A ray is a continuous one-to-one image of the half-line $[0,\infty)$.


If $f:[0,\infty)\to \mathbb R ^2$ is continuous and one-to-one, then we say that the ray $X=f[0,\infty)$ limits onto itself if for each $x\in X$ there exists a sequence $(a\_n)\in [0,\infty)^\omega$ such that $a\_n\to\infty$ and $f(a\_n)\to x$. This is equivalent to saying that each initial segment of the ray is nowhere dense in the ray.


A connected space is said to be indecomposable if it cannot be written as the union of two proper closed connected subsets.



> 
> **Question.** Is every plane ray which limits onto itself indecomposable?
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F.B. Jones has shown that every locally connected plane ray is locally compact. Note that a ray limiting onto itself must be first category and therefore not locally compact. So a ray limiting onto itself is not locally connected.


S. Curry has shown that if the closure of the ray is one-dimensional and non-separating, then the ray (as well as its closure) is indecomposable.


If it helps, assume that the ray is nowhere dense in the plane. Then we at least know that the closure is one-dimensional.


*Jones, F. Burton*, [**One-to-one continuous images of a line**](http://dx.doi.org/10.4064/fm-67-3-285-292), Fundam. Math. 67, 285-292 (1970). [ZBL0192.60102](https://zbmath.org/?q=an:0192.60102).


*Curry, Stephen B.*, [**One-dimensional nonseparating plane continua with disjoint (\epsilon)- dense subcontinua**](http://dx.doi.org/10.1016/0166-8641(91)90014-D), Topology Appl. 39, No. 2, 145-151 (1991). [ZBL0718.54042](https://zbmath.org/?q=an:0718.54042).



A plane ray which limits onto itself
It seems like there are decomposable rays which limit onto themselves.


Let $E=\bigcup\_{n\geq0}E\_n$ and $O=\bigcup\_{n\geq0}O\_n$, with $E\_n=[2n,2n+1]$ and $O\_n=[2n+1,2n+2]$. We want to construct a ray $f:[0,\infty)$ which limits onto itself and such that $f(E),f(O)$ are closed in the ray $X$ and connected.


For $f(E)$ to be connected it will be enough to ensure that $\forall x\in f(E)\;\forall\varepsilon>0\;\exists N$ such that $\forall n>N$, $d(x,f(E\_n))<\varepsilon$. Indeed, if that happens any clopen of $f(E)$ which contains $x$ will have to contain $f(E\_n)$ for $n$ big enough, so it will contain all $f(E)$.


For $f(E)$ to be closed in $X$ it will be enough to assign a small tubular neighborhood $B\_n$ to the set $f(O\_n^o)$ (where $O\_n^o$ is the interior of $O\_n$) such that $f(E)$ doesn't intersect $B\_n$.


The construction will be inductive. Set $f([0,2])$ as some arbitrary smooth curve. Now given $n\geq1$, suppose that we have defined $f$ in $[0,2n]$ injectively and we have nhoods $A\_i$ of $f(E\_i^o)$ and $B\_i$ of $f(O\_i^o)$, for $i=1,\dots, n-1$, such the sets $A\_i$ don't intersect $f(O\cap[0,2n])$ and the $B\_i$ don't intersect $f(E\cap[0,2n])$. Suppose also that the $A\_i$ are pairwise disjoint and bounded by simple closed curves (and the same for the $B\_i$).


This means that the set $R\_n=\mathbb{R}^2\setminus (f([0,2n])\cup \cup\_{i=1}^{n-1}\overline{B\_i})$ is homeomorphic to the annulus $(0,1)\times\mathbb{S}^1$, and its boundary is a non injective curve formed by the $f(E\_i)$ and the boundaries of the $B\_i$ (similarly for $S\_n=\mathbb{R}^2\setminus (f([0,2n])\cup \cup\_{i=1}^{n-1}\overline{A\_i})$).


We will define $f(E\_n)$ as a smooth injective curve in $R\_n$ (except for the initial point $f(2n)$). We can define this curve so that it passes at distance $<2^{-n}$ of every point of the set $f(E\cap[0,2n])$, which is contained in the boundary of $R\_n$. Similarly, we define $f(O\_n)$ as a smooth curve in $S\_n\setminus f(E\_n)$ (which is homeomorphic to an annulus) which passes at distance $<2^{-n}$ of every point of $f(O\cap[0,2n])$. Then define $A\_n,B\_n$ to be small tubular neighborhoods of $f(E\_n^o)$ and $f(O\_n^o)$ such that the conditions above are satisfied.


This could be modified to make the curve nowhere dense: at each step we can enlarge $A\_n$ and $B\_n$ so that they contain some point $p\_n$ which is not in $f([0,2n+2])$. If we take a sequence $p\_n$ dense in the plane, the ray won't intersect a nhood of each $p\_n$, so it will be nowhere dense. If for some $n$, $p\_n$ is already in $\overline{A\_i}$ for some $i<n$, we wouldn't need to enlarge $A\_n$, and the same for $B\_n$ (this is to keep the $A\_i$ pairwise disjoint).


