Notes - Metric Spaces MT23, Compactness

Every sequence of elements of $X$ has a convergent subsequence.

Sequential compactness implies closed and bounded.

Any closed subset of a sequentially compact metric space is sequentially compact.

Suppose:

• $X$ is a metric space
• $U$ is an open subset of $X$
• $K$ is a sequentially compact subset of $X$
• $K \subseteq U$

Then $\exists \varepsilon > 0$ such that

\[\bigcup_{z \in K} B(z, \varepsilon)\]

is contained in $U$.

Suppose for a contradiction that $\forall \varepsilon > 0$, there exists some $x \in K$ such that $B(x, \varepsilon) \not\subseteq U$.

Consider the sequence $\varepsilon _ n = 1/n$, and pick $x _ n$ as above, so that $B(x _ n, \varepsilon _ n) \not\subseteq U$. Then there exists some $y _ n \in U^C$ such that $d(x _ n, y _ n) < 1/n$. Since $K$ is sequentially compact, there exists a subsequence $(x _ {n _ k})^\infty _ {k = 1}$ that converges to an element $p \in K \subseteq$. But since $U^C$ is closed, $y _ n$ must also converge to $p$ and this would imply $p \in U^C$, a contradiction.

The image of a sequentially compact metric space under a continuous map is also sequentially compact.

Sequential compactness is a homeomorphism invariant.

Suppose:

• $f : X \to \mathbb R$
• $f$ is continuous
• $X$ is sequentially compact

Then $f$ is uniformly continuous.

Suppose $f$ were not uniformly continuous. Then $\exists \varepsilon > 0$ such that $\forall n \in \mathbb N$, there exists $a _ n, b _ n \in X$ such that $d(a _ n, b _ n) < 1/n$ but $d(f(a _ n), f(b _ n)) \ge\varepsilon$.

Since $X$ is sequentially compact, $a _ n$ has a subsequence $a _ {n _ k}$ such that $(a _ {n _ k})$ converges to a point $\ell$. Then $(b _ {n _ k})$ also converges to $\ell$ as $d \to \infty$.

Relabelling to avoid double subscripts, we can assume that we have sequences $a _ n$ and $b _ n$ with limit $\lim _ {n \to \infty} a _ n = \lim _ {n \to \infty} b _ n = \ell$ and $d(f(a _ n), f(b _ n)) \ge \varepsilon$ for all $n$.

Since $f$ is continuous at $\ell$, there exists $\delta > 0$ such that $\forall x \in X$ where $d(\ell, x) < \delta$, $d(f(\ell), f(x)) < \varepsilon / 2$.

If $n$ is sufficiently large, $d(\ell, a _ n), d(\ell, b _ n) < \delta$.

Hence

\[\begin{aligned} \varepsilon &\le d(f(a_n), f(b_n)) \\\\ &\le d(f(a_n), f(\ell)) + d(f(\ell), f(b_n)) \\\\ &< \varepsilon / 2 + \varepsilon/2 \\\\ &= \varepsilon \end{aligned}\]

This is a contradiction.

It is also sequentially compact.

Any closed and bounded subset of $\mathbb R^n$ is sequentially compact.

Sequential compactness implies it is complete and bounded.

(In fact, we have the stronger condition that we have sequeuntial compactness iff the metric space is complete and totally bounded)

$f _ n$ sequence of fixed-slope “tent” functions on $\mathbb R$ where the spike gets closer and further and further from origin. Then any difference between two terms is at least the height of the tent, so cannot have any Cauchy sequence, so not sequentially compact.

Total boundedness.

For all $\varepsilon > 0$, it may be covered by finitely many open balls of radius $\varepsilon$.

A metric space is sequentially compact if and only if it’s complete and totally bounded.

Consider $f _ n(x) = n$, it has no convergent subsequence with respect to the sup norm.

Take $X = [0, 1]$ and define $(f _ n)$ by

\[f_n(x) = \begin{cases}0 &\text{if } x \notin \left(\frac 1 {n+1}, \frac 1 n\right) \\\\ 1 &\text{if }x = \frac 1 2\left(\frac 1 {n+1}+ \frac 1 n\right) \\\\ \text{?} &\text{piecewise linear elsewhere}\end{cases}\]

Has no convergent subsequence.

