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Section 5.3 compact sets and sequences

What good is the notion of compactness? We'll start with how it interacts with sequences.

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.

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 \(x_{n_k}\to y\text{.}\)

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:

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.

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.