6
$\begingroup$

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 $\inf\limits_{x\in M} f(x) >0$.

($\Rightarrow$)

$\quad$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 $\inf f(M)>0$.

($\Leftarrow$)

$\quad$ 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. $\textbf{(why isn't there a infimum postive?)}$

$\square$

$\endgroup$
1
$\begingroup$

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 $\inf \{f(x): x \in N\}=f(x_0)>0$.

$\endgroup$
4
  • $\begingroup$ 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
  • 1
    $\begingroup$ @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

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.