Mathematical Analysis

If \((x_n)_{n=1}^\infty\subseteq [a,b]\text{,}\) there is a subsequence \((x_{n_k})_{k=1}^\infty\subseteq (x_n)_{n=1}^\infty\) converging to some \(x\in \mathbb{R}\text{.}\) (from Bolzano-Weierstrass Theorem, every bounded sequence has a convergent subsequence). Since \(a\leq x_{n_k}\leq b\) for all \(k\geq 1\) and \(x=\lim_{k\rightarrow\infty}x_{n_k}\text{,}\) we have \(a\leq x\leq b\text{.}\) So \(x\in[a,b]\text{.}\)

If \(A\) is not closed let \(x\in\overline{A}\setminus A\text{.}\) Then exists \((x_n)_{n=1}^\infty\subseteq A\) with \(x=\lim_{n\rightarrow\infty}x_n\text{.}\) Any subsequence of \((x_n)_{n=1}^\infty\) converges to \(x\text{,}\) so no subsequence of \((x_n)_{n=1}^\infty\) converges to a point in \(A\text{.}\)

Further, suppose \(A\subseteq X\) is closed and \(X\) is compact. Let \((x_n)_{n=1}^\infty\subseteq A\) be a sequence. There is a subsequence \((x_{n_k})_{k=1}^\infty\) which converges in \(X\text{.}\) Since \(A\) is closed, \(\lim_{k\rightarrow\infty}x_{n_k}\in A\text{.}\)

Suppose \((x_n)_{n=1}^\infty\subseteq X\) is Cauchy. Since \(X\) is compact, \((x_n)_{n=1}^\infty\) has a convergent subsequence. Now, every Cauchy sequence with a convergent subsequence converges. So \((x_n)_{n=1}^\infty\) converges.

Set \(x=\lim_{n\rightarrow\infty}x_n, y=\lim_{n\rightarrow\infty}y_n\text{.}\) Then \(d(x,y)\leq d(x,x_n)+d(x_n,y_n)+d(y_n,y)\text{.}\) So \(d(x,y)\leq \liminf_{n\rightarrow\infty}d(x_n,y_n)\text{.}\)

Similarly, \(d(x_n,y_n)\leq d(x_n,x)+d(x,y)+d(y,y_n)\text{.}\) So \(\limsup_{n\rightarrow\infty}d(x_n,y_n)\leq d(x,y)\text{.}\)

Suppose \(X\) is an unbounded metric space, so \(\sup_{x,y\in X}d(x,y)=\infty\text{.}\) Then for all \(n\geq 1, \exists x_n,y_n\in X\) with \(d(x_n,y_n)\geq n\text{.}\) If \(X\) is compact, there exist convergent subsequences \((x_{n_k})_{k=1}^\infty\) and \((y_{n_k})_{k=1}^\infty\) of \((x_n)_{n=1}^\infty,(y_n)_{n=1}^\infty\text{.}\) But then \((d(x_{n_k},y_{n_k}))_{k=1}^\infty\) converges and hence is bounded, but \(d(x_{n_k},y_{n_k})\geq n_k\) for all \(k\geq 1\text{.}\)

(\(\impliedby\)) Let \((x_k)_{k=1}^\infty\) be a sequence in \(A\subseteq \mathbb{C}^n\) and write \(x_k=(x_k^{(1)},x_k^{(2)},\dots,x_k^{(n)}),x_k^{(j)}\in\mathbb{C}\text{.}\) Note that for any \(x=(x^{(1)},\dots, x^{(n)})\in \mathbb{C}^n\text{,}\) and any \(j=1,\dots, n\text{,}\) we have \(\left|{x^{(j)}}\right|=(\left|{x^{(j)}}\right|^2)^{\frac{1}{2}}\leq (\sum_{i=1}^n\left|{x^{(i)}}\right|^2)^{\frac{1}{2}}=\left|{\left|{{x}}\right|}\right|^2\text{.}\) In particular, since \((x_k)_{k=1}^\infty\subseteq\mathbb{C}^n\) is bounded, \((x_k^{(1)})_{k=1}^\infty \subseteq \mathbb{C}\) is also bounded. By the Bolzano-Weierstrass theorem, \((x_k^{(1)})_{k=1}^\infty\) has a convergent subsequence. Call it \((x_{k(1,l)}^{(1)})_{l=1}^\infty\subseteq \mathbb{C}\) (so \(\mathbb{Z}_{\geq 1}\rightarrow \mathbb{Z}_{\geq 1}\) by \(l\mapsto k(1,l)\) strictly increasing.) Let \(x^{(1)}=\lim_{l\rightarrow\infty}x_{k(1,l)}^{(1)}\text{.}\) Then \((x_{k(1,l)})_{l=1}^\infty\subseteq \mathbb{C}\) is a bounded sequence and hence \((x_{k(1,l)}^{(2)})_{l=1}^\infty\subseteq \mathbb{C}\) is bounded. So \((x_{k(1,l)}^{(2)})_{l=1}^\infty\subseteq \mathbb{C}\) has a convergent subsequence \((x_{k(2,l)}^{(2)})_{l=1}^\infty\) converging to some \(x^{(2)}\in\mathbb{C}\text{.}\)

