MHB Is Every Continuous Function on a Metric Space Bounded?

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Here is this week's POTW:

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Show that if a metric space $X$ has the property that every real-valued continuous function on $X$ is bounded, then $X$ is compact.
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Remember to read the http://www.mathhelpboards.com/showthread.php?772-Problem-of-the-Week-%28POTW%29-Procedure-and-Guidelines to find out how to http://www.mathhelpboards.com/forms.php?do=form&fid=2!

 

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Hi MHB community,

I've removed one of the directions of the original problem. The edited problem statement you see above is the "meat" of the original statement. Good luck and thanks to all who participate!

 

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No one answered this week's problem. You can read my solution below.

Spoiler

Proof by contraposition. If $X$ is not compact, then there is an infinite sequence $\{x_n\}_{n\in \Bbb N}$ which has no convergent subsequence. The set $E = \{x_n : n\in \Bbb N\}$ has no point of accumulation, so it is a closed and discrete subset of $X$. Define a function $f : E\to \Bbb R$ by setting $f(x_n) = n$ if $x_n$ is unique and $0$ otherwise. Since $E$ is discrete, $f$ is continuous. By Tietze's extension theorem, $f$ has a continuous extension to all of $X$, call it $F$. As $E$ has infinitely many distinct points, $F(x_n) = n$ for infinitely many $n$. So $F$ is unbounded.