k)$, for $k\in\mathbb{N}_0$, and $\inf\{|x|\,:\,x\in C_k\}\rightarrow +\infty$ as $k\rightarrow\infty$. Observe that, for each $k \in \mathbb{N}_0$, $f(B_k^+)$ contains either $B_{k+1}^+$ or $B_{k+1}^-$, and $f(B_k^-)$ contains either $B_{k+1}^+$ or $B_{k+1}^-$. Thus, if $C_0 \in \mathfrak{B}$, then there exists at least one covering sequence starting at $C_0$.

Suppose that there exists more than one covering sequence starting at each element of $\mathfrak{B}$. Choose any interval $C_0$ of $\mathfrak{B}$ that satisfies $C_0\subset (R,+\infty)$. It is easy to see that there are uncountably many different covering sequences starting at $C_0$. We can apply Lemma~\ref{lemm:sequence} (with $n_k=1$, for $k\in\mathbb{N}$) to obtain an uncountable collection of points in $C_0\cap I(f)$. The lemma follows on choosing $J=C_0$.

Let us assume instead that there is a set $C_0$ in $\mathfrak{B}$ for which there is a unique covering sequence, $C = (C_k)_{k\in\mathbb{N}_0}$ say, starting at $C_0$. By starting from $C_1$ instead of $C_0$, if necessary, we can assume that $C_0$ is bounded.  If $f(C_0) \ne C_1$, then $f(C_0)$ contains two elements of $\mathfrak{B}$, so there is more than one covering sequence starting at $C_0$, contrary to assumption. Hence $f(C_0) = C_1$, and for the same reason $f(C_k) = C_{k+1}$, for $k\in\mathbb{N}$.

Let $f^{a}(x)$ and $f^b(x)$ be the end points of $C_0$, where $0\leq a<b$. Then $f^{a+k}(x)$ and $f^{b+k}(x)$ are the end points of $C_k$, for $k\in\mathbb{N}_0$. Suppose, in order to reach a contradiction, that $f^{b+k}(x)<f^{a+k}(x)$, for $k\in\mathbb{N}_0$. Let $m=b-a$. By choosing $k=qm+r$, where $r\in\{0,1,\dots,m-1\}$, for values $q=0,1,2,\dotsc$ in turn, we see that
\[
f^{a+r}(x)>f^{a+m+r}(x)>f^{a+2m+r}(x)>\dotsb.
\]
Therefore the sequence $(f^n(x))_{n\in\mathbb{N}_0}$ is bounded above, which is a contradiction. Thus, on the contrary, there is an integer $k\in\mathbb{N}_0$ for which $f^{a+k}(x)<f^{b+k}(x)$, so that $C_k=[f^{a+k}(x),f^{b+k}(x)]$. By replacing $x$ with $f^{a+k}(x)$, and relabelling our covering sequence, we can assume that $C_0=[x,f^m(x)]$. (Making this change increases the size of a finite number of the intervals from the collection $\mathfrak{B}$.) Let $g=f^m$, and let $D_k=C_{mk}$, for $k\in\mathbb{N}_0$. It follows that $D_k = [g^k(x),g^{k+1}(x)]$ and $g(D_k) = D_{k+1}$, for $k\in\mathbb{N}_0$. Notice that $I(f)=I(g)$.

We can assume that the set of positive fixed points of $g$ is unbounded above, as otherwise $(K,+\infty)\subset I(f)$, for some number $K$. Let $(k_n)_{n\in\mathbb{N}_0}$ be an increasing sequence of positive integers such that $D_{k_n}$ contains a fixed point $t_n$ of $g$, for $n\in\mathbb{N}_0$.

Let $\underline{s} = s_0 s_1 s_2 \ldots$ be an infinite sequence of $0$s and $1$s. There are uncountably many such sequences. We construct a sequence $(E_n)_{n\in\mathbb{N}_0}$, inductively, for use in Lemma~\ref{lemm:sequence}. Set $E_k = D_k$, for $0~\leq~k~\leq~k_0$. Note that
\[
[t_0, g^{k_0+1}(x)] \cup D_{k_0+1} \subset g([t_0, g^{k_0+1}(x)]) \subset g(D_{k_0}).
\]
If $s_0 = 0$, then define
\[
E_{k_0+1}=[t_0, g^{k_0+1}(x)]\quad\text{and}\quad E_{k_0+2} = D_{k_0+1},
\]
and if $s_0=1$, then define
\[
E_{k_0+1} = D_{k_0+1}.
\]

Suppose now that the values $s_0,\dots,s_p$ have been used to define intervals $E_0,\dots,E_n$, where $E_n=D_q$ for some positive integer $q$. Let $j\in\mathbb{N}$ be such that $q < k_{j}$. Set $E_{n+k} = D_{q+k}$, for $0 < k \leq k_{j} - q$. Note that
\[
[t_j, g^{k_j+1}(x)] \cup D_{k_j+1} \subset g([t_j, g^{k_j+1}(x)]) \subset g(D_{k_j}).
\]
If $s_{p+1} = 0$, then define
\[
E_{n+k_{j} - q+1} = [t_j, g^{k_j+1}(x)]\quad\text{and}\quad E_{n+k_{j} - q+2} = D_{k_j+1},
\]
and if $s_{p+1}=1$, then define
\[
E_{n+k_{j} - q+1} = D_{k_j+1}.
\]
This completes the definition of the sequence of compact intervals $(E_n)_{n\in\mathbb{N}_0}$.

Applying Lemma~\ref{lemm:sequence}, we obtain a point $y$ in $E_0$ such that $g^n(y) \in E_n$, for $n\in\mathbb{N}$. Because $g^n(y)\to+\infty$ as $n\to\infty$, we see that $y \in D_0 \cap I(f)$. Each sequence $\underline{s} = s_0 s_1 \ldots$ generates a different sequence of intervals $(E_n)_{n\in\mathbb{N}_0}$, and thereby gives rise to a different point in $D_0\cap I(f)$. Therefore $D_0 \cap I(f)$ is uncountable. It follows that $D_k \cap I(f)$ is uncountable, for $k\in\mathbb{N}$, so the lemma is established on choosing $J=D_k$, for a sufficiently large positive integer $k$.
\end{proof}

\begin{proof}[Proof of Theorem~\ref{theo:R1uncountable}]
Suppose that $f : \mathbb{R} \to \mathbb{R}$ is a continuous function, and that $x \in I(f)$. Let $U$ be a neighbourhood of $x$. We can assume that $U$ meets $\mathbb{R}\setminus I(f)$, as otherwise there is nothing to prove. Choose a point $v$ in $U \cap (\mathbb{R}\setminus I(f))$. Then there exists $R>0$ such that $|f^{n_k}(v)|<R$, for some increasing sequence of positive integers $(n_k)_{k\in\mathbb{N}}$. We know that $|f^{n_k}(x)|\to+\infty$ as $k\to\infty$, and, after conjugating $f$ by the function $x\mapsto -x$ if necessary, we can assume that $f^{m_k}(x)\to +\infty$ as $k\to\infty$, for some subsequence $(m_k)_{k\in\mathbb{N}}$ of $(n_k)_{k\in\mathbb{N}}$. By Lemma~\ref{jdf}, there is a compact interval $J\subset (R,+\infty)$ such that $J\cap I(f)$ is uncountable. Let $L$ denote the closed interval with end points $v$ and $x$. Then we can choose a sufficiently large positive integer $k$ such that $J\subset f^{m_k}(L)$. For each point $y$ in $J \cap I(f)$, there exists a point $z$ in $L$ such that $f^{m_k}(z) = y$, so $z \in I(f)$. Hence $U \cap I(f)$ is uncountable, as required.
\end{proof}

\begin{proof}[Proof of Theorem~\ref{theo:R1open}]
Let $f : \mathbb{R} \to \mathbb{R}$ be a continuous function. Suppose, in order to reach a contradiction, that one of the components of the complement of $I(f)$ is \emph{not} closed. Then, after conjugating $f$ by $x\mapsto -x$ if necessary, we can assume that there is an open interval $(a,b)$ in $\mathbb{R}\setminus I(f)$ such that $b\in I(f)$. Since $a \notin I(f)$, there is a sequence $(n_k)_{k\in\mathbb{N}}$ of positive integers and $R > 0$ such that $|f^{n_k}(a)| \leq R$, for $k\in\mathbb{N}$. On the other hand, we know that $|f^{n_k}(b)|\to +\infty$ as $k\to\infty$. By passing to a subsequence, we can in fact suppose that either $f^{n_k}(b)\to+\infty$ as $k\to\infty$ or $f^{n_k}(b)\to -\infty$ as $k\to\infty$; let us assume the former (the other case can be handled in a similar way). Now, for values of $k$ sufficiently large that $f^{n_k}(b)>R$, we have
\[
[R,f^{n_k}(b)) \subset f^{n_k}([a,b)) \subset \mathbb{R}\setminus  I(f).
\]
This is a contradiction, because for sufficiently large values of $k$, the intervals  $[R,f^{n_k}(b))$ contain iterates of $b$ (so they intersect $I(f)$). Therefore, contrary to our assumption, all components of the complement of $I(f)$ are closed.
\end{proof}

\section{examples}\label{sec:examples}

In this section we provide examples that demonstrate differences between the class of $I_1$-sets and the class of $I_2$-sets. Many of our examples are of sets that appeared in Figure~\ref{fig:sets} from the introduction. We shall see that sets (a), (c) and (e) from that figure \emph{are} $I_2$-sets, and sets (b), (d) and (f) are \emph{not} $I_2$-sets.

