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I already proved the first statement:

If $M$ is compact $\Rightarrow$ every positive continuous function $f:M\rightarrow \mathbb{R}$ has positive infimum.

Now, I need to prove the converse: If $M$ is a metric space such that every positive continuous function $f:M\rightarrow \mathbb{R}$ has positive infimum, so $M$ is compact.

I found this question here: $M$ is compact iff $f:M\to\mathbb{R}$ has a positive infimum.

But I want to prove this without using pseudocompactness, and this question uses. Can someone just give me some hints? I really don't want the answer itself.

gHem
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Mateus Rocha
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  • Maybe something like the second answer of this question should work changing $\inf$ by $\sup$ or $1/n$ by $n$ in the definition of $g$: https://math.stackexchange.com/questions/1244557/existence-of-a-continuous-function-which-does-not-achieve-a-maximum – Bias of Priene Nov 28 '18 at 03:29

1 Answers1

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If there is a sequence $\{x_n\}$ with no convergent subsequence the $E=\{x_1,x_2,\cdots\}$ is a closed set. Define $f:E\to (0,1) $ by $f(x_n)=\frac 1 n$. Then $f$ is continuous. By (one form of) Tietze Extension Theorem there exists a continuous function $F:X \to (0,1)$ such that $F=f$ on $E$. This continuous function does not have a positive infimum. [I can provide more information on Tietze's Theorem if needed].

Kavi Rama Murthy
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