# Show that $C(X)$ is metrizable if and only if $X = \cup_n K_n$.

This is Example. 2.2 in Chapter 4 of Conway's book on Functional Analysis :

If X is completely regular and $$C(X)$$ is the set of all continuous functions from $$X$$ to $$\mathbb{F}$$ then $$C(X)$$ is metrizable if and only if $$X = \cup_n K_n$$ where each $$K_n$$ is compact, $$K_1 \subset K_2 \subset \dots$$, and if $$K$$ is any compact subset of $$X$$, then $$K \subset K_n$$ for some $$n$$.

I don't recall a theorem related to this neither could I find anything about this on the Internet. How to start this i.e. which method I can use for proving this?

I should add that there is a question

compact-open metrizability

However, here, one direction is claimed to be "straight forward exercise" and, for the other direction, the answer just points to a proposed exercise in page 68 of Robert A. McCoy, Ibula Ntantu, Topological Properties of Spaces of Continuous Functions, Springer Lecture Notes in Math, Volume 1315 (1988). So actually no answer is presented.

My Attempt:

If $$X = \cup_n K_n$$ where each $$K_n$$ is compact, $$K_1 \subset K_2 \subset \dots$$, and if $$K$$ is any compact subset of $$X$$, then $$K \subset K_n$$ for some $$n$$, then considering the family $$\{p_{K_n} \}$$ of seminorms I proved that for any $$K$$ compact, there is a $$n$$ such that $$p_K \leq p_{K_n}$$. So, using the family $$\{p_{K_n} \}$$, we can dominated all other seminorms $$p_K$$. It follows that the $$\{p_{K_n} \}$$ defined the topology of $$C(X)$$ and then $$C(X)$$ is metrizable. I am not sure how to make the argument really rigorous and I think it may be wrong because I did not use that $$X = \cup_n K_n$$ where each $$K_n$$ is compact, $$K_1 \subset K_2 \subset \dots$$.

If $$C(X)$$ is metrizable, I started with an arbitrary compact $$K_1$$ and I defined by induction $$K_{n+1} = K_n \cup A_n$$ where $$A_n$$ is a compact subset of $$\{x \in X: p_{K_n}(x) \leq 1\} \setminus K_n$$. But I could not prove that $$X = \cup_n K_n$$ where each $$K_n$$ is compact, $$K_1 \subset K_2 \subset \dots$$.

1. In Conway's book all spaces are assumed to be Hausdorff unless the contrary is specified. (see Example 1.6 in chapter III and the paragraph after definition 1.2 in chapter IV).

2. Example 1.5. Let $$X$$ be a completely regular space and let $$C(X) =$$ all continuous functions from $$X$$ into $$\Bbb F$$. If $$K$$ is a compact subset of $$X$$, define $$p_K(f)=\sup \{|f(x)| : x \in K\}$$. Then $$\{p_K: K \text{ compact in } X\}$$ is a family of seminorms that make $$C(X)$$ into a LCS.

3. (From proposition 2.1)A LCS is metrizable if and only if its topology is determined by a countable family of seminorms.

Now, we want to prove:

(Example 2.2) $$C(X)$$ is metrizable if and only if $$X= \bigcup_{n=1}^\infty K_n$$, where $$K_n$$ is compact, $$K_1 \subseteq K_2 \subseteq \cdots$$, and if $$K$$ is any compact subset of $$X$$, then $$K \subseteq K_n$$ for some $$n$$.

Proof:

$$(\Leftarrow)$$ Let $$F = \{p_K: K \text{ compact in } X\}$$ and $$G=\{p_{K_n}: n\in \{1, \dots \} \}$$. Let $$\tau_F$$ and $$\tau_G$$ be the topologies generated by $$F$$ and $$G$$ respectively. Since $$G \subseteq F$$, we have that $$\tau_G \subseteq \tau_F$$.

Now, given any $$p_K \in F$$, there is $$n$$ such that $$K \subseteq K_n$$. It means that $$p_{K_n} \in G$$ and $$p_K \leq p_{K_n}$$. So, for all $$f_0 \in C(X)$$, $$\{f: p_{K_n}(f-f_0)<\varepsilon \} \subseteq \{f: p_K(f-f_0)<\varepsilon \}$$ It follows immetiately that $$\tau_F \subseteq \tau_G$$. So, $$\tau_F = \tau_G$$. So the topology of $$C(X)$$ (that is $$\tau_F$$) is determined by a countable family of seminorms (that is $$G$$). So, by proposition 2.1, $$C(X)$$ is metrizable.

