# $(X,d)$ is metric space. $(X,d)$ is compact if and only if any continuous function on $X$ has a maximum.

$$(X,d)$$ is metric space. $$(X,d)$$ is compact if and only if any continuous function on $$X$$ has a maximum.

I dont know whether these functions real valued or not but only real valued functions may make sense, I think.

In that case

$$\Rightarrow$$ is easy.

About $$\Leftarrow$$

My 1st attempt:

I thought some special functions that I can use e.g. $$\varphi:X\to \mathbb R\\ \varphi(x)=d(x_0,x)$$ for some fix $$x_0\in X$$

Since this is a continuous function then it attains its maximum on $$X$$, that implies $$X$$ is bounded .

With this motivation I can define $$\psi:X\to \mathbb R \\\psi(x)=d(x,X)=\inf\limits_{a\in X}d(x,a)$$

since it attains its maximum so $$X=\overline X$$

So I have boundness and closedness but these dont really imply compactness in metric spaces.

My 2nd attempt was:

Considering any sequences in $$X$$ and using $$f(x_{k_n})$$ is convergent iff $$x_{k_n}$$ is convergent. By showing every sequence has convergent subsequence and using continuity of any $$f:X\to\mathbb R$$.

I am stuck is there hint or answer you can give me?

• "I dont know whether these functions real valued or not but only real valued functions may make sense, I think." They are most likely talking about real-valued functions with the standard order and metric on $\Bbb R$, yes. That ought to be specified, though. Jan 21, 2021 at 20:48
• Here is a proof. Assuming (X,d) is not compact, you can explicitly construct a continuous function without a maximum.
– user460426
Jan 21, 2021 at 21:28

HINT: Prove the contrapositive: assume that $$X$$ is not compact, and construct a continuous real-valued function that does not attain a maximum. Perhaps the simplest approach is to show that if $$X$$ is not compact, it has a countably infinite closed discrete subset $$D=\{x_n:n\in\Bbb N\}$$. Define $$f:D\to\Bbb N:x_n\mapsto n$$ and apply the Tietze extension theorem.

Added. It’s more work, but you can actually avoid the Tietze extension theorem. Every subset of $$D$$ is closed, and $$X$$ is normal, so we can carry out the following construction. There are open sets $$U_0$$ and $$V_0$$ such that $$x_0\in U_0$$, $$D\setminus\{x_0\}\subseteq V_0$$, and $$\operatorname{cl}U_0\cap\operatorname{cl}V_0=\varnothing$$. Suppose that $$n\in\Bbb N$$, and we already have an open set $$V_n$$ containing every $$x_k$$ with $$k>n$$; then there are open subsets $$U_{n+1}$$ and $$V_{n+1}$$ of $$V_n$$ such that $$x_{n+1}\in U_{n+1}$$, $$x_k\in V_{n+1}$$ for each $$k>n+1$$, and $$\operatorname{cl}U_{n+1}\cap\operatorname{cl}V_{n+1}=\varnothing$$. In this way we recursively construct open sets $$U_n$$ with pairwise disjoint closures such that $$x_n\in U_n$$ for each $$n\in\Bbb N$$.

Let

$$\mathscr{U}=\left\{X\setminus\bigcup_{n\in\Bbb N}U_n\right\}\cup\{U_n:n\in\Bbb N\}\,;$$

$$\mathscr{U}$$ is an open cover of $$X$$. $$X$$ is a metric space, so it is paracompact, and $$\mathscr{U}$$ therefore has a locally finite open refinement $$\mathscr{R}$$. For each $$R\in\mathscr{R}$$ there is a $$U_R\in\mathscr{U}$$ such that $$R\subseteq U_R$$. For each $$U\in\mathscr{U}$$ let $$V_U=\bigcup\{R\in\mathscr{R}:U_R=U\}$$, and let $$\mathscr{V}=\{V_U:U\in\mathscr{U}\}$$. It is not hard to show that $$\mathscr{V}$$ is locally finite, and it is clearly an open refinement of $$\mathscr{U}$$, since $$V_U\subseteq U$$ for each $$U\in\mathscr{U}$$.

For $$n\in\Bbb N$$ let $$V_n=V_{U_n}$$. Note that $$U_n$$ is the only member of $$\mathscr{U}$$ containing $$x_n$$, so $$x_n\in V_n$$. $$X$$ is completely regular, so for each $$n\in\Bbb N$$ there is a continuous function $$f_n:X\to[0,1]$$ such that $$f(x_n)=1$$ and $$f_n[X\setminus V_n]=\{0\}$$. For $$x\in X$$ let

$$f(x)=\sum_{n\in\Bbb N}nf_n(x)\,;$$

clearly $$f$$ is a function from $$X$$ to the non-negative reals, and $$f(x_n)=n$$ for each $$n\in\Bbb N$$, so $$f$$ is unbounded above. It’s also clear that $$f\upharpoonright V_n=f_n$$ for each $$n\in\Bbb N$$, so $$f$$ is continuous at each point of $$\bigcup_{n\in\Bbb N}V_n$$.

Finally, let $$x\in X\setminus\bigcup_{n\in\Bbb N}V_n$$; clearly $$f(x)=0$$. If $$x$$ has an open nbhd $$W$$ disjoint from $$\bigcup_{n\in\Bbb N}V_n$$, then $$f[W]=\{0\}$$, and $$f$$ is continuous at $$x$$. If not, then $$x\in\operatorname{cl}\bigcup_{n\in\Bbb N}V_n$$. $$\mathscr{V}$$ is locally finite, so $$x$$ has an open nbhd $$W$$ such that $$F=\{n\in\Bbb N:W\cap V_n\ne\varnothing\}$$ is finite. Then

$$x\in\operatorname{cl}\bigcup_{n\in F}V_n=\bigcup_{n\in F}\operatorname{cl}V_n\,,$$

and the sets $$\operatorname{cl}V_n$$ are pairwise disjoint, so there is a unique $$n(x)\in\Bbb N$$ such that $$x\in\operatorname{cl}V_{n(x)}$$. I’ll leave the very last step to you: use this fact to show that $$f$$ is continuous at $$x$$, thereby completing the proof that $$f$$ is continuous.

• Is this the contrapositive? It seems that they want to prove the other direction.
– user460426
Jan 21, 2021 at 21:26
• @Flowsnake: You’re absolutely right; fixed now. Jan 21, 2021 at 21:31
• @BrianM.Scott Thank you very much for this great answer, but can you give me hint to do the problem without using any extention theorem? I wonder there is another approach? Jan 24, 2021 at 22:48
• @Jale'dejaleuffnejale: I found a way to do so; it’s more work, and it does require that you know that a metric space is paracompact. Jan 24, 2021 at 23:48