$C(K, K)$ equipped with certain topology compact?

Question

Let $K$ be a compact Hausdorff space, let $C(K)$ be the space of continuous functions $K \rightarrow \mathbb C$ equipped with the norm $\left\Vert \cdot\right\Vert _{\infty}$ and let $C(K, K)$ be the space of all continuous functions $K \rightarrow K$. We define a topology on $C(K, K)$ in the following way:

A net $(g_i)_{i\in I}\subseteq C(K, K)$ shall converge to an Element $g \in C(K, K)$ iff $f\circ g_i \rightarrow f\circ g$ in $C(K)$ for every $f \in C(K)$.

As I defined this topology myself my first question is if it is (well-) known or/and has a name. Further I'm wondering if $C(K, K)$ is compact in this topology.

Answer 1

This is not a complete answer but if $K$ is a compact subset of $\mathbb C$ your topology seems to be the same as the topology of $\|\cdot \|_\infty$:

• With $f=id \in C(K)$ convergence for your topology implies converge for $\|\cdot \|_\infty$.
• If $\|g_n-g\|_\infty \to 0$ let $f \in C(K)$. Let us show that $\|f \circ g_n - f\circ g \|_\infty \to 0$.

Let $\epsilon >0$. By Heine, as $f$ is continuous on the compact $K$, $f$ is uniformly continuous. So there exists $\delta >0$ such that $|x-y| < \delta \implies |f(x)-f(y)|<\epsilon$. But for $n$ sufficiently large, $\forall x \in K$, $|g_n(x)-g(x)|<\delta$.

So we obtain: for all $\epsilon >0$, for $n$ sufficiently large: $$\|f \circ g_n - f\circ g \|_\infty < \epsilon$$ i.e $\|f \circ g_n - f\circ g \|_\infty \to 0$.

Answer 2

I don't know if that notion of convergence properly defines a topology, but still, I don't expect $C(K, K)$ to be compact.

Indeed, reasoning as in Delta-u's answer, we see that when $K=[0, 1]$ the space $C([0, 1], [0, 1])$ coincides with the unit ball of $C([0,1])$. (By the way, here $C(X)$ denotes the Banach space of real-valued continuous functions on $X$, equipped with the uniform norm). This unit ball is not compact.

It is probably the case that $C(K, K)$ is compact only in the trivial case in which $K$ is a finite set.