If it is bounded as a subset of $\mathcal F$ with respect to the sup norm, i.e.

\[\exists M \text{ s.t. } |f(x)| \le M \text{ } \forall x \in X \text{ and } \forall f \in \mathcal F\]

\[\forall \varepsilon > 0 \text{ } \exists \delta >0 \text{ s.t. } \forall f \in \mathcal F \text{ } \forall x, y \in X \text{ } d(x, y) < \delta \implies d(f(x), f(y)) < \varepsilon\]

i.e. the choice of delta is uniform in the choice of $f$.

Suppose:

• $X$ is a sequentially compact metric space
• $\mathcal F \subseteq C(X)$
• $\mathcal F$ is equicontinuous
• $\mathcal F$ is uniformly bounded
• $\mathcal F$ closed

Then:

• $\mathcal F$ is sequentially compact.

We have a sequentially compact metric space $X$ and we want to find out when a set of functions $\mathcal F \subseteq C(X)$ is also sequentially compact.

Suppose $X$ were unbounded. Fix some $x _ 0 \in X$ and consider a sequence $(x _ n)$ where $d(x _ 0, x _ n) \ge n$ for all $n$. But then $(x _ n)$ must have some convergenct subsequence $x _ {n _ k} \to x$. Then $\exists N$ such that $\forall n > N$, $d(x _ {n _ k}, x) < 1$. But then

\[d(x_0, x) \ge d(x_0, x_{n_k}) - d(x_{n_k}, x) \ge n_k-1\]

for sufficiently large $n$. But $d(x _ 0, x)$ is some finite number, a contradiction.

Suppose it were not closed, then $\overline Y \setminus Y$ is non-empty. Let $y$ be a point in this set, and let $(y _ n)$ be a sequence of points in $Y$ that converge to $y$. Any convergent subsequence must also converge to $y$, so no subsequence can converge to an element of $y$, which contradicts the assumption $Y$ is sequentially compact.

• Continuous functions preserve limits, so can you translate one convergent subsequence to another.
• Alternatively, you can use the fact that the preimage of an open set by a continuous function is also open, and use this to translate open covers from one space to the other

Each $U _ i$ is open and

\[X \subseteq \bigcup_{i \in I} U_i\]

If $J \subseteq I$, then $\{U _ i : i \in J\}$ is a subcover, and if $ \vert J \vert < \infty$, then $J$ is a finite subcover.

Every open cover has a finite subcover.

\[\mathbb R = \bigcup_{n \in \mathbb N} (-n, n)\]

but this has no finite subcover.

If $(S _ i)$ is a sequence of nonempty closed subsets of $X$ with

\[S_1 \supseteq S_2 \supseteq \cdots\]

Then

\[\bigcap^\infty_{n = 1} S_n \ne \varnothing\]

Suppose:

• $X$ is compact
• $(S _ i)$ is a sequence of non-empty, closed subsets of $X$
• $S _ 1 \supseteq S _ 2 \supseteq \cdots$

Then:

\[\bigcap^\infty_{n = 1} S_n \ne \varnothing\]

Useful because it can be used to prove that compactness implies sequential compactness. Proof by contradiction, suppose the intersection is empty. Then the sequence $(S _ 1^C, S _ 2^C, \cdots)$ forms an open cover which by compactness implies there exists some $n$ such that $(S _ 1^C, S _ 2^C, \cdots, S _ n^C)$ is a finite open cover. But since these are nested (the other way around), it means $S _ n^C = X$, but then $S _ n^C = \emptyset$, a contradiction.

$\bar A _ k$, the closure of the $k$-th tail of the sequence.

Let $A _ k$ be the $k$-th tail of this sequence. Then we get $\bar A _ 1 \supseteq \bar A _ 2 \subseteq \cdots$, a sequence of non-empty closed sets in $X$. From the lemma that we are allowed to use, the intersection of these are non-empty. Let $a$ be a point in the intersection, and inductively construct $(x _ {n _ k})$ where each $x _ {n _ k}$ is within $1/k$ of $a$ as follows:

Suppose we have already constructed $n _ 1, \cdots, n _ k$. Then $a$ lies in $\bar A _ {n _ k + 1} = \overline{\{x _ {n _ k + 1}, x _ {n _ k + 2}, \cdots\}\,}$, so there exists some point where $x _ j \in A _ {n _ k + 1}$ is within $1/(k+1)$ of $a$. Take $j$ as our value for $n _ {k+1}$.

