For any topological space $(X,\tau)$, let $\text{End}(X)$ denote the set of continuous functions $f:X\to X$. We say that ${\cal C}\subseteq \text{End}(X)$ *covers* $\text{End}(X)$ if for every $f\in \text{End}(X)$ there is $h\in {\cal C}$ and $x\in X$ such that $f(x) = h(x)$. 


I am interested in *minimal* covers (that is, covers that have the property that when you remove a member, it is no longer a cover). A boring example is the set of all constant functions which always has the same cardinality as $X$ - and taking away one member of the collection of all constant functions destroys the "covering" property.


Note that $\text{End}(\mathbb{R})$ does have a countable cover: For $z\in\mathbb{Z}$ let $f\_z:\mathbb{R}\to\mathbb{R}$ be defined by $r \mapsto r+z$. Let $${\cal C} = \{f\_z: z\in\mathbb{Z}\}\cup\{{\bf 0}\}, $$ where ${\bf 0}$ denotes the constant zero function. Clearly this cover does not have a minimal subcover.


**Question.** Does ${\mathbb R}$ have a countable minimal cover?



Minimal covering sets of continuous endomorphisms
Here is an example:


**Example.** Let $g$ be a continuous strictly increasing function such that $\lim\_{x \to -\infty} g(x) = -1$ and $\lim\_{x \to \infty} g(x) = 1$; for example
$$g(x) = \tfrac{2}{\pi}\arctan(x).$$
For $n \in \mathbf Z$, let $f\_n(x) = n$ and $g\_n(x) = g(x) + n + \tfrac{1}{2}$. Then $\{f\_n\} \cup \{g\_n\}$ form a covering: if $h$ is continuous and does not intersect any $f\_n$, then there exists $n \in \mathbf Z$ such that $n < h(x) < n+1$ for all $x$, so $h$ intersects $g\_n$ as $\lim\_{x \to -\infty} g\_n(x) < n$ and $\lim\_{x \to \infty} g\_n(x) > n + 1$.


On the other hand, if we remove any $g\_n$, then the constant function $h(x) = n + \tfrac{1}{2}$ lies strictly between $f\_n$ and $f\_{n+1}$ and strictly between $g\_{n-1}$ and $g\_{n+1}$, so does not intersect any member. Similarly, if we remove $f\_n$, then the function $h(x) = g(x) + n = g\_n(x) - \tfrac{1}{2}$ lies strictly between $f\_{n-1}$ and $f\_{n+1}$ and strictly between $g\_{n-1}$ and $g\_n$, so does not intersect any member. $\square$


**Remark.** There are no *finite* covers: if $\{f\_1,\ldots,f\_n\}$ is a finite collection of continuous functions, then the function
$$g = \max(f\_1,\ldots,f\_n) + 1$$
is continuous (since $\max \colon \mathbf R^n \to \mathbf R$ is continuous) and does not intersect any of the $f\_i$.