By Theorem~\ref{theo:R1open}, it is not possible for an $I_1$-set to have an open complementary component. The following example shows that this is not true of $I_2$-sets.

\begin{example}\label{example.halfplane}
Let $f : \mathbb{R}^2 \to \mathbb{R}^2$ be the continuous function given by
\[
f(x,y) =
\begin{cases}
(x, y+1),      &\qfor x \geq 0, \\
(x, y),          &\qfor x \leq -1,\\
(x(x + 2), y + x + 1), &\qfor -1 < x < 0.
\end{cases}
\]
It can be shown by an elementary calculation that $I(f)$ is the closed half-plane $\{(x, y) : x \geq 0 \}$.
\end{example}

In contrast to this example, we observe that the escaping set of a transcendental entire function cannot be closed. This follows from the observation that if, on the contrary, $f$ is a transcendental entire function and $I(f)$ is closed, then the Julia set of $f$, which is equal to $\partial I(f)$, is contained in $I(f)$, which contradicts the well-known fact that the Julia set contains periodic points of $f$.

It follows from Theorem~\ref{theo:R1uncountable} that $\mathbb{Z}$ is not an $I_1$-set. In the next two examples we will show that both  $\mathbb{Z} \times \{0\}$ and its complement are $I_2$-sets.

\begin{example}\label{example.Z}
Let $f : \mathbb{R}^2 \to \mathbb{R}^2$ be the continuous function given by
\[
f(x, y) =
\begin{cases}
  (x+1, |2y| + \sin^2 (\pi x)), &\qfor                       y \leq 1, \\
  (x + 2 - y, (2-y)\sin^2 (\pi x) + 2), &\qfor              1 < y \leq 2, \\
  (x, y), &\qfor y > 2.
\end{cases}
\]
Since $f(n,0) = (n+1,0)$, for $n\in\mathbb{Z}$, we have $\mathbb{Z} \times \{0\} \subset I(f)$. Now suppose that $(x, y) \notin \mathbb{Z} \times \{0\}$. Then $f(x,y)$ lies in the upper half-plane $\{(x, y) : y \geq 0 \}$. It can then be seen, by a calculation, that $f^{k+1}(x,y) = f^k(x,y)$, for all sufficiently large values of $k$, so $(x,y) \notin I(f)$. Therefore $I(f)=\mathbb{Z} \times \{0\}$.
\end{example}

\begin{example}\label{example.Zcomp}
Let $g : \mathbb{R}^2 \to \mathbb{R}^2$ be the continuous function given by
\[
g(x, y) = f(x, y) - (1, 0),
\]
 where $f$ is the function in Example~\ref{example.Z}. It can be shown, using similar methods to those of  Example~\ref{example.Z}, that $I(g) = \mathbb{R}^2 \setminus (\mathbb{Z} \times \{0\})$.
\end{example}

The $I_2$-set $\mathbb{R}^2 \setminus (\mathbb{Z} \times \{0\})$ from the previous example is shown in Figure~\ref{fig:sets}(a). The set $\bigcup_{n\in\mathbb{Z}} D(n,\tfrac14)$, which is shown in Figure~\ref{fig:sets}(b), is not an $I_2$-set, because it does not satisfy the conditions of Theorem~\ref{theo:hasnounboundedcomponent}.

The next two examples are each about classes of sets that \emph{do} satisfy the conditions of Theorem~\ref{theo:hasnounboundedcomponent}. One of the sets from Example~\ref{example.open} is shown in Figure~\ref{fig:sets}(c): we will see that this \emph{is} an $I_2$-set. (Actually, Figure~\ref{fig:sets}(c) is a reflection in the $x$-axis of one of the sets from Example~\ref{example.open}, but this qualification is insignificant.) Example~\ref{example.open} shows that, in contrast to the one-dimensional case, there are open $I_2$-sets without unbounded components. One of the sets from Example~\ref{example.nastyone} is shown in Figure~\ref{fig:sets}(d): we will see that this \emph{is not} an $I_2$-set.

\begin{example} \label{example.open}
In this example (and only in this example), we use polar coordinates: $(r,\theta)$ represents the point in the plane with modulus $r$ and argument $\theta$. We represent the origin by $(0,\theta)$, for any value of $\theta$.

Let $(\rho_n)_{n\in\mathbb{N}}$, $(\phi_n)_{n\in\mathbb{N}}$, and $(\phi'_n)_{n\in\mathbb{N}}$ be increasing sequences of positive real numbers such that $\rho_n\rightarrow +\infty$ as $n\rightarrow\infty$, and
\[
0 < \phi_1 < \phi'_1 < \phi_2 < \phi'_2 < \dots < 2\pi.
\]
Let
\[
A_n = \{ (r, \theta) : 0< r < \rho_n, \ \phi_n < \theta < \phi'_n \}, \quad\qfor n\in\mathbb{N};
\]
these are disjoint, bounded open sectors. Let $S = \bigcup_{n\in\mathbb{N}} A_n$. We will construct a continuous function $f:\mathbb{R}^2\to\mathbb{R}^2$ such that $I(f)=S$, so that $S$ is an $I_2$-set.

First define $f(x) = 0$, for $x \notin S$, so $I(f) \subset S$. Next choose any number $\varepsilon$ in $(0, 1/2)$. The idea is to define $f$ in $S$ such that the following three properties hold: first,
\[
f(A_n) \subset A_{n+1}, \quad\qfor n\in\mathbb{N};
\]
second,
\[
\quad f\left(\frac{\rho_n}{1+\varepsilon}, \frac{\phi_n + \phi'_n}{2}\right) = \left(\frac{\rho_{n+1}}{1+\varepsilon}, \frac{\phi_{n+1} + \phi'_{n+1}}{2}\right), \quad\qfor n\in\mathbb{N};
\]
and third, the orbit of each point in $S$ eventually joins the orbit
\[
\left(\frac{\rho_1}{1+\varepsilon}, \frac{\phi_1 + \phi'_1}{2} \right)\to \left(\frac{\rho_2}{1+\varepsilon}, \frac{\phi_2 + \phi'_2}{2}\right) \to \cdots \to \left(\frac{\rho_n}{1+\varepsilon}, \frac{\phi_n + \phi'_n}{2} \right) \to \cdots.
\]
It then follows that $S = I(f)$.

To complete the definition of the function $f$, we use the function $h : [0,1] \to [0, 1/(1+\varepsilon)]$ given by
\[
h(t) =
\begin{cases}
t/\varepsilon, &\qfor 0 \leq t \leq \varepsilon/(1+\varepsilon), \\
1/(1+\varepsilon), &\qfor \varepsilon/(1+\varepsilon) < t < 1/(1+\varepsilon), \\
(1-t)/\varepsilon, &\qfor 1/(1+\varepsilon) \leq t \leq 1.
 \end{cases}
\]
Clearly $h$ is continuous, and $h(0) = h(1) = 0$. Moreover, if $x \in (0,1)$, then we have that $h^k(x) = 1/(1+\varepsilon)$, for all sufficiently large values of $k$.

For each $n\in\mathbb{N}$, and for $(r, \theta) \in A_n$, we define
\[
f(r, \theta) = \left((1+\varepsilon)\rho_{n+1}h\left(\frac{r}{\rho_n}\right)h\left(\frac{\theta - \phi_n}{\phi'_n - \phi_n}\right), \frac{\phi_{n+1} + \phi'_{n+1}}{2} \right).
\]

It is a straightforward calculation to see that $f : \mathbb{R}^2 \to \mathbb{R}^2$ is continuous, and it satisfies the first two of the three properties described above. To see that the third property also holds, choose any point $x$ in $S$. Without loss of generality we can assume that $x \in A_1$. Write $x = (r_1, \theta_1)$ and, in general, write $f^n(x) = (r_{n+1}, \theta_{n+1})$, for $n\in\mathbb{N}$, where $r_n\geq 0$ and $0\leqslant \theta_n<2\pi$. Observe that $\theta_n = (\phi_n + \phi'_n)/2$, for $n > 1$. Let $\alpha_n = r_n / \rho_n$, for $n\in\mathbb{N}$. Then $\alpha_{n+1} = h(\alpha_n)$, for $n>1$. Therefore, for sufficiently large values of $n$,  we have $\alpha_{n} = 1/(1+\varepsilon)$, and hence $r_n = \rho_n/(1+\varepsilon)$, as required. This completes our construction of the function $f$.
\end{example}

Our next example shows that an open set can satisfy the conditions of Theorem~\ref{theo:hasnounboundedcomponent} without being the escaping set of a continuous function.

\begin{example}\label{example.nastyone}
Let $D$ be a bounded plane domain, and let $(U_n)_{n\in\mathbb{N}}$ be a sequence of bounded plane domains such that
\begin{enumerate}[label=(\roman*)]
\item $\overline{U_n} \cap \overline{U_m} = \emptyset$, for $n \ne m$;
\item $U_n \cap D = \emptyset$, for $n\in\mathbb{N}$;
\item for each $n\in\mathbb{N}$ there exists a point $p_n$ such that $\partial U_n\cap \partial D = \{p_n\}$;
\item  $\sup\{|x|\,:\, x \in U_n\}\to+\infty$ as $n\rightarrow\infty$.
\end{enumerate}
Let
\[
S = D \cup \bigcup_{n\in\mathbb{N}} U_n.
\]
Clearly $S$ satisfies the conditions of Theorem~\ref{theo:hasnounboundedcomponent}. A picture of such a set is given in Figure~\ref{fig:sets}(d).

