Let $(X,d)$ be a metric space.

Assume every continuous function on $X$ is bounded.

Prove that $X$ is compact.

Well, i don't know which continuous function should i fix to start an argument.

I tried to fix a function $f(x)=d(x,x_0)$, but i think it doesn't work.

How do i prove this?


By contradiction. Suppose $X$ is not compact; therefore one can pick an infinite sequence $x_1,x_2,...$ that does not accumulate at any point in $X$.

Then pick an unbounded function on that sequence, say $f(x_n)=n$, and turn it into a continuous one. This is possible because topology of a metric allows interpolation function between disjoint close subsets.

For example, you can enclose each $x_n$ into balls $x_n\in B_{\epsilon_n}$ such as $B_{\epsilon_n}$ are pairwise disjoint, then set $f(x)=0,x\in (X-\cup B_{\epsilon_n})$ and $f(x)=n(1-\frac{d(x,x_n)}{\epsilon_n}),x\in B_{\epsilon_n}$

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