Theorem 5.3.1.
If \(A\) is compact and \((x_n)_{n\in\mathbb{N}}\) is Cauchy, then \(x_n\) converges to a limit in \(A\text{.}\)
Section 5.3 compact sets and sequences
Proof.
Notice that, because \(A\) is closed, the limit of the sequence (if it exists) must lie in \(A\text{.}\) Further, we proved long ago that a Cauchy sequence can have at most one subsequential limit.
Consider the set \(\tilde{X}=\left\{x_n\middle\vert n\in\mathbb{N}\right\}\) and its closure \(\overline{\tilde{X}}\text{.}\)
Because \(\overline{\tilde{X}}\subseteq A\) and \(\overline{\tilde{X}}\) is closed, we see that by Proposition 5.2.23 \(\overline{\tilde{X}}\) is compact.
If \(x_n\) fails to converge, we know that \(\forall a\in \overline{\tilde{X}},\exists\epsilon>0:\forall K,\exists n>K:d(a,x_n)\geq \epsilon\text{.}\) Call this \(\epsilon_a\) and consider the open cover for \(\overline{\tilde{X}}\) given by:
\begin{equation*}
\mathcal{U}=\left\{B_{\epsilon_{x_n}}(x_n)\middle\vert n\in\mathbb{N}\right\}
\end{equation*}
By compactness of \(\overline{\tilde{X}}\text{,}\) this set has a finite subcover. That is: there are finitely many
\begin{equation*}
B_{\epsilon_{x_{n_1}}}(x_{n_1}),B_{\epsilon_{x_{n_2}}}(x_{n_2}),\ldots, B_{\epsilon_{x_{n_N}}}(x_{n_N})
\end{equation*}
which suffice to cover \(\overline{\tilde{X}}\text{.}\)
Now consider \(K=\max\{n_1,\ldots,n_N\}\text{;}\) by assumption that \((x_n)_{n\in\mathbb{N}}\) fails to converge, there is \(n\gt K\) so that \(x_n\) lies in none of the listed sets. This is a contradiction.
Theorem 5.3.2.
Let \(A\subseteq M\) be a compact subset of a metric space and \(\left(x_n\right)_{n\in\mathbb{N}}\) a sequence. Then \(\left(x_n\right)_{n\in\mathbb{N}}\) has a convergent subsequence which converges to some \(a\in A\text{.}\)
Proof.
Consider the open cover \(\mathcal{U}_{1}=\left\{B_{1}(y)\middle\vert y\in A\right\}\text{.}\) There is a finite subcover for this cover; that is, there are \(y^1_1,\ldots,y^1_K\) so that every point of \(A\) is within \(1\) of one of them. By the pigeonhole principle, infinitely many terms of \((x_n)_{n\in\mathbb{N}}\) fall into one \(B_{1}(y^1_j)\text{;}\) call that one \(y_1\text{.}\) Pick \(x_{n_1}\) to be one of the terms that lies in \(B_1(y_1)\text{.}\)
Now consider the open cover \(\mathcal{U}_{2}=\left\{B_{\frac{1}{2}}(y)\middle\vert y\in A\right\}\text{.}\) Again, there is a finite subcover. Applying the pigeonhole principle to the infinitely many terms of \((x_n)_{n\in\mathbb{N}}\) which lay in \(B_1(y_1)\text{,}\) there must be one of this finite subcover which contains infinitely many terms of \((x_n)_{n\in\mathbb{N}}\text{;}\) call the center of that one \(y_2\text{.}\) In particular, there is some \(n_2\gt n_1\) with \(x_{n_2}\in B_{\frac{1}{2}}(y_2)\text{.}\)
Observe that \(d(y_1,y_2)\leq \frac{3}{2}\text{,}\) \(d(x_{n_1},y_1)\lt 1\text{,}\) and \(d(x_{n_2},y_2)\lt \frac{1}{2}\text{.}\)
Lather, rinse, repeat to construct a sequence \(y_k\) and a subsequence \((x_{n_k})_{k\in\mathbb{N}}\) so that \(d(y_k,y_{k+1})\lt \frac{3}{2^k}\) and \(d(x_{n_k},y_k)\lt \frac{1}{2^{k-1}}\text{.}\)
The condition on the \(y_k\) tells us that \((y_k)_{k\in\mathbb{N}}\) is Cauchy, hence by Theorem 5.3.1, \(y_k\to y\) for some \(y\text{.}\) But then it’s clear that \(x_{n_k}\to y\text{.}\)
Checkpoint 5.3.3.
Give a proof of Theorem 5.3.2 by contraposition: assume \(\left(x_n\right)_{n\in\mathbb{N}}\) has no convergent subsequences and produce an open cover for \(A\) which has no finite subcovers.
Definition 5.3.4.
The condition on a set \(A\) that for any sequence in \(A\text{,}\) there is a convergent subsequence which converges to a point of \(A\) is very nice; we say in this case that \(A\) is sequentially compact.
So Theorem 5.3.2 can be rephrased by saying that compact implies sequentially compact. The converse is also true:
Theorem 5.3.5. General Bolzano-Weierstraß Theorem.
Example 5.3.6.
Consider \(S^\infty=\left\{(x_1,\ldots,x_n,\ldots)\middle\vert\ \lVert x\rVert_\infty =1\right\}\text{,}\) the unit sphere in \(\mathbb{R}^\infty\text{.}\) This set is closed and bounded. However, the sequence
\begin{equation*}
(1,0,0,\ldots), (0,1,0,\ldots),(0,0,1,\ldots),\ldots
\end{equation*}
has no convergent subsequences. Thus the unit sphere in \(\mathbb{R}^\infty\) is not compact.
Checkpoint 5.3.7.
Consider the hemisphere cover of \(S^\infty\) whose elements are, for each \(n\in\mathbb{N}\text{,}\)
\begin{equation*}
U_n^+=\left\{x\in \mathbb{R}^\infty\middle\vert x_n\gt 0\right\}, U_n^-=\left\{x\in \mathbb{R}^\infty\middle\vert x_n\lt 0\right\}
\end{equation*}
(You should check that indeed this is an open cover -- try drawing the corresponding pictures for \(S^1\subseteq \mathbb{R}^2\) and \(S^2\subseteq\mathbb{R}^3\text{.}\))
Show that no finite subcollection covers \(S^\infty\text{.}\)
Remark 5.3.8.
Confusingly, sometimes Theorem 5.3.5 is called the General Heine-Borel Theorem or not named at all.