We claim that $S$ is not an $I_2$-set. To see why this is so, suppose that $S = I(f)$ where $f : \mathbb{R}^2 \to \mathbb{R}^2$ is a continuous function. As we saw earlier, if $W$ is a component of $I(f)$, then there is another, different component $W'$ of $I(f)$ such that $f(W) \subset W'$ and $f(\partial W) \subset \partial W'$. Therefore, after relabelling the sets $U_i$, we can assume that $f(\partial D) \subset \partial U_0$, in which case $f(p_0) \in \partial U_0$. We can also assume that  $f(\partial U_0) \subset \partial U_1$, and hence that $f(p_0) \in \partial U_1$. This is a contradiction, because $\partial U_1$ is disjoint from $\partial U_0$. Therefore, contrary to our assumption, $S$ is not an $I_2$-set.
\end{example}

Our last two examples are illustrated by Figures~\ref{fig:sets}~(e) and~(f), in that order. Both are examples of unbounded open sets from which infinity is not accessible; the first is an $I_2$-set and the second is not an $I_2$-set. In the first example we use the notation $\pi_1(\xi)$ and $\pi_2(\xi)$ to denote the $x$- and $y$-coordinates, respectively, of a point $\xi\in\mathbb{R}^2$.

\begin{example}\label{example:snake}
Let  $A_n = \sum_{k=0}^n 1/(k+1)$, for $n\in\mathbb{N}_0$. We define an unbounded curve $\Gamma : [0,+\infty) \to \mathbb{R}^2$, as follows. Given any number $t\in[0,+\infty)$, we define $n\in\mathbb{N}_0$ and $\lambda\in[0,1)$ by the equation $t=n+\lambda$. Then we set
\[
\Gamma(t) =
\begin{cases}
   \left(\lambda A_{n/2}, (2-\lambda)/2^{n+1}\right), &\text{for $n$ even}, \\
   \left((1-\lambda)A_{(n-1)/2}, (2-\lambda)/2^{n+1}\right),& \text{for $n$ odd}.
\end{cases}
\]
One can check that $\Gamma$ satisfies the inequality
\begin{equation}\label{kdl}
|\Gamma(s)-\Gamma(t)|<4A_n|s-t|,\quad\text{when}\quad s,t\in[n,n+2],\quad n\geq 0.
\end{equation}

For simplicity we denote the image set $\Gamma([0,+\infty))$ by $\Gamma$, and we follow similar conventions for other unbounded curves in this example. The image set $\Gamma$ consists of a sequence of line segments, which join the points
\[
\left(0,1\right), \left(A_0,1/2\right), \left(0, 1/4\right), \left(A_1, 1/8\right), \ldots,  \left(0,1/2^{2n}\right), \left(A_n, 1/2^{2n+1}\right), \left(0, 1/2^{2n+2}\right), \ldots,
\]
in that order. The ``right-hand'' points, which are of the form $\left(A_n, 1/2^{2n+1}\right)$, correspond to odd integer values of $t$. The ``left-hand'' points, which are of the form $\left(0, 1/2^{2n}\right)$, correspond to even integer values of $t$. If $y \in (0, 1]$, then there exists a unique $x \geq 0$ such that $(x, y) \in \Gamma$. We denote this unique value $x$ by $\phi(y)$.

Let $U$ be the unbounded domain in $\mathbb{R}^2$ given  by
\[
U = \{ (x, y) : y \in (0, 1) \text{ and } |x - \phi(y)| < y \}.
\]
The set $U$ is enclosed by three curves: the horizontal line segment $\{(x,1) : -1\leq x\leq 1 \}$, an unbounded curve $\Gamma_l$ to the left of $\Gamma$, and an unbounded curve $\Gamma_r$ to the right of $\Gamma$. The two curves $\Gamma_l$ and $\Gamma_r$ both zigzag in the same fashion as $\Gamma$. We parameterise them in the same way as $\Gamma$, in the sense that, if $t\geq 0$, then
\[
\pi_2(\Gamma(t)) = \pi_2(\Gamma_r(t)) =\pi_2(\Gamma_l(t)),
\]
and
\[
\pi_1(\Gamma(t)) = \pi_1(\Gamma_r(t)) -  \pi_2(\Gamma(t)) = \pi_1(\Gamma_l(t)) + \pi_2(\Gamma(t)).
\]
The set $U$ is shown in Figure~\ref{haf}. It is unbounded, and infinity is not accessible from $U$.

\begin{figure}[ht]
\centering
\includegraphics{figI}
\caption{An unbounded open set that is an $I_2$-set}
\label{haf}
\end{figure}

We must now define a continuous function $f : \mathbb{R}^2 \to \mathbb{R}^2$ with $I(f) = U$. To begin with, we define $f$ to be the identity function on the set
\[
V = \{ (x, y) : y\notin (0,1) \}.
\]

Next we define $f$ on $U$. To do this, we first define a function $\psi$ on $\Gamma \cup \Gamma_l \cup \Gamma_r$ that satisfies $\psi(\Gamma) = \Gamma$, $\psi(\Gamma_l) = \Gamma_l$ and $\psi(\Gamma_r) = \Gamma_r$. Roughly speaking, the function $\psi$ will be chosen so that it maps each ``zigzag'' of $\Gamma$ (and $\Gamma_l$ and $\Gamma_r$) downwards to the zigzag below. We will then define  $f$ on $U$ by interpolating between $\psi$ on $\Gamma$ and $\psi$ on $\Gamma_l \cup \Gamma_r$.

 Let $\beta : [0,+\infty)\to[0,+\infty)$ be the continuous function given by
\[
\beta(t) =
\begin{cases}
n + 1, &\qfor t \in [n - 1/n, n],\\
(1+1/n)t, &\qfor t \in [n, n+1-1/(n+1)],
\end{cases}
\]
for $n\in\mathbb{N}$.  One can check that $\beta$ satisfies the inequality
\begin{equation}\label{iab}
|\beta(t)-(t+1)|\leq \frac{1}{n}, \quad\text{when}\quad t\in[n,n+1],\quad n\in\mathbb{N}_0.
\end{equation}
We also record two other important properties of $\beta$, namely that (a) $\beta(n) = n+1$, for $n\in\mathbb{N}$, and (b) if $t>0$, then $\beta^k(t) \in\mathbb{N}$, for all sufficiently large values of $k$.

Let
\[
h_1(t) =
\begin{cases}
4t, &\qfor t \in [0, 1/2], \\
2t+1, & \qfor t \in [1/2,+\infty),
\end{cases}
\]
and define
\[
\psi(\Gamma(t)) = \Gamma(h_1\circ \beta\circ h_1^{-1}(t)), \quad\text{for}\quad t\in[0,+\infty).
\]

The function $\psi$ is continuous on $\Gamma$, and it maps $\Gamma$ into itself. Using property (a) from above, we can see that $\psi$ maps each \emph{right-hand} vertex $\left(A_n, 1/2^{2n+1}\right)$ of $\Gamma$ to the next right-hand vertex $\left(A_{n+1}, 1/2^{2n+3}\right)$, for $n\in\mathbb{N}$. Next, by property (b), if $\xi \in \Gamma$, then $\psi^k(\xi)$ is one of the right-hand vertices of $\Gamma$, for all sufficiently large  $k$. We deduce that $\psi^k(\xi) \rightarrow \infty$ as $k\rightarrow\infty$, for $\xi \in \Gamma$. Finally, by combining inequalities \eqref{kdl} and \eqref{iab}, it can be shown that $|\psi(\Gamma(t))-\Gamma(t+2)|\to 0$ as $t\to\infty$, and hence that $|\psi(\Gamma(t))-\Gamma(t)|\to 0$ as $t\to\infty$; we omit the detailed calculations.

The definition of $\psi$ on $\Gamma_l \cup \Gamma_r$ is similar to that on $\Gamma$. We define $\psi$ on $\Gamma_l$ and describe its properties; the definition and properties on $\Gamma_r$ are much the same but with $\Gamma_r$ replacing $\Gamma_l$. Let $h_2(t)=2t$, and define
\[
\psi(\Gamma_l(t)) = \Gamma_l(h_2\circ\beta\circ h_2^{-1}(t)\big), \quad\text{for}\quad t\in[0,+\infty).
\]

Then $\psi$ is continuous on $\Gamma_l$, and it maps $\Gamma_l$ into itself. It maps each \emph{left-hand} vertex of $\Gamma_l$ to the left-hand vertex below, and if $\xi \in \Gamma_l$, then $\psi^k(\xi)$ is one of the left-hand vertices of  $\Gamma_l$, for all sufficiently large $k$. Hence $\psi^k(\xi) \rightarrow (0, 0)$ as $k\rightarrow\infty$, for $\xi \in \Gamma_l$. Finally, we observe that $|\psi(\Gamma_l(t))-\Gamma_l(t)|\to 0$ as $t\to\infty$.

Next we define $f$ on $U$ in such a way that $f(x,y)=\psi(x,y)$ when $(x,y)\in \Gamma\cup\Gamma_l\cup\Gamma_r$ and $y<1/2$. To do this, it  is slightly easier to use an alternative coordinate system. Suppose that $(x,y)$ is a point with $0<y\leq 1$. Then, by the definitions of $\Gamma$ and $\phi$, there exists a unique real number $\eta$ such that $x = \phi(y) + \eta y$. We write $(x, y) = \floor*{\eta, y}$, and say that $\eta$ and $y$ are the \emph{$U$-coordinates} of the point $(x,y)$. We also say that $\eta$ is the \emph{$\eta$-coordinate} of the point $(x,y)$. Note that $(x,y) \in \Gamma$ if and only if $(x, y) = (\phi(y), y) = \floor*{0,y}$. Note also that $(x,y) \in \Gamma_l$ if and only if $(x, y) = \floor*{-1,y}$, and $(x,y) \in \Gamma_r$ if and only if $(x, y) = \floor*{1,y}$.

