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Show that if any continuous real-valued function on a metric space $X$ attains its maximum value, then $X$ is totally bounded.

So, want to show that for every $\epsilon > 0$ there exists a finite subset $\{x_1,...,x_n\}$ of $X$ s.t. $X\subset\cup_{k=1}^nB(x_k,\epsilon)$.

I'm thinking to prove it using contrapositive, that if $X$ is not totally bounded then there exists a continuous real valued function on $X$ that is unbounded above, but I'm stuck!!

Any help would be appreciated.

rose
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  • wait what? $f:\mathbb{R}\to\mathbb{R}$ given by $f(x)=e^{-x^2}$ is a continuous function that attains a maximum at $x=0$ but there is no way that $\mathbb{R}$ is contained inside of finitely many epsilon balls for any $\varepsilon>0$ – C Squared Jun 05 '21 at 16:18
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    @CSquared We need every continuous function to attain a maximum so $f(x)=x$ wouldn't attain a maximum. – daruma Jun 05 '21 at 16:20
  • @daruma i assume for clarity then, it should read ‘if $\texit{every}$ continuous real-valued...’ – C Squared Jun 05 '21 at 16:23
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    @ArcticChar Specifically : the second case in this answer is a solution to the OP – SolubleFish Jun 05 '21 at 16:31

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