Week 3, Stepan Malkov

Compactness and Completeness

Recall that a sequence $(a_n)_{n=1}^\infty$ is Cauchy if for every $\epsilon>0$ there exists an $N$ such that $d(a_n,a_m) < \epsilon$ whenever $n,m \geq N.$ Today, we will cover compact, sequentially compact, and complete metric spaces.

Definition:

A metric space $(X,d)$ is called compact if for any collection of open sets $\{U_i\}_{i \in I}$ such that $K \subseteq \bigcup_{i \in I} U_i,$ there exist $i_1,...,i_n$ such that $K \subseteq \bigcup_{k=1}^n U_{i_k}.$ In other words, every open cover has a finite subcover. If $X' \subseteq X$ and $(X',d)$ is compact, we call the set $X'$ compact.

Definition

A metric space $(X,d)$ is called sequentially compact if every sequence has a convergent subsequence with its limit in $X.$ If $X' \subseteq X$ and $(X',d)$ is compact, we call the set $X'$ compact.

Definition:

A metric space $(X,d)$ is called \textbf{complete} if every Cauchy sequence has a limit in $X.$

We consider several examples of compact and complete sets.

Example:

Any closed and bounded subset of $\mathbb{R}^n$ is compact. As we will show later today, every compact set in $\mathbb{R}^n$ is closed and bounded.
Any metric space $(X,d)$ where $X$ has finitely many elements is compact.
$\mathbb{Q}$ with the usual metric is not complete. Let $(a_n)_{n=1}^\infty \to \sqrt{2},$ where $a_n \in \mathbb{Q}$ gives the first $n$ digits of $\sqrt{2}.$ Then, $(a_n)_{n=1}^\infty$ is Cauchy but does not converge to an element in $\mathbb{Q},$ i.e. $\mathbb{Q}$ is not complete.
$C([a,b])$ with the metric of uniform convergence is complete, since continuous functions on a closed and bounded interval are uniformly continuous, and the limit of a uniformly convergent sequence is uniformly continuous.

We now prove some important results that relate these three categories.

Proposition

A set $A$ in a metric space $(X,d)$ is compact if and only if it is sequentially compact.

Proof:

( $\implies$ ) Suppose $A$ is compact, and let $(x_n)_{n=1}^\infty$ be a sequence in $A.$ Note that $(x_n)$ has a convergent subsequence in $A$ if and only if there is an $x \in A$ that is a limit point of $(x_n).$ For sake of contradiction, suppose the sequence has no limit points. Then, for every $a \in A, a \not = x_n$ for some $n,$ there exist an $r_a >0$ such that $B(a,r_a)$ contains no elements of the sequence, and for each $n$ there exists an $r_n$ such that $B(x_n,r_n)$ does not contain any element of the sequence other than $x_n.$ But then, if

$U=\bigcup_{a \in A \setminus \{x_1,x_2,...\}} B(a,r_a), \quad V= \bigcup_{n} B(x_n,r_n),$

$U \cup V$ is an open cover of $A,$ so it has a finite subcover $U'$ . Note that for all $n,$ $x_n \not \in U,$ and since the subcover $U'$ is finite, there exists an $N$ such that $B(x_N,r_N) \not \in U'.$ But by construction, this implies $x_N \not \in U',$ so $U'$ does not cover $A,$ which is a contradiction.
( $\impliedby$ ) Suppose $A$ is sequentially compact. We first show that there exists an $r>0$ such that for every $a \in A$ and every open cover $\{U_i\}$ of $A,$ $B(a,r) \in U_k$ for some $k.$ Suppose there isn’t such a $\delta.$ Then, for every $n>0,$ there exists an $a_n \in A$ such that $B(a_n,\frac{1}{n})$ is not fully contained in any $U_i.$ Extract a convergent subsequence $(a_{n_k}) \to a \in A.$ Then, $B(a,\epsilon) \in U_j$ for some $\epsilon>0.$ But for $n$ such that $d(a,a_{n_k}) < \frac{\epsilon}{2}$ and $n_k > \frac{2}{\epsilon},$

$B\left(a_{n_k},\frac{1}{n_k}\right) \subseteq B(a,\epsilon) \subseteq U_j,$

which is a contradiction. Now, suppose there exists an open cover $\{U_i\}_{i \in I}$ of $A$ which does not have a finite subcover. Using the lemma, pick $u_1 \in U_{i_1}$ such that for some $r>0$ , $B(u_1,r) \subseteq U_{i_1}.$ Since $\{U_{i_1}\}$ is finite, it does not cover $A,$ so there is a $u_2 \in A \setminus U_{i_1}$ with $B(u_2,r) \subseteq U_{i_2}.$ Proceeding in this fashion, we construct a sequence of elements $(u_n)_{n=1}^\infty.$ Then, since $d(u_i,u_j) \geq r$ for all $i,j,$ no subsequence of $(u_n)$ is Cauchy, so no subsequence of $(u_n)$ converges, which is a contradiction.

Proposition

A compact set $A \subseteq X$ is closed, bounded, and complete.

Proof:

Completeness is easy: if for all $\epsilon>0,$ there exist $k,N$ such that $d(u_n,u_m) < \frac{\epsilon}{2}$ for $n,m \geq N$ and $d(u_{n_k},u)<\frac{\epsilon}{2},$ then

$d(u_n,u) \leq d(u_n,u_{n_k}) + d(u_{n_k},u) < \epsilon$

for $n_k \geq N,$ so $u_n \to u.$ Similarly, if $u_n \to u \in X$ and $u_{n_k} \to a \in A,$ then $u = a \in A,$ i.e. $A$ is closed. If $A$ is not bounded, it is possible to construct a sequence $(u_n)_{n=1}^\infty$ such that

$u_n \in B(x,r_{n+1}) \setminus B(x,r_n)$

for some $x \in A$ and $r_0 = 1, r_{n+1}>r_n+1$ for all $n$ (otherwise, $A$ would be bounded). Then, $d(u_n,u_m) \geq 1,$ so $(u_n)_{n=1}^\infty$ has no convergent subsequence, i.e. $A$ is not compact.