Let $\alpha : [0,1] \to [0, 1]$ be the continuous function given by
\[
\alpha(t) =
\begin{cases}
0, &\qfor t \in [0, 1/2], \\
2t-1, & \qfor t \in [1/2,1].
\end{cases}
\]

We recall the definition $\pi_2(x,y)=y$. Now, choose any point $(x,y)$ in $U \cup \Gamma_l \cup \Gamma_r$. We write $(x,y)$ in the form $\floor*{\varepsilon\eta, y}$, where $\eta\in[0,1]$ and $\varepsilon$ is either $-1$ or $1$. Let
\[
f(\floor*{\varepsilon\eta, y})= \floor*{\varepsilon\eta', y'},
\]
where
\[
\eta'=(1-\alpha(y))\alpha(\eta)+\alpha(y)\eta,
\]
and
\[
y'=(1-\alpha(y))\big((1-\eta)\pi_2(\psi(\floor*{0, y}))+\eta\pi_2(\psi(\floor*{\varepsilon, y}))\big)+\alpha(y)y.
\]
This is a well-defined function, because $\eta'\in[0,1]$ and $y'\in (0,1]$. Clearly it is continuous on $U\cup\Gamma_l\cup\Gamma_R$. When $\eta=0$, we have $\eta'=0$, so $f$ maps $\Gamma$ into itself, and similarly it maps $\Gamma_l$ and $\Gamma_r$ into themselves. What is more, if $\eta<1$, then $\eta'<1$, so $f$ maps $U$ into itself.

Next we prove that $f$ is continuous on $\overline{U}\cup V$. When $y=1$, we have $y'=1$ and $\eta'=\eta$, so $f$ coincides with the identity function. Observe that if $y<1/2$, then the definition of $f$ simplifies to
\[
f(\floor*{\varepsilon\eta, y})= \floor*{\varepsilon\alpha(\eta), (1-\eta)\pi_2(\psi(\floor*{0, y}))+\eta\pi_2(\psi(\floor*{\varepsilon, y}))},
\]
in which case $y'<1/2$ also. Notice in particular that $f(x,y)=\psi(x,y)$ when $(x,y)\in \Gamma\cup\Gamma_l\cup\Gamma_r$ and $y<1/2$.  Now suppose that $\big((x_n,y_n)\big)_{n\in\mathbb{N}}$ is a sequence of points in $U\cup\Gamma_l\cup\Gamma_r$ that converges to a point $(u,0)$. Let $\varepsilon_n\eta_n\in[-1,1]$ be the $\eta$-coordinate of $(x_n,y_n)$, where $\varepsilon_n$ is $-1$ or $1$ and $\eta_n\in[0,1]$, so that $x_n=\phi(y_n)+\varepsilon_n\eta_ny_n$. We know that $x_n\to u$ and $y_n\to 0$ as $n\to\infty$, and it follows that $\phi(y_n)\to u$ and $\phi(y_n)+\varepsilon_n y_n\to u$ as $n\to\infty$. Hence $\floor*{0, y_n}\to (u,0)$ and $\floor*{\varepsilon_n, y_n}\to (u,0)$ as $n\to\infty$. It then follows that $\psi(\floor*{0, y_n})\to (u,0)$ and $\psi(\floor*{\varepsilon_n, y_n})\to (u,0)$ as $n\to\infty$. Furthermore, if we define
\[
(x_n',y_n') = (1-\eta_n)\psi(\floor*{0, y_n})+\eta_n\psi(\floor*{\varepsilon_n, y_n}),
\]
then $(x_n',y_n')\to (u,0)$ as $n\to\infty$, also. Let $\eta_n'\in[-1,1]$ be the $\eta$-coordinate of $(x_n',y_n')$, so that $x_n'=\phi(y_n')+\eta_n'y_n'$. Then we see that $\phi(y_n')\to u$ as $n\to\infty$. Hence
\[
f(x_n,y_n) = (\phi(y_n')+\varepsilon_n\alpha(\eta_n)y_n',y_n') \to (u,0)\quad\text{as}\quad n\to\infty.
\]
This completes our argument to show that $f$ is continuous on $\overline{U}\cup V$.

Let us now prove that $U\subset I(f)$ and $(\Gamma_l\cup\Gamma_r)\cap I(f)=\emptyset$. Choose any point  $\floor*{\varepsilon\eta, y}$ in $U\cup\Gamma_l\cup\Gamma_r$, where, as usual, $\varepsilon$ is $-1$ or $1$ and $\eta\in[0,1]$, and for the moment we assume that $1/2<y<1$. If $1/2<y<3/4$, then one can check that $y'<1/2$. If $3/4\leq y<1$, then one can check that $y'<\alpha(y)$. It follows that the $y$-component of $f^k(\floor*{\varepsilon\eta, y})$ is less than  $1/2$, for a sufficiently large positive integer $k$. Since $f$ coincides with $\psi$ on $\Gamma\cup\Gamma_l\cup\Gamma_r$ when $y<1/2$, we see that $(\Gamma_l\cup\Gamma_r)\cap I(f)=\emptyset$ and $\Gamma\subset I(f)$. Now, if  $\floor*{\varepsilon\eta, y}\in U$ and $y<1/2$, then $\eta'=\alpha(\eta)$. It follows that $f^k(\floor*{\varepsilon\eta, y})\in \Gamma$, for a sufficiently large positive integer $k$.  Hence $U\subset I(f)$.

It remains to define $f$ in the set
\[
W = \{(x, y) \notin \overline{U} : 0<y<1\}.
\]
Points $(x,y)$ in this set can also be written in $U$-coordinates, in the form $(x,y)=\floor*{\varepsilon\eta,y}$, where $\eta>1$ and $\varepsilon$ is $-1$ or $1$. We define
\[
f(\floor*{\varepsilon\eta,y}) = f(\floor*{\varepsilon,y}) + \floor*{\eta - \varepsilon, 0}.
\]
 We leave it for the reader to check that $f$ is continuous on $\mathbb{R}^2$, and that $I(f) = U$.
\end{example}

We finish, as promised, with an example of an unbounded open set that is not an $I_2$-set.

\begin{example}\label{example:failsnake}
Let $E = \{ (x, y) : y < 0, (x-5)^2 + y^2 < 1 \}$ and define $S = U \cup E$, where $U$ is the domain from Example~\ref{example:snake}. A stylised image of $S$ is shown in Figure~\ref{hxf}.

\begin{figure}[ht]
\centering
\includegraphics{figH}
\caption{An unbounded open set that is not an $I_2$-set}
\label{hxf}
\end{figure}

Let us suppose, in order to reach a contradiction, that $S=I(f)$ for some continuous function $f:\mathbb{R}^2\to\mathbb{R}^2$. As $f(S) \subset S$, we must have either $f(E)\subset E$ or $f(E)\subset U$. In fact, the former case cannot arise, because $E$ is bounded and lies in the escaping set. The closure of $E$ is compact, and $E$ is connected, so $f(E)$ is a bounded, connected subset of $U$. It follows that there is a positive constant $k$ such that
\[
f(E) \subset \{(x,y)\in U\,:\, y \geq k\}.
\]
Choose $\varepsilon>0$ such that if $\text{dist}((x,y),E)\leq \varepsilon$, then the $y$-component of $f(x,y)$ is greater than $k/2$. Let us assume also that $\varepsilon<k/2$.

Now choose a vertical cross section $X$ of $U$, in the obvious way. The set $X$ is simply a small vertical line segment, contained wholly within $U$, which splits $U$ into two components: a bounded component $L$ and an unbounded component $R$. We can choose $X$ so that each point of $X$ lies within a distance $\varepsilon$ of $E$.

Let $\gamma:[0,1]\to U$ be a continuous path such that $\gamma([0,1)) \subset L$, such that $\gamma(1)\in X$, and finally such that $f^n(\gamma(0))\in X\cup R$, for $n\in\mathbb{N}$; it is easy to see that such a path exists. Let $|\gamma|$ denote the image of $\gamma$ in $U$. We call sets $|\gamma|$ that arise in this way \emph{restricted paths}.

Suppose that $\gamma$ is a restricted path, and $p=\gamma(0)$ and $q=\gamma(1)$. We know that the $y$-component of $f(q)$ is greater than or equal to $k/2$, so $f(q)\in L$. Let $N$ be the largest positive integer such that $f^N(q)\in L$. Notice that $f^n(p)\in X\cup R$, for $n\in\mathbb{N}$. Consider the path $\delta(t)=(f^N\circ\gamma)(1-t)$. It satisfies $\delta(0)\in L$ and $\delta(1)\in X\cup R$. Let $s\in (0,1]$ be the smallest positive value such that $\delta(s)\in X$. Finally, let
\[
\phi:[0,1]\to U,\quad \phi(t) = \delta(st).
\]
It follows by the definition of the integer $N$ that $|\phi|$ is a restricted path, and that
\[
|\phi| \subset |\delta|=f^N(|\gamma|).
\]
We have shown that \emph{given any restricted path $|\gamma|$, there is another restricted path $|\phi|$ and a positive integer $N$ such that $|\phi|\subset f^N(|\gamma|)$.}

Using this observation, we can construct a sequence of restricted paths $(\alpha_i)_{i\in\mathbb{N}}$ and a sequence of positive integers $(n_i)_{i\in\mathbb{N}}$ such that
\[
|\alpha_{i+1}|\subset f^{n_i}(|\alpha_{i}|), \quad\qfor i\in\mathbb{N}.
\]
As the sets $|\alpha_i|$ are compact, it follows from Lemma~\ref{lemm:sequence} that there is a point $x$ in $|\alpha_1|$ such that the sequence of iterates $(f^i(x))_{i\in\mathbb{N}}$ has a bounded subsequence. This is a contradiction, because $x \in I(f)$. Thus, contrary to our earlier assumption, $S$ is not an $I_2$-set.
\end{example}

\itshape{Acknowledgement:} \normalfont We are grateful to John Osborne for a number of helpful comments.
\bibliographystyle{acm}


\section{Introduction}

The $\delta$~Scuti class of variables are dwarfs or giants with spectral types
between A2 and F5.  They lie on an extension of the Cepheid instability strip
with periods in the range 0.02--0.3~d.  Most of the pulsational driving in
these stars is by the $\kappa$~mechanism in the He{\sc II} partial  ionization
zone.  A large number of  $\delta$~Sct stars have
been identified in photometric time series data from the {\it Kepler}
spacecraft.  A study of these stars by \citet{Balona2011c} has revealed that
even in the center of the instability strip, no more than half the stars
pulsate as $\delta$~Sct variables.  This is a surprising finding; the reason
for damping of pulsations in constant stars in the instability strip is
presently not known.

