everybody! Could you give me feedback if my statements below $ \Leftrightarrow $ are correct or I need to change them?

$\quad$Show that a metric space $M$ is compact iff for every continuous function $f:M\rightarrow \mathbb{R}$, such that $f(x)>0$ for all $x\in M$, we have $\inf\limits_{x\in M} f(x) >0$.


$\quad$If $A$ is a bounded subset of $\Bbb R$, and if $s=\sup A$ then for each $\epsilon >0$, $s−\epsilon$ is not an upper bound of $A$ (since s is the least upper bound of $A$) and therefore, there is some $a\in A$ such that $a>s−\epsilon$. But, since $s$ is an upper bound of $A$, $a\leq s$. So, $|s−a|<\epsilon$. Since this occurs for every $\epsilon>0$, $s\in \overline{A}$. For a similar reason, $\textbf{inf}$ $A \in \overline{A}$. And, since $\inf f(M)\in f(M)\subset(0,\infty)$, there is some $m\in M$ such that $\inf f(M)>0$ and, since $f(M)>0$, this proves that $\inf f(M)>0$.


$\quad$ If there is a sequence $\{x_n\}$ with no convergent subsequence the $E=\{x_1,x_2,⋯\}$ is a closed set. Define $f:E\rightarrow (0,1)$ by $f(x_n)=\frac 1 n$. Then $f$ is continuous. By (one form of) Tietze Extension Theorem there exists a continuous function $F:X\rightarrow (0,1)$ such that $F=f$ on $E$. This continuous function does not have a positive infimum. $\textbf{(why isn't there a infimum postive?)}$



The second part is correct but the first one looks complicated and vague. It is not even clear as to where compactness comes in.

If $M$ is compact and $f$ is continuous then $f(M)$ is compact. The infimum of any compact subset of $\mathbb R$ is attained. So there exists $t \in f(M)$ such that $t \leq f(x)$ for all $x \in M$. Since $t \in f(M)$ we can write $t=f(x_0)$ for some $x_0$. Now $\inf \{f(x): x \in N\}=f(x_0)>0$.

  • $\begingroup$ Hi @kavi-rama-murthy, Which Theorem, Lemma or proposition did you use? Could you share it with me? $\endgroup$ Dec 19 '20 at 6:23
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    $\begingroup$ @FernandoSousa I sued two basic theorems: 1. Continuous image of a compact space is compact. 2. The infimum of a compact subset of $\mathbb R$ is attained (so the infimum is actually the minimum of the set). $\endgroup$ Dec 19 '20 at 6:26
  • $\begingroup$ The Extreme Value Theorem??? Suppose that $K$ is a compact subset of $\mathbb{R}^n$, and that $f:K \to \mathbb{R}$ is continuous. Then the set $f(K)=\{f(x):x \in K\}$ is compact, and there exists $x_∗$ and $x_∗$ in $K$ such that $f(x_∗)=\sup\{f(x):x \in K\},f(x_∗)=\inf \{f(x):x \in K\}$ $\endgroup$ Dec 19 '20 at 6:48
  • $\begingroup$ @FernandoSousa Yes, that is correct. $\endgroup$ Dec 19 '20 at 7:13

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