Suppose $C \subseteq \mathbb R^n$. Then $C$ is compact if and only if $C$ is closed and bounded.

The Heine-Borel theorem, since it says $C \subseteq \mathbb R^n$ is compact if and only if $C$ is closed and bounded.

Overall idea: consider the set $S$ of $x$ such that $[a, x]$ is covered by finitely many elements of $S$. Then using supremums, you can show that $[a, x]$ is bounded above by $b$, and then you can actually show that it must actually attain the supremum.

Let $U = \{U _ i \mid i \in I\}$ be an open cover of $[a, b]$ (the $U _ i$ are open in $\mathbb R$).

Let $S$ be the set of all $x \in [a, b]$ for which $[a, x]$ is covered by some finite subcollection of $U _ i$.

$S \ne \emptyset$, since $a \in S$, and $S$ is bounded above by $b$, so $S$ certainly has a supremum $c = \sup S$ and $c \in [a, b]$.

$c > a$, since $a \in U _ i$ for some open $U _ i$, then $U _ i$ contains an interval $[a - \eta, a + \eta]$ where $\eta > 0$ so $a + \eta \in S$.

Assume that $c < b$. Since $U$ is an open cover of $[a, b]$, $c$ lies in some set $U _ i$. Since $U _ i$ is open, some open interval $[c - \varepsilon, c + \varepsilon]$ with $\varepsilon > 0$ is contained in $U _ j$.

Assume that $\varepsilon$ is sufficiently small that $c - \varepsilon > a$ and $c + \varepsilon < b$.

Then $c - \varepsilon$ is contained in $S$, or else it would be an upper bound for $S$, smaller than $c$.

So $[a, c - \varepsilon]$ is covered by finitely many sets from $U$.

These sets, together with $U _ j$ give a covering of $[a, c + \varepsilon]$ by a finite subcollection. This contradicts the fact that $c$ is an upper bound for $s$.

So $c = b$. Since $b \in U _ i$, for some $i$ then $U _ i$ contains some interval $[b - \delta, b + \delta]$ for $\delta > 0$. Since $c = \sup S = b$, $b - \delta \in S$, so $[a, b - \delta]$ is covered by a finite subcollection of $U$.

This subcollection, together with $U _ i$ gives a finite subcover of $[a, b]$, so $[a, b]$ is compact.

Alternative proof (sketch): use the fact that in metric spaces, compactness and sequential compactness are the same. Consider any sequence consisting of elements of $[a, b]$, then apply/prove the “scenic viewpoint theorem” to show that there exists a monotonic subsequence. Then this is a bounded monotonic subseqeuence, so must converge to some limit.

The following are equivalent:

• $X$ is compact
• $X$ is sequentially compact
• $X$ is complete and totally bounded

Pick a subsequence $S _ 1$ such that the first coordinate converges, then pick a subsequence $S _ 2$ of $S _ 1$ such that the second coordinate converges, etc.

Then take the diagonal sequence $T$ is the $i$-th coordinate is the $i$-th coordinate of $S _ i$. Then this sequence converges in $T$.

• Heine-Borel says that a subset of $\mathbb R^n$ is closed and bounded iff it is compact
• Bolzano-Weierstrass says that if a subset of $\mathbb R^n$ is closed and bounded implies it is sequentially compact
• So Heine-Borel says the condition is necessary and sufficient, Bolzano-Weierstrass says that it is only sufficient

Let $(x _ n, y _ n) \in (X \times Y)^{\mathbb N}$. Then $\exists s _ n$ s.t. $x _ {s _ n} \to x \in X$. But then $(y _ {s _ n}) \in Y^\mathbb N$, so $\exists k _ n$ s.t. $y _ {s _ {k _ n}\,} \to y \in Y$. Hence the overall subsequence $(x _ {s _ {k _ n}\,}, y _ {s _ {k _ n}\,}) \to (x, y)$, so it has a convergent subsequence.