Prior to the {\it MOST, CoRoT} and {\it Kepler} space missions, it was
thought that the frequency spectra in $\delta$~Sct stars may become very
dense as the detection threshold is lowered.  The reason is that modes of
high degree, $l$, which are not seen from the ground because of the low
amplitudes due to cancellation effects, should be easily seen at the
precision level attained for these space missions \citep{Balona1999}.
Indeed, this expectation appears to have been fulfilled in {\it CoRoT}
observations of HD~50844 \citep{Poretti2009}.  This star appears to have a
very high mode density over the whole frequency range, indicating that modes
of relatively high $l$ are seen.  However, in a study of {\it Kepler}
$\delta$~Sct stars, \citet{Balona2011c} find that, in general, the mode
density is quite moderate and that HD~50844 is probably an exception.

In order to study mode density, one needs to be sure that frequencies are
correctly extracted from the data.  This is not an issue for the high
amplitude peaks in the periodogram, but since the number of modes is
expected to increase with decreasing amplitude, one needs to estimate the
probability that a given peak is due to noise (the false-alarm probability).
The significance level of a frequency can be estimated easily only for the
case of an equally-spaced time series, when the Lomb-Scargle criterion can
be used \citep{Scargle1982}.  When the sampled times are not uniformly spaced,
the problem is intractable and can only be solved by numerical means
\citep{Frescura2008}.  Ground-based data are seldom equally spaced and the
``four-sigma'' rule is often used to estimate the significance level
\citep{Breger1993}.  This rule states that a peak is  deemed significant if
its amplitude exceeds the background noise level of the periodogram by a
factor of four.  This is a purely empirical rule with no statistical
foundation, but does seem to be reasonable in many cases
\citep{Kuschnig1997, Koen2010}.  The {\it CoRoT} data are uniformly spaced
and it is therefore important to apply the correct statistics.

 For a time series of duration, $t$, the width of a peak in the
power spectrum is inversely proportional to $t$.  Thus for a time series of a
given duration, there is a maximum limit on the number of peaks per frequency
interval that can be measured. Since the mode density is expected to
increase with decreasing amplitude, blending of peaks due to the finite
frequency resolution becomes very important. Thus the number of frequencies
that can be extracted, and hence the mode density, depends not only on a
correct estimation of the false alarm probability, but also on effects
related to the finite frequency resolution.  These effects have not been
taken into account in estimates of mode density by \citet{Poretti2009}
and \citet{Balona2011c}.

From the above considerations, it is clear that one needs to carefully reconsider
the question of significance in frequency extraction for {\it CoRoT} and {\it
Kepler} data and, in particular, the importance of frequency resolution and
close frequencies.  In this paper we use simulations at various mode
densities to determine the effect of mode density on frequency extraction.
We also compare results obtained with the Lomb-Scargle false alarm
probability (FAP) criterion with those calculated using the four-sigma
rule.  It turns out that neither of these criteria are useful if the
number of modes increases with decreasing amplitude.  Finally, we compare
the amplitude density in the periodogram of HD\,50844 with those of {\it
Kepler} $\delta$~Scuti stars.

\section{The data and noise properties}

The {\it CoRoT} observations of HD~50844 consist of $n = 155827$ white light
data points obtained between HJD~2452590.0684 and HJD~2452647.7817 (57.7~d)
with a cadence of 30~s.  The time series is almost uninterrupted, except for
data taken at the southern magnetic anomaly and some other points which were
rejected because they were clearly anomalous.

We first of all need to understand the noise properties of these data.  One
way to do this is to select stars which appear to be constant, or vary the
least.  The noise level in the periodogram of such a star may be used to
calibrate the noise level in any other star in a purely empirical way.

The definition of what is meant by the noise level in the periodogram needs
to be clarified.  In visual estimates of the noise level, this is generally
taken as the height of the majority of peaks.  The peaks in the periodogram
may be likened to blades of grass on a lawn and the mean height of the peaks
may be called the ``grass'' level.  This loose definition has been placed on
a firmer footing by \citet{Balona2011a} who defined $\sigma_G = 2.5\sigma$,
where $\sigma$ is the median height of the periodogram peaks in a region
free of signals.  In the four-sigma rule, a peak with amplitude $A$ may be
considered significant if $A > 4\sigma_G$.  One may also define the mean
background noise level as just the average of the power or the amplitude.
Since the background noise level typically increases towards low frequencies,
and since the backgound is often difficult to estimate in crowded regions of
the spectrum, the exact meaning of ``mean background level'' is often not
clear.

It can be shown that the mean amplitude, $A$, of the periodogram of pure
white noise with variance $\sigma_0$ is given by $A = 2\sigma_0/\sqrt{n}$,
where $n$ is the number of points in the time series (see \citet{Kendall1976}).
From the high-frequency tail of the periodogram of HD~50844 we find $A =
0.0035$~mmag which gives $\sigma_0 = 0.691$~mmag.  Cast in terms of apparent
magnitude, $V$, and duration of the time series, $\Delta t$, the noise level
in the periodogram can be written as
$$\log \sigma_G = a + \frac{1}{5}V - \frac{1}{2}\log \Delta t,$$
where $a$ is a constant.  By fitting 175 of the least variable stars in the
A--F range in the {\it Kepler} data, \citet{Balona2011a} found
$$\log \sigma_G = -0.93 + 0.21V - 0.47\log \Delta t,$$
where $\sigma_G$ is measured in ppm and $\Delta t$ in days, confirming the
expected relationship.

In the case of the {\it CoRoT} data, we do not have access to a large number
of constant or nearly constant A--F stars for a completely independent
estimate of the noise level.  However, we do have data for HD~292790, also
measured by \citet{Poretti2009}.  This star is variable at low frequencies,
but the noise level is well defined at the higher frequencies relevant for
$\delta$~Sct stars.  The magnitude of HD~292790 is $V = 9.48$ and the
time-series duration $\Delta t = 54.66$~d.  For the $\delta$~Scuti star
HD~50844, $V = 9.09$ and $\Delta t = 57.70$~d.  Fig.~\ref{period}
shows the periodograms for these two stars.  Fig.~\ref{pnoise} shows an
expanded region to show the noise levels.

The periodogram of HD~292790 can be understood in terms of rotation of a
spotted star with frequency $f = 0.4386$~d$^{-1}$ \citep{Poretti2009}.
The periodogram contains a signal at 13.9668~d$^{-1}$ and its harmonics
which is the orbital frequency of the satellite.  For this star we
estimate $\sigma_G = 5$ ppm, which allows us to determine the constant
$a$ in the above equation.  We  obtain
$$\log \sigma_G = -0.33 + 0.2V - 0.5\log \Delta t,$$
which means that {\it Kepler} data are about four times more precise than
{\it CoRoT} data.  This formula gives $\sigma_G = 4.0$~ppm for HD~50844,
which is indeed the noise level of the high frequency tail in Fig.~\ref{pnoise}.
Thus, according to the four sigma rule, we may assume that for HD~50844 only
frequencies  with amplitudes exceeding approximately 16~ppm are likely to be real.

The spectral window of the {\it CoRoT} data for HD~50844 is shown in
Fig.~\ref{specwin}. The FWHM of the central peak is about 0.02~d$^{-1}$,
which means that frequencies separated by less than about 0.01~d$^{-1}$ are
not likely to be resolved.

\begin{figure}
\centering
\includegraphics{period.ps}
\caption{Periodograms of HD~50844 (top) and HD~292790 (bottom).  The
estimated noise level for HD~50844 is $\sigma_G = 4.0$\,ppm while for
HD~292790 it is $\sigma_G = 5.0$\,ppm (see Fig.\,\ref{pnoise} for an expanded
view). }
\label{period}
\end{figure}

\begin{figure}
\centering
\includegraphics{pnoise.ps}
\caption{Periodograms of HD~50844 (top) and HD~292790 (bottom) with expanded
amplitude scale to show the noise level.}
\label{pnoise}
\end{figure}

\begin{figure}
\centering
\includegraphics{specwin.ps}
\caption{Spectral window of the HD~50844 {\it CoRoT} data.}
\label{specwin}
\end{figure}

\section{Tests of significance}

Although the four-sigma test of significance is widely used because it is
simple to apply, it is by no means a rigorous test.  For example,
\citet{Koen2010} finds that the actual significance levels corresponding to
the four sigma limit may vary by orders of magnitude, depending on the exact
data configuration.  He finds that the number and time spacing of the
observations have little influence on the significance levels.  On the other
hand, the total duration of the time series, the frequency range that is
searched and previous prewhitening of the data greatly affect the significance
level of the four sigma rule. In particular, prewhitening removes too much
power at the given frequency, hence the estimated mean noise level is lower
than it should be, decreasing the false alarm probability.  This means that
repeated application of the rule on successively prewhitened data will tend
to assign significance to peaks which are probably noise.  \citet{Koen2010}
also find that the four sigma rule, when applied to a single frequency peak,
is likely to underestimate its significance.  In other words, there are
peaks of lower amplitude which are significant, but which are considered to
be noise by the four-sigma rule.

