LEC CAL1 8

Converse of the extreme value theorem

Theorem 177.1(Extreme value theorem).

Let $X$ be a metric space. Then $X$ is compact if and only if every continuous real valued function on $X$ takes a maximum and a minimum value.

Proof.

We have already shown the forward implication here.

$(⟸)$ Assume every real valued function on $X$ takes a minimum and maximum value. To show that $X$ is compact, it is necessary and sufficient to show that it is complete and totally bounded.

Argue by contradiction to show $X$ is totally bounded. If $X$ is not totally bounded, there exists $r>0$ and a countably infinite subset of $X$ which we may enumerate as $\{x_n{\}}_{n=1}^{\infty }$, for which the collection of open balls $\{B_r(x_n)\}$ is disjoint. For each natural number $n$, define $f_n$ by

$$f_n=\{\begin{array}{ll}r/2-d(x,x_n)& d(x,x_n)\le r/2\\ 0& \text{otherwise}.\end{array}$$

Define the function $f:X\to \mathbb{R}$ by

$$f(x)=\sum _{n=1}^{\infty }n{f}_n(x)\text{ for all }x\in X.$$

Since each $f_n$ is continuous and vanishes outside $B_{r/2}(x_n)$ and the collection $\{B_r(x_n)\}$ is disjoint, $f$ is well-defined and continuous. But for each natural number $n$, $f(x_n)=nr/2$, and hence $f$ is unbounded above. This is a contradiction.

It remains to show that $X$ is complete. Let $(x_n)$ be a Cauchy sequence in $X$. For each $x\in X$, we infer form the triangle inequality that $\{d(x,x_n)\}$ is a Cauchy sequence in $\mathbb{R}$, which converges since $\mathbb{R}$ is complete. Thus, we can define $f:X\to \mathbb{R}$ by

$$f(x)=\underset{n\to \infty }{\lim }d(x,x_n).$$

$f$ is continuous: $d(x,y)<ϵ$ $⟹$ $|d(x,x_n)-d(y,y_n)|\le d(x,y)<ϵ$ $⟹$ $|f(x)-f(y)|<ϵ$. By assumption, there is a point $x\in X$ at which $f$ takes a minimum value. Since $(x_n)$ is Cauchy, the infimum of $f$ on $X$ is $0$. Therefore $f(x)=0$ and hence $(x_n)\to x$. Thus $X$ is complete.□

Lebesgue covering Lemma

Theorem 177.2(Lebesgue covering Lemma).

Let $\{{\mathcal{O}}_{\lambda }{\}}_{\lambda \in \Lambda }$ be an open cover of a compact metric space $X$. Then there exists $ϵ>0$, called a Lebesgue number for the cover, such that for each $x\in X$, the ball $B_{ϵ}(x,X)$ is contained in some member of the cover.

Proof.

Assume there is no such Lebesgue number. Then for every$ϵ>0$, there exists $x\in X$ such that $B_{ϵ}(x)$ is not contained in any ${\mathcal{O}}_{\lambda }$. Let $(x_n)$ be the sequence of such points generated by the $ϵ$ sequence $\left(\frac{1}{n}\right)$. Since $X$ is compact, there exists a subsequence $(x_{n_k})$ of $(x_n)$ which converges, say to a point $x_0\in X$. Now there exists some ${\lambda }_0\in \Lambda$ for which $x_0\in {\mathcal{O}}_{{\lambda }_0}$. Since ${\mathcal{O}}_{{\lambda }_0}$ is open, there exists an open ball $B_{r_0}(x_0)\subseteq {\mathcal{O}}_{{\lambda }_0}$. We may choose an index $k$ such that $d(x_0,x_{n_k})<r_0/2$ and $1/n_k<r_0/2$. Then, $B_{1/n_k}(x_{n_k})\subseteq {\mathcal{O}}_{{\lambda }_0}$, which is a contradiction.□

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