$$(\Rightarrow)$$ Suppose that $$C(X)$$ is metrizable. Then, repeating the steps in the proof of proposition 2.1, suppose $$\rho$$ is a metric of $$C(X)$$. For each $$n \geq 1$$, let $$U_n=\{f: \rho(f,0)< \frac1n\}$$. Then, there are $$C_1, \cdots C_r$$ compact subsets of $$X$$ and positive numbers $$\varepsilon_1, \cdots \varepsilon_r$$ such that $$\bigcap_{i=1}^r\{f: p_{C_i}(f)<\varepsilon_i \} \subseteq U_n$$

Define $$K_0=\emptyset$$ and let $$K_n=C_1 \cup \cdots \cup C_r \cup K_{n-1}$$ and $$\varepsilon = \min \{\varepsilon_1, \dots , \varepsilon_r\}$$. So, we have,

$$\{f: p_{K_n}(f)<\varepsilon \} \subseteq \bigcap_{i=1}^r\{f: p_{C_i}(f)<\varepsilon_i \} \subseteq U_n$$

As in the proof of proposition 2.1, we can show that $$G=\{p_{K_n}: n\geq 1 \}$$ determines the topology of $$C(X)$$, that is, $$\tau_F$$.

So, given $$K$$ is any compact subset of $$X$$ and any $$\varepsilon >0$$, since $$p_K \in F$$, we have that there are $$p_{K_1}, \cdots p_{K_s} \in G$$ and positive numbers $$\varepsilon_1, \cdots \varepsilon_s$$ such that

$$\bigcap_{i=1}^s\{f: p_{K_i}(f)<\varepsilon_i \} \subseteq \{f: p_K(f)<\varepsilon \}$$

Let $$\delta = \min \{\varepsilon_1, \dots , \varepsilon_s\}$$. Since $$K_1 \subseteq \cdots \subseteq K_s$$, we have

$$\{f: p_{K_s}(f)<\delta \} = \bigcap_{i=1}^s\{f: p_{K_i}(f)<\varepsilon_i \} \subseteq \{f: p_K(f)<\varepsilon \} \tag{1}$$

Since $$X$$ is a completely regular space, we have that $$K \subset K_s$$. In fact, if there is $$x_0 \in K \setminus K_s$$, then there is a continuous function $$h$$ such that $$h(x_0)=1$$ and $$h|_{K_s}=0$$. Take $$a>\varepsilon$$, then $$ah$$ is a continuous function, $$p_K(ah) \geq a > \varepsilon$$ and $$p_{K_s}(ah)=0<\delta$$. So $$ah \in \{f: p_{K_s}(f)<\delta \}$$ and $$ah \notin \{f: p_K(f)<\varepsilon \}$$, which contradicts $$(1)$$. So, $$K \subset K_s$$.

So we proved that for any $$K$$ compact subset of $$X$$, there is $$s \geq 1$$ such that $$K \subseteq K_s$$.

Now, let us prove that $$X= \bigcup_{n=1}^\infty K_n$$. For any $$x\in X$$, $$\{x\}$$ is compact, so by what we have just proved, there is $$s \geq 1$$ such that $$\{x\} \subseteq K_s$$, that is $$x \in K_s$$. So, $$X \subseteq \bigcup_{n=1}^\infty K_n$$. But, for all $$n\geq 1$$, $$K_n \subseteq X$$. So, $$\bigcup_{n=1}^\infty K_n \subseteq X$$. So, $$X = \bigcup_{n=1}^\infty K_n$$. $$\square$$

Remark: Note that in the $$(\Rightarrow)$$ part, we don't need to use directly that $$X= \bigcup_{n=1}^\infty K_n$$, where $$K_n$$ is compact, $$K_1 \subseteq K_2 \subseteq \cdots$$. The reason behind it is clear at the last paragraph of the second part of the answer:

The condition "if $$K$$ is any compact subset of $$X$$, then $$K \subseteq K_n$$ for some $$n$$" (called hemicompacity) implies the condition "$$X= \bigcup_{n=1}^\infty K_n$$, where $$K_n$$ is compact, $$K_1 \subseteq K_2 \subseteq \cdots$$" (called $$\sigma$$-compacity). The proof is exactly the last paragraph of the second part of the answer.

Proof: For any $$x\in X$$, $$\{x\}$$ is compact, so by what we have just proved, there is $$s \geq 1$$ such that $$\{x\} \subseteq K_s$$, that is $$x \in K_s$$. So, $$X \subseteq \bigcup_{n=1}^\infty K_n$$. But, for all $$n\geq 1$$, $$K_n \subseteq X$$. So, $$\bigcup_{n=1}^\infty K_n \subseteq X$$. So, $$X = \bigcup_{n=1}^\infty K_n$$. $$\square$$