Is there a lighter restriction on $X$ that makes the topology of compact convergence on $C(X)$ metrizable?

Question

I have recently reading a theorem involved compact convergence. Another definition can be found here.

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Let $(X, \mathcal{T})$ be a topological space and $\left(Y, d_{Y}\right)$ be a metric space. A sequence of functions $$ f_{n}: X \rightarrow Y, n \in \mathbb{N}, $$ is said to converge compactly as $n \rightarrow \infty$ to some function $f: X \rightarrow Y$ if, for every compact set $K \subseteq X$, $$ \left.\left.f_{n}\right|_{K} \rightarrow f\right|_{K} $$ uniformly on $K$ as $n \rightarrow \infty$. This means that for all compact $K \subseteq X$, $$ \lim _{n \rightarrow \infty} \sup _{x \in K} d_{Y}\left(f_{n}(x), f(x)\right)=0 $$

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Some properties (from the same Wikipedia page):

• If $f_{n} \rightarrow f$ uniformly, then $f_{n} \rightarrow f$ compactly.
• If $(X, \mathcal{T})$ is a compact space and $f_{n} \rightarrow f$ compactly, then $f_{n} \rightarrow f$ uniformly.
• If $(X, \mathcal{T})$ is a locally compact space, then $f_{n} \rightarrow f$ compactly if and only if $f_{n} \rightarrow f$ locally uniformly.

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Let $X$ be compact and $Y:=\mathbb R$. If we consider the space $C(X)$ of real-valued continuous functions on $X$, then compact convergence reduces trivially to uniform convergence which is metrizable.

We keep working with $C(X)$. Is there a lighter restriction on $X$ that makes the topology of compact convergence metrizable?

Answer 1

If $X$ is $\sigma-$ compact, i.e. a countable union of compact sets $K_n$ then $d(f,g)=\sum\limits_{k=1}^{\infty}\frac {\|f-g\|_n} {2^{n}(1+\|f-g\|_n)}$ define a metric which gives uniform convergence on compact sets. Here $\|f-g\|_n=\sup \{|f(x)-g(x)|: x \in K_n\}$.