More precise significance tests for periodograms have been discussed by,
among others, \citet{Scargle1982, Horne1986, Koen1990} and
\citet{Schwarzenberg1998}.  Somewhat different methods are discussed by
\citet{Reegen2007, Reegen2008} and \citet{Baluev2008}.  An excellent discussion
of the problem is presented by \citet{Frescura2008}.  The problem is clearly a
very difficult one for data which is not uniformly sampled.  For uniformly-sampled
data Scargle's significance test \citep{Scargle1982} is easy to calculate and
gives the false-alarm probability of a periodogram peak given the power
level and the noise variance of the data.  Since the {\it CoRoT} and {\it Kepler}
data are equally sampled, except for occasional small gaps, it is clear that
this test is to be preferred to the four-sigma rule.

For equally-spaced data, \citet{Scargle1982} shows that the false-alarm
probability (FAP), $P(z)$, of a peak with power $z$ is given by
$$P(z) = 1 - \left\{1 - \exp(-z/\sigma^2_0)\right\}^{N_i},$$
where $\sigma^2_0$ is the noise variance of the data and $N_i$ is the number of
independent frequencies.  We call this the Lomb-Scargle significance
criterion.  It should be noted that in some references the power is normalized
by the noise variance, while in others it is not.  We use the un-normalized definition.
This relationship can be inverted to give the power for any FAP,
$$z = -\sigma^2_0\ln\left\{1 - \left(1 - P(z)\right)^{1/N_i}\right\}.$$
The number of independent frequencies is well defined only for
equally-spaced data and is given by $N_i = n/2$, where $n$ is the number of
data points \citep{Frescura2008}.  Astronomers usually prefer the periodogram
to be a function of amplitude rather than power.  For our definition of power,
$z$, the amplitude is given by $A = 2\sqrt{z/n}$.  Given a certain FAP,
$P_A$, one may then calculate an amplitude, $A_A$, which corresponds to this
probability,
$$A_A = 2 \sqrt{ -\frac{\sigma^2_0}{n}\ln\left\{1 - \left(1 -
P_A\right)^{1/N_i}\right\}}.$$
For {\it CoRoT} observations of HD~50844, we find $A_A = 0.014$~mmag for
$P_A = 0.01$ and $A_A = 0.015$~mmag for $P_A = 0.001$.  The four-sigma rule
suggests that only amplitudes above 0.016~mmag are significant, which is
certainly true, but is very stringent and corresponds to a FAP of less than
0.001, whereas a FAP of $P_A = 0.01$ is often sufficient.

\section{Simulations}

Before we apply the Lomb-Scargle criterion to the {\it CoRoT} data of HD~50844,
it is important to test it on synthetic data.  These tests take the actual
times of {\it CoRoT} observations of this star to generate a time series
comprising of many sinusoidal variations with randomly generated
frequencies, amplitudes and phases.  The simulations consist of uniformly
distributed frequencies in the range 0--30~d$^{-1}$, which is roughly the
frequency range of pulsations in HD~50844.  In $\delta$~Sct stars, the number of
modes increases sharply with decreasing amplitude.  To roughly mimic this,
the amplitudes in our simulations are exponentially distributed.  A noise error
with a Gaussian distribution and standard deviation of 0.500~mmag was added to
each point of the synthetic time series.  This leads to a mean periodogram
height of 0.0025~mmag or a grass noise level $\sigma_G = 0.006$~mmag,
roughly similar to that found in HD~50844.  The aim of these simulations is
to determine the effect of mode density on frequency extraction and to test
the reliability of these frequencies in data with a high signal-to-noise (S/N)
ratio.  We also wish to investigate the efficiency of the four-sigma rule and
the Lomb-Scargle significance criterion.

The time series was analyzed using the Lomb periodogram and the algorithm
described by \citet{Press1989} for fast implementation.  The peak of maximum
amplitude is located and its frequency measured by fitting a quadratic to
points around maximum amplitude.  This frequency, together with up to 20
previous frequencies, is used in a simultaneous least-squares Fourier fit to
the data.Once the limit of 20 frequencies has been reached, the original time
series is replaced by the prewhitened time series and the process repeated
until the peak of highest amplitude is no longer significant.

We start with the sparsest frequency set of 30 frequencies.  In this case
we are able to extract 65 frequencies for which FAP $>$ 0.01 (or 47
frequencies using the four-sigma limit) even though the simulated data has
only 30 frequencies.  Of these frequencies, the first 28 of highest
amplitude match the known frequencies.  The two missing frequencies do
appear, but are far down on the list since they have very low amplitudes.
The question that needs to be asked is where do the additional 35 frequencies
which have significant amplitudes come from?

Fig.\,\ref{a30} shows the periodogram in a specific frequency region and also
a schematic periodogram of the same region which shows that the large peak
actually consists of two unresolved closely-spaced frequencies.  These
frequencies and amplitudes are  $f_1 = 19.3458$~d$^{-1}$, $A_1 = 7.118$~mmag,
$f_2 = 19.3300$~d$^{-1}$, $A_2 = 0.574$~mmag.  The figure also shows a
schematic periodogram of the extracted frequencies.  Instead of extracting
just a single frequency, the code finds numerous, roughly equally-spaced
frequency components of relatively high amplitudes.  A third frequency at
$f_3 = 19.4475$~d$^{-1}$, $A_3 = 1.855$~mmag also shows fictitious
components.

\begin{figure}
\centering
\includegraphics{a30.ps}
\caption{Schematic periodogram of known frequency components (with positive
amplitudes) and extracted components (negative amplitudes) in a simulation.}
\label{a30}
\end{figure}

The origin of these fictitious components are easy to understand.  They come
about because the extracted frequency of the unresolved peak is
19.3465~d$^{-1}$, which differs from the true frequency.  Even though the
difference is only 0.0007~d$^{-1}$, the prewhitened data still contains
significant signal because the amplitude of the unresolved peak is so high.
Prewhitening, in fact, is equivalent to adding a fictional signal to the
data.  When the prewhitening frequency and amplitude is sufficiently close
to the true values, both are removed through destructive interference.  If the
prewhitening frequencies and amplitudes not quite correct, what remains is a
sequence of equally-spaced frequencies of diminishing amplitude (an
interference pattern).  If the frequency of interest has a low amplitude,
the residual interference pattern may be below the noise level, which is
nearly always the case in ground-based observations.  Space data have such
high signal-to-noise ratio that the amplitudes of the residual interference
pattern is well above the noise level.  Thus far more frequencies are
extracted than actually exist.  Provided that the S/N level is sufficiently
high, successive frequency extraction will lead to an ever increasing number
of frequencies at low amplitudes which do not actually exist.

\begin{table}
\centering
\caption{Results of frequency extraction using the Lomb periodogram on
simulated data. Random frequencies uniformly distributed in the range
0--30~d$^{-1}$ and random amplitudes exponentially distributed in the
range 0--10~mmag were used. $N_0$ is the number of generated frequencies,
$N$ is the number of these frequencies with amplitudes above the FAP = 0.01
threshold.  $N_{\rm ex}$ is the number of extracted frequencies with
FAP $<$ 0.01 using the Lomb-Scargle periodogram.  The frequencies extracted
using the four-sigma criterion is given by $N_4$.  The two last columns give
the numbers of significant frequencies using non-linear optimization.
$L_{\rm ex}$ refers to the numbers using the Lomb-Scargle FAP and
$L_4$ to the four-sigma criterion.}
\begin{tabular}{rrrrrr}
\hline
\hline
 $N_0$ & $N$ & $N_{\rm ex}$& $N_4$ & $L_{\rm ex}$  & $L_4$ \\
\hline
  30 &   30 &   65 &   47 &   30 &   29 \\
  50 &   50 &  137 &   87 &   57 &   54 \\
 100 &   98 &  220 &  150 &  132 &  112 \\
 200 &  198 &  537 &  350 &  402 &  320 \\
 300 &  298 &  879 &  585 &  729 &  596 \\
 500 &  493 & 1611 & 1141 & 1439 & 1197 \\
 700 &  693 & 2101 & 1545 & 1976 & 1687 \\
1000 &  998 & 3711 & 2215 & 2566 & 2283 \\
2000 & 1982 & 3405 & 2805 & 3108 & 2797 \\
3000 & 2969 & 3599 & 2968 & 3488 & 3187 \\
\\
\hline
\end{tabular}
\label{syn}
\end{table}

Results of simulations using an increasing density of frequencies are shown
in Table~\ref{syn}.  We note that the four-sigma criterion is more stringent
than the Lomb-Scargle FAP criterion, though it is clear that these criteria
are irrelevant in identifying the true frequencies.  Note also that the
number of extracted frequencies increases only slowly for $N > 1000$.  The
reason for this is due to the finite resolution imposed by the limited
duration of the time series.

There are two problems that come to light.  The first problem consists in
unavoidable errors in blended frequencies with high amplitudes, resulting in
a cascade of low-amplitude (but highly significant) fictitious frequencies.
The second problem is one of frequency resolution which increases the number
of blended frequencies and compounds the problem.  It might be possible to
recognize close frequencies using nonlinear optimization.  To test this
possibility we used the Levenberg-Marquard algorithm in combination with the
Lomb periodogram to optimize the frequencies, grouping closely-spaced
frequencies in a simultaneous optimization scheme.  Results are shown in
Table\,\ref{syn}.  While nonlinear optimization gives much better results
for low mode densities, it still fails badly as the density increases.

Another symptom which arises from continuous prewhitening of data with very
high S/N is the development of a plateau in the periodogram consisting of a
very large number of blended peaks of almost equal amplitude.  An example of
this can be seen in the simulations arising from the simulation with $N_0 =
1000$ frequencies.  Such a plateau occurs in other simulations, but is most
prominent in those with a high mode density.  The development of a plateau
can be seen in Fig.\,3 of \citet{Poretti2009}.

\begin{figure}
\centering
\includegraphics{plateau.ps}
\caption{Periodogram of simulated data with 1000 actual frequencies after
prewhitening by 2000 frequencies showing the characteristic plateau.}
\label{plateau}
\end{figure}

From these simulations we conclude that one should be very careful in
extracting frequencies with small amplitudes in stars with very high S/N data
containing several high-amplitude peaks.  In particular, no trust should be
placed in the very large number of frequencies  which results from
successive prewhitening in such stars.  This is undoubtedly the case in
HD\,50844.

