Minimal covering sets of continuous endomorphisms

Question

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?

Answer 1

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 $f_i$.