10. Compactness

10.1 Open Covers and the Definition of Compactness

Open Cover

Let $X$ be a metric space with $\mathcal{U}=\{U_i:i\in I\}$ a collection of open subsets of $X$

$\mathcal{U}$ is an open cover of $X$ if

$$X=\bigcup _{i\in I}{U}_i$$

Note that for subspaces $Y\subseteq X$ then need $Y\subseteq {\bigcup }_{i\in I}{U}_i$

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Subcover

Let $X$ be a metric space with open cover and let

$$\mathcal{U}=\{U_i:i\in I\}\text{ be an open cover of }X$$

If $J\subseteq I$ then
sub-collection $\{U_i:i\in J\}$ is a subcover of $\mathcal{U}$ if

$$X=\bigcup _{i\in J}{U}_i$$

If $|J|<\infty$ then $\{U_i:i\in J\}$ is a finite subcover

Note that for subspaces $Y\subseteq X$ then need $Y\subseteq {\bigcup }_{i\in J}{U}_i$

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10.2 Compactness implies Sequential 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 $Y$

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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

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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 $x_{n_k+1}$

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10.3 Sequential Compactness implies Compactness

Sequentially Compactness implies Compactness

Let $X$ be a metric space

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

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16 - Equivalence Statements for Compactness

Equivalence Statements for Compactness

Let $X$ be a metric space, then the following are equivalent

1. $X$ is compact

2. $X$ is sequentially compact

3. $X$ is complete and totally bounded

Proof

As a consequence of

1. Compactness implies sequentially compactness
2. Sequentially Compactness Characterisation Theorem

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17 - Heine-Borel

Heine-Borel

Let $X$ be a subset of ${\mathbb{R}}^n$

$$X\text{ is compact}⟺X\text{ is closed and bounded}$$

Proof

Proof $⟹$
Apply

1. Compactness implies sequentially compactness
2. Closed and boundedness of sequentially compact subspaces

Proof $⟸$
Apply

1. Bolzano-weierstrass
2. ^6eefad

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