Sequentially compact implies complete and totally bounded:

Completeness: The general idea is to show that a Cauchy sequence must actually converge to the limit of one of its convergent subsequences. Take $(x _ n)^\infty _ {n=1}$, a Cauchy sequence in $X$. Since $X$ is sequentially compact, $(x _ n)^\infty _ {n=1}$ has a convergent subsequence, say $(x _ {n _ k})^\infty _ {n = 1}$. Suppose that $\lim _ {k \to \infty} x _ {n _ k} = a$. Let $\varepsilon > 0$.

Since $(x _ n)^\infty _ {n = 1}$ is Cauchy, there is some $N$ such that for all $n, m \ge N$, $d(x _ n, x _ m) \le \varepsilon / 2$. Since $\lim _ {k \to \infty} x _ {n _ k} = a$, we may find $k$ such that $\forall n _ k \ge N$, $d(x _ {n _ k}, a) \le \varepsilon / 2$. But then if $n \ge N$, we have

\[d(x_n, a) \le d(x_n, x_{n_K}) + d(x_{n_K}, a) \le \varepsilon / 2 + \varepsilon / 2 = \varepsilon\]

Total boundedness: Suppose that it is not totally bounded, and let $\varepsilon$ be such that there is no way to cover $X$ by finitely many balls of radius $\varepsilon$.

We can construct a squence $(x _ n)^\infty _ {n=1}$ such that all elements are seperated by at least $\varepsilon$, i.e. $d(x _ i, x _ j) \ge \varepsilon$ whenever $i \ne j$.

Suppose that $x _ 1, \cdots, x _ n$ have already been selected.

By assumption, the balls $B(x _ i, \varepsilon)$ do not cover $X$, so we can select a point $x _ {n+1} \in X$ which does not lie in any of the balls, therefore $d(x _ i, x _ {n+1}) \ge \varepsilon$ for all $1, \cdots, n$.

This sequence clearly does not have a convergent subsequence, a contradiction. So $X$ must be totally bounded.

Complete and totally bounded implies sequentially compact:

Let $\sigma$ be a sequence of elements of $X$. Use the total boundedness assumption for $\varepsilon = 1, 1/2, 1/4, \cdots$.

Then for each nonnegative integer $m$, there is a finite collection of open balls

\[B_1^{(m)}, \cdots, B_{k_m}^{(m)}\]

of radius $2^{-m}$ which cover $X$.

Start with the balls $B _ 1^{(0)}, \cdots B^{(0)} _ {k _ 0}$ of radius $1$. At least one of these balls contains infinitely many elements of $\sigma$. Write $B _ 0$ for the ball with this property. Let $\sigma^{(0)}$ be the ball with this property.

Let $\sigma^{(0)}$ be the infinite subsequence of elements of $\sigma$ contained in this ball.

Now consider the balls $B _ 1^{(1)} \cdots B _ {k _ 1}^{(1)}$ of radius $1/2$. One of these contains infinitely many elements of $\sigma^{(0)}$.

Write $B _ 1$ for this ball, and similarly let $\sigma^{(1)}$ be the infinite subsequence of elements contained in it.

Continue inductively, at each step producting $\sigma^{(r)}$ contained in $B _ r$ and a subsequence of $\sigma^{(r-1)}$.

Consider $\sigma^\star$, formed by a diagonal argument. Take $\sigma^\star _ i$ to be $\sigma _ i^{(i)}$. Then $\sigma^\star$ is a subsequence of $\sigma$.

By construction, $\sigma^\star _ i \in B _ r$ for each $i \ge r$.

Now $\sigma _ i^\star$ is a Cauchy sequence. Let $\varepsilon > 0$. Then let $N$ be such that $2^{-N} < \varepsilon / 2$. Then if $n, m \ge N$, then both $x _ n, x _ m$ both lie in $B _ N$, which is a ball of radius $2^{-N}$.

Hence

\[d(x_n, x_m) < \varepsilon / 2 + \varepsilon / 2 = \varepsilon\]

as required. Since $X$ is complete, $\sigma^\star$ converges, and so we have found a subsequence of $\sigma$ which converges.

(Note that completeness is only used right at the very end, this proof actually shows that any totally bounded space has a Cauchy subsequence).

• Consider an arbitrary sequence in the closed subset
• By the sequential compactness of the overall space, it must also have a convergent subsequence
• Since it is closed, this subsequence must converge to an element of the subset

• $X = \mathbb R$
• $Y = [-1, 1]$
• $f(x) = \sin(x)$

Todo.

Todo.