23 - Compactness

Compactness

Let $X$ be a metric space then

$X$ is compact if every open cover has a finite subcover

For subspaces $Y\subseteq X$ definition holds identical for $Y$

Cantor's Intersection Theorem lemma

Suppose that $X$ is a compact metric space with nested sequence

$$S_1\supseteq S_2\supseteq S_3\supseteq \cdots$$

where $S_1,S_2,\cdots$ are non-empty closed subsets of $X$ then

$$\bigcap _{n=1}^{\infty }{S}_n\ne \varnothing$$

so the intersection is nonempty

Proof

Suppose intersection is empty

So by complements $S_i^c$ (which are open sets) are an open cover of $X$
Hence as the intersection is empty then the complements are an open cover of $X$

By compactness, then there exists a finite subcover
So for some $n$ the sets $S_1^c,\cdots ,S_n^c$ cover $X$

However as $S_1^c\subseteq S_2^c\cdots \subseteq S_n^c$ then $S_n^c$ covers (and equals) $X$

However this is a contradiction as $S_n$ is non-empty

Compactness implies Sequential Compactness

Let $X$ be a metric space

$$X\text{ is compact }⟹X\text{ is sequentially compact}$$

Proof

Suppose $X$ is compact and suppose $(x_n{)}_{n=1}^{\infty }$ is a sequence of elements of $X$

For each natural number $n\in \mathbb{N}$ define

$$A_n:=\{x_n,x_{n+1},x_{n+2},\cdots \}$$

So we have

$$A_1\supseteq A_2\supseteq A_3\supseteq \cdots$$

Hence

$${\overline{A}}_1\supseteq {\overline{A}}_2\supseteq {\overline{A}}_3\supseteq \cdots$$

Applying Cantor’s intersection theorem then

$$\bigcap _{n=1}^{\infty }\text{is nonempty}$$

Let $a$ be a point in the intersection
Inductively construct a subsequence $(x_{n_k}{)}_{k=1}^{\infty }$ such that

$$d(x_{n_k},a)<\frac{1}{k}\text{ for all }k$$

Hence as $k\to \infty$ the $x_{n_k}\to a$

Suppose $n_1,\cdots ,n_k$ have already been constructed
As $a$ lies in ${\overline{A}}_{n_k+1}$, the closure of set $\{x_{n_k+1},x_{n_k+2},\cdots \}$

In particular there is some element of sequence at distance less than $\frac{1}{(k+1)}$ from $a$
So take that as $x_{n_k+1}$

Sequentially Compactness implies Compactness

Let $X$ be a metric space

$$X\text{ is sequentially compact}⟹X\text{ is compact }$$