\section{Amplitude distribution}

The simulations described above show that the effect of finite frequency
resolution is very important.  In fact, it is likely to be of far greater
importance well before the amplitude threshold appropriate to a given false
alarm probability.  A question that is directly relevant to estimation of
the mode density is the distribution of amplitudes.  In order to estimate
the mode density for the lowest detectable amplitudes, one needs to
calculate the amplitude distribution, i.e., the number of modes within a
given amplitude range.  From the results of the previous section, it is
clear that repeated prewhitening will lead to an ever increasing number of
false modes for low amplitudes.  In this section we investigate the
amplitude distribution using simulations.

For this purpose, we generated a number of time series with random
frequencies uniformly distributed in the frequency range 0--30~d$^{-1}$
with uniformly distributed amplitudes in the amplitude range 0--10~mmag.
This differs from the analysis in the previous section where we used an
approximately exponential amplitude distribution.  A uniform amplitude
distribution is not expected in $\delta$~Sct stars, but it allows the
known and extracted amplitude distributions to be more easily compared.

Table\,\ref{uni} shows the number of simulated frequencies and the number
of extracted frequencies using the Lomb periodogram.  We did not attempt to
apply non-linear optimization of the frequencies, since this does not
resolve the frequency resolution problem discussed in the previous section.
Note that the number of extracted frequencies is much larger than the number
of real frequencies.  This is not surprising because, in general, the
amplitudes in this simulation are much larger than in the exponential
distribution which increases the cascading effect that arises when
frequencies are unresolved.

\begin{table}
\centering
\caption{Results of frequency extraction using the Lomb periodogram on
simulated data with uniform amplitude distribution.  $N_0$ is the total
number of simulated frequencies and $N$ is the number of
simulated frequencies with amplitudes above the FAP = 0.01 threshold.
$N_{\rm ex}$ is the number of extracted frequencies with FAP $<$ 0.01
using the Lomb-Scargle periodogram.}
\begin{tabular}{rrr}
\hline
\hline
   $N_0$ & $N$ & $N_{\rm ex}$ \\
\hline
  30 &     30 &   105 \\
  50 &     50 &   238 \\
 100 &     99 &   490 \\
 200 &    199 &  1226 \\
 300 &    299 &  1833 \\
 500 &    499 &  2684 \\
 700 &    699 &  3217 \\
1000 &    998 &  3711 \\
2000 &   1998 &  3711 \\
3000 &   2996 &  5212 \\
\\
\hline
\end{tabular}
\label{uni}
\end{table}

\begin{figure}
\centering
\includegraphics{amphist.ps}
\caption{Distribution of amplitudes extracted from the Lomb periodogram for
a selection of simulated data.  The solid histogram shows the relative number
of extracted frequencies as a function of amplitude (mmags).  The dashed
histogram is the true distribution.  The panels are labeled according to the
total number of known frequencies.}
\label{amphist}
\end{figure}

\begin{table*}
\centering
\caption{Significant frequencies and amplitudes in HD~50844.  The frequency $f$ is
in d$^{-1}$, the amplitude $A$ in mmillimags and the phase $\phi$ in radians.
This is a fit to $A\cos(2\pi f(t - t_0) + \phi)$ with $t_0 = 2590.0000$.
The second and third columns list mode identifications from spectroscopy,
$(l,m)$, and radial velocity amplitudes, $A_{\rm RV}$ (km\,s$^{-1}$) from
\citet{Poretti2009}.}
\begin{tabular}{lrrrrr}
\hline
\hline
 Term   & (l,m)  & $A_{\rm RV}$   &\multicolumn{1}{c}{$f$ (d$^{-1}$)}
&\multicolumn{1}{c}{$A$ (mmag)}  & \multicolumn{1}{c}{$\phi$ (rad)}\\
\hline
$f_1 $   & (0,0) & 0.72 & $ 6.92527 \pm 0.00007$ & $16.73 \pm 0.01$ & $ 0.575 \pm 0.001$ \\
$f_2 $   & (3,1) & 0.39 & $11.21672 \pm 0.00008$ & $ 6.76 \pm 0.01$ & $ 1.559 \pm 0.001$ \\
$f_3 $   & (5,1) &      & $11.25803 \pm 0.00010$ & $ 4.37 \pm 0.01$ & $ 0.773 \pm 0.002$ \\
$f_4 $   & (3,3) & 0.52 & $12.84848 \pm 0.00010$ & $ 3.55 \pm 0.01$ & $-2.281 \pm 0.002$ \\
$f_5 $   &       & 0.45 & $12.23840 \pm 0.00009$ & $ 3.53 \pm 0.01$ & $-0.010 \pm 0.002$ \\
$f_6 $   & (4,3) &      & $14.44681 \pm 0.00008$ & $ 3.08 \pm 0.01$ & $ 1.986 \pm 0.003$ \\
$f_7 $   & (3,2) &      & $13.27339 \pm 0.00008$ & $ 2.87 \pm 0.01$ & $-2.871 \pm 0.003$ \\
$f_8 $   & (5,3) & 0.11 & $13.35871 \pm 0.00008$ & $ 2.51 \pm 0.01$ & $ 1.803 \pm 0.003$ \\
$f_9 $   & (11,7)& 0.05 & $11.75098 \pm 0.00010$ & $ 1.58 \pm 0.01$ & $ 1.585 \pm 0.005$ \\
$f_{10}$ & (2,-2)& 0.08 & $ 5.26680 \pm 0.00011$ & $ 1.24 \pm 0.01$ & $ 1.362 \pm 0.007$ \\
$f_{11}$ &       & 0.20 & $14.46124 \pm 0.00011$ & $ 1.24 \pm 0.01$ & $ 0.071 \pm 0.007$ \\
$f_{12}$ & (4,2) & 0.10 & $14.43221 \pm 0.00011$ & $ 1.08 \pm 0.01$ & $-2.167 \pm 0.008$ \\
$f_{13}$ &       & 0.10 & $ 9.95266 \pm 0.00011$ & $ 1.02 \pm 0.01$ & $ 2.798 \pm 0.008$ \\
         &       &      & $ 0.00497 \pm 0.00011$ & $ 1.26 \pm 0.04$ & $-2.483 \pm 0.029$ \\
$f_{14}$ &       &      & $11.98370 \pm 0.00011$ & $ 0.94 \pm 0.01$ & $ 1.148 \pm 0.009$ \\
$f_{15}$ &       &      & $ 6.55652 \pm 0.00011$ & $ 0.82 \pm 0.01$ & $ 2.895 \pm 0.010$ \\
$f_{17}$ &       &      & $13.56876 \pm 0.00011$ & $ 0.85 \pm 0.01$ & $-3.090 \pm 0.010$ \\
$f_{16}$ &       &      & $ 7.40484 \pm 0.00011$ & $ 0.85 \pm 0.01$ & $ 1.412 \pm 0.010$ \\
$f_{18}$ & (5,0) &      & $10.26038 \pm 0.00011$ & $ 0.79 \pm 0.01$ & $-0.864 \pm 0.010$ \\
$f_{19}$ &       & 0.08 & $ 5.78209 \pm 0.00011$ & $ 0.74 \pm 0.01$ & $-1.964 \pm 0.011$ \\
         &       &      & $11.70934 \pm 0.00012$ & $ 0.69 \pm 0.01$ & $-1.332 \pm 0.012$ \\
$f_{20}$ &       &      & $ 6.62851 \pm 0.00012$ & $ 0.71 \pm 0.01$ & $ 1.919 \pm 0.012$ \\
         &       &      & $12.12271 \pm 0.00012$ & $ 0.70 \pm 0.01$ & $ 1.160 \pm 0.012$ \\
         &       &      & $ 7.63511 \pm 0.00012$ & $ 0.66 \pm 0.01$ & $ 2.999 \pm 0.013$ \\
         &       &      & $27.29546 \pm 0.00012$ & $ 0.66 \pm 0.01$ & $ 2.046 \pm 0.013$ \\
$f_{22}$ & (3,1) &      & $14.47801 \pm 0.00011$ & $ 0.68 \pm 0.01$ & $ 1.201 \pm 0.013$ \\
         &       &      & $14.23452 \pm 0.00011$ & $ 0.56 \pm 0.01$ & $ 2.125 \pm 0.015$ \\
$f_{27}$ & (4,-2)&      & $ 5.67451 \pm 0.00012$ & $ 0.61 \pm 0.01$ & $-1.642 \pm 0.014$ \\
$2f_1$   &       &      & $13.84992 \pm 0.00012$ & $ 0.59 \pm 0.01$ & $-2.488 \pm 0.014$ \\
         &       &      & $ 4.66075 \pm 0.00012$ & $ 0.53 \pm 0.01$ & $ 0.390 \pm 0.016$ \\
         &       &      & $ 1.28065 \pm 0.00012$ & $ 0.51 \pm 0.01$ & $ 3.041 \pm 0.016$ \\
$f_{30}$ &    -  & 0.08 & $ 5.41839 \pm 0.00012$ & $ 0.53 \pm 0.01$ & $-2.480 \pm 0.016$ \\
$f_{32}$ & (5,-2)&      & $ 5.49082 \pm 0.00012$ & $ 0.51 \pm 0.01$ & $-0.946 \pm 0.016$ \\
         &       &      & $ 3.48107 \pm 0.00012$ & $ 0.47 \pm 0.01$ & $ 0.403 \pm 0.018$ \\
         &       &      & $11.33797 \pm 0.00012$ & $ 0.48 \pm 0.01$ & $-0.782 \pm 0.017$ \\
         &       &      & $ 1.17976 \pm 0.00012$ & $ 0.46 \pm 0.01$ & $-2.041 \pm 0.018$ \\
$f_{44}$ & (6,4) &      & $14.60065 \pm 0.00012$ & $ 0.46 \pm 0.01$ & $ 0.842 \pm 0.018$ \\
         &       &      & $14.26363 \pm 0.00012$ & $ 0.46 \pm 0.01$ & $ 0.472 \pm 0.018$ \\
         &       &      & $18.72302 \pm 0.00012$ & $ 0.45 \pm 0.01$ & $ 1.073 \pm 0.019$ \\
         &       &      & $ 7.25400 \pm 0.00012$ & $ 0.44 \pm 0.01$ & $ 2.297 \pm 0.019$ \\
         &       &      & $10.62982 \pm 0.00012$ & $ 0.43 \pm 0.01$ & $-1.538 \pm 0.019$ \\
         &       &      & $14.06179 \pm 0.00012$ & $ 0.42 \pm 0.01$ & $ 2.927 \pm 0.020$ \\
         &       &      & $15.75242 \pm 0.00012$ & $ 0.42 \pm 0.01$ & $-0.870 \pm 0.020$ \\
$f_{43}$ & (4,-2)& 0.12 & $ 5.04317 \pm 0.00012$ & $ 0.42 \pm 0.01$ & $-1.560 \pm 0.020$ \\
         &       &      & $ 0.32135 \pm 0.00012$ & $ 0.42 \pm 0.01$ & $-0.776 \pm 0.020$ \\
         &       &      & $13.06176 \pm 0.00013$ & $ 0.40 \pm 0.01$ & $ 3.040 \pm 0.021$ \\
$f_{50}$ &(12,10)&      & $15.22420 \pm 0.00012$ & $ 0.39 \pm 0.01$ & $-2.349 \pm 0.021$ \\
$f_{46}$ &(14,12)&      & $19.75031 \pm 0.00012$ & $ 0.39 \pm 0.01$ & $-0.179 \pm 0.021$ \\
         &       &      & $ 1.23137 \pm 0.00013$ & $ 0.38 \pm 0.01$ & $-1.064 \pm 0.022$ \\
         &       &      & $ 3.38306 \pm 0.00013$ & $ 0.37 \pm 0.01$ & $ 2.196 \pm 0.023$ \\
         &       &      & $ 0.93230 \pm 0.00013$ & $ 0.37 \pm 0.01$ & $ 0.115 \pm 0.023$ \\
$f_{51}$ & (8,5) & 0.08 & $11.64339 \pm 0.00013$ & $ 0.36 \pm 0.01$ & $ 1.352 \pm 0.023$ \\
         &       &      & $ 2.22293 \pm 0.00013$ & $ 0.35 \pm 0.01$ & $-2.960 \pm 0.024$ \\
         &       &      & $ 3.79566 \pm 0.00013$ & $ 0.34 \pm 0.01$ & $-0.829 \pm 0.024$ \\
         &       &      & $ 3.57994 \pm 0.00013$ & $ 0.33 \pm 0.01$ & $-0.908 \pm 0.025$ \\
         &       &      & $ 6.57856 \pm 0.00013$ & $ 0.35 \pm 0.01$ & $ 1.095 \pm 0.024$ \\
         &       &      & $15.56830 \pm 0.00013$ & $ 0.33 \pm 0.01$ & $ 2.248 \pm 0.025$ \\
         &       &      & $ 2.85138 \pm 0.00013$ & $ 0.32 \pm 0.01$ & $-1.230 \pm 0.026$ \\
         &       &      & $ 8.56364 \pm 0.00013$ & $ 0.31 \pm 0.01$ & $ 1.316 \pm 0.027$ \\
\\
\hline
\end{tabular}
\label{hd}
\end{table*}

In Fig.\ref{amphist} we show the distribution of amplitudes for four
selected simulations.  In all cases, the true distribution is flat (uniform
amplitude distribution) and extends from 0--30~d$^{-1}$.  The amplitude
distributions derived from the extracted frequencies are very different and
always show a large number of low-amplitude frequencies (out of scale in the
Figure).  In addition, there is considerable spillage to high amplitudes
due to two or more frequencies which are unresolved.  The larger the number
of frequencies the greater the number of frequencies at very low and very
high amplitudes.

It is clear that high mode density leads to a grossly distorted amplitude
distribution obtained by successive prewhitening.  In fact, numerical
experiments show that if successive prewhitening is performed until the
false alarm probability of the highest peak reaches a reasonable level ($P_A
= 0.01 or 0.001$), the resulting amplitude distribution is practically the same,
independently of the actual distribution that is chosen for the simulation.
The distribution is dominated by the low-amplitude peaks which are mostly
false peaks arising from unresolved frequencies.

\section{Avoiding prewhitening}

Given the problems associated with continuous prewhitening, it is clear that
another method of extracting frequencies is desirable for data with very
high S/N.  One possibility is simply to locate the peaks in the periodogram
of the raw (i.e. non-prewhitened) data.  This has to be done with care
and in combination with prewhitening of the most dominant peaks because much
of the structure in the periodogram is a result of the window function.

\section{Frequencies in HD~50844}

If we apply the Lomb-Scargle criterion to the 30-s cadence data of HD~50844
we find that there are about 1800 frequencies with amplitudes in excess of
0.014~mmag  ($P_A = 0.01$) and 1700 frequencies in excess of 0.015~mmag
($P_A = 0.001$), mostly  within the range 0--30~d$^{-1}$.  This is
equivalent to 60 frequencies per d$^{-1}$ which is at the limit of
frequency resolution.  As we have seen, these numbers are likely to be
considerably larger than actually present in the star.

In Table\,\ref{hd} we list the frequencies of highest amplitude.  Since
all these frequencies have quite large amplitudes, there is no danger of
them being fictitious.  Our frequencies and amplitudes agree very well with
those of \citet{Poretti2009} and we have adopted the same numbering scheme for
the frequencies.  There are very few combination frequencies; we find
$f_{10} = 4f_1 - 2f_2$  and $f_{12} = 2f_4-f_3$, but these are most probably
coincidences.  The harmonic $2f_1$ is clearly visible.

\section{Comparison with {\it Kepler} $\delta$~Scuti stars}

One question that is of interest is whether HD~50844 has a higher mode
density than other $\delta$~Scuti stars.  We cannot really answer this
question because it is not possible to extract frequencies below a certain
threshold level.  Also, we have seen that the finite time resolution and the
prewhitening technique provide severe limitations on the frequencies that
can be reliably extracted.  What can be answered is whether the amplitude
density in the periodogram is larger than in other $\delta$~Scuti stars.
To do this we need to compare the amplitude density in HD~50844 with those
of other stars.

The {\it Kepler} data provide by far the most homogeneous sample of light
curves of $\delta$~Sct stars.  Nearly all $\delta$~Sct stars observed by
{\it Kepler} were discovered in the data from Quarters 0,1 and 2 which are
in the public domain.  However, for most of these stars the length of the
time series is about 30~d.   Since frequency resolution is important for
determining mode density, we need to ensure that the length of the time
series for each star is the same.  We decided to truncate the time series of
all the {\it Kepler} stars to 25~d in order to obtain the largest sample of
$\delta$~Scuti stars.  Also, we limit the data to stars observed with short
cadence (exposure times of about 60~s).  While considerably more {\it
Kepler} $\delta$~Sct stars have been observed with long cadence (exposure
time about 30~min), these cannot be used because the highest frequency
that can be detected is only about 24~d$^{-1}$, whereas most $\delta$~Sct
stars still have modes with frequencies as high as 50~d$^{-1}$.

There are 357 $\delta$~Scuti stars observed in short-cadence mode in the
{\it Kepler} database.  Since all these stars were observed for the same
length of time, we can use the area of the periodogram above a certain
amplitude level as proportional to the number of modes with amplitudes higher
than this level.  Fig.\,\ref{moden} shows this plot for the {\it Kepler}
$\delta$~Sct stars.  The top panel shows the plot for HD\,50844; the star
is clearly not exceptional in this regard.

\begin{figure}
\centering
\includegraphics{moden.ps}
\caption{Relative amplitude density as a function of amplitude for {\it Kepler}
$\delta$~Scuti stars.  The top panel shows the same plot expanded to show
HD~50844 (thick line).}
\label{moden}
\end{figure}

\section{Conclusion}

Using simulations, we show that it is not possible to extract frequencies
reliably down to the expected noise level using successive prewhitening.
The reason is due to unresolved peaks in the periodogram which become more
numerous as the amplitude decreases.  Such unresolved peaks lead to a
multitude of erroneous frequencies. Prewhitening by a frequency which differs
from the true frequency by a significant amount leaves many frequency residuals
of lower amplitude.  In the high S/N data from space missions, these spurious
residual frequencies often have significant amplitudes, leading to a cascading
effect of spurious frequencies.  Simulations show that frequency extraction
using the Lomb-Scargle or four sigma criterion can lead to a gross over-estimate of
the true number of frequencies.  Using nonlinear optimization does not solve
this problem.  It appears that the extraordinary high mode density of the
{\t CoRoT} star HD\,50844 found by \citet{Poretti2009} is not real, but a
result of this phenomenon.

Because of the effect described above, it is not possible to distinguish
between frequencies present in the star and spurious low-amplitude frequencies
caused by prewhitening of unresolved frequency groups.  Thus it is
not possible to count the number of individual modes with any degree of
certainty below a certain amplitude level.  One can, however, compare the
power in the periodogram above a given amplitude for different stars.  We
made such a comparison for HD\,50844 with 357 $\delta$~Scuti stars in the
{\it Kepler} public archive.  It turns out that HD\,50844 is not exceptional
in this regard.

\bibliographystyle{mn2e}