Note that \((x_{k(2,l)}^{(1)})_{l=1}^\infty\) is a subsequence of \((x_{k(1,l)}^{(1)})_{l=1}^\infty\text{,}\) so \(\lim_{l\rightarrow\infty}x_{k(2,l)}^{(1)}=x^{(1)}\) and \(\lim_{l\rightarrow\infty}x_{k(2,l)}^{(2)}=x^{(2)}\text{.}\)

Continue in this way choosing \((x_{k(j+1,l)})_{l=1}^\infty\) to be a subsequence of \((x_{k(j,l)})_{l=1}^\infty\) for \(j=1,\dots,n-1\text{.}\) To arrange for \((x_{k(n,l)})_{l=1}^\infty\) to be convergent and set \(x^{(j)}=\lim_{l\rightarrow\infty}x_k^{(i)}\) for \(i=1,\dots,n\text{.}\) Then set \(k_l=k(n,l)\geq 1\) for each \(l\geq 1\text{.}\) Then \((x_{k_l})_{l=1}^\infty\) is a subsequence of \((x_k)_{k=1}^\infty\) (in fact a subsubsub...subsequence (\(n\) “subs”)).

So \((x_{k_l}^{(j)})_{l=1}^\infty\) converges to \(x^{(j)}\in\mathbb{C}\) for all \(j=1,\dots,n\text{.}\) Then if \(x=(x^{(1)},\dots,x^{(n)}\text{,}\) we have \(\lim_{l\rightarrow\infty}x_{k_l}=x\text{.}\) Since \(A\) is closed and each \(x_{k_l}\in A\text{,}\) we have \(x\in A\text{.}\) So every sequence in \(A\) has a subsequence converging in \(A\text{.}\) This shows that \(A\) is compact.

(\(\implies\)) same as for \(\mathbb{C}^n\text{.}\)

(\(\impliedby\)) \(\mathbb{R}^n\subseteq \mathbb{C}^n\) is closed, and so if \(A\) is closed in \(\mathbb{R}^n\) then \(A\) is also closed in \(\mathbb{C}^n\text{.}\) So \(A\) is compact by the Heine-Borel Theorem.

If \(\varepsilon=\frac{1}{2}\text{,}\) any set \(A\subseteq X\) with \(\text{diam}(A)<\varepsilon\) contains at most one point. So if we could write \(X=\cup_{i=1}^n A_i\) with \(\text{diam}(A_i)<\frac{1}{2}\text{,}\) then \(X\) would have at most \(n\) points. So \(X\) would be finite.

Let \(A_0=X\text{.}\) Assuming \(k\geq 1, A_{k-1}\) has been constructed, note that \(A_{k-1}\) is totally bounded (subset of totally bounded set is totally bounded). So \(\exists B_1,\dots,B_n\subseteq A_{k-1}\) with \(A_{k-1}=\cup_{i=1}^n B_i\) and \(\text{diam}(B_i)\leq \frac{1}{k}\text{.}\) Since \(\{{m\geq 1}\,:\,{x_m\in A_{k-1}}\}=\cup_{i=1}^n\{{n\geq 1}\,:\,{x_m\in B_i}\}\) is infinite, there is an \(i_0\in\{1,\dots,n\}\) such that \(\{{m\geq 1}\,:\,{x_m\in B_{i_0}}\}\) is infinite. Set \(A_k=B_{i_0}\text{.}\) Now construct \((x_{n_k})_{k=1}^\infty\) as follows: Since \(\{{n\geq 1}\,:\,{x_n\in A_1}\}\) is infinite, it's not empty, so there exist \(n_1\geq 1\) with \(x_{n_1}\in A_1\text{.}\) Now if \(n_k\) has been constructed, then \(\{{n\geq 1}\,:\,{x_n\in A_{k+1}}\}\) is infinite, and hence \(\{{n\geq 1}\,:\,{x_n\in A_{k+1}}\}\setminus\{1,2,\dots,n_k\}\) is non-empty. Choose \(n_{k+1}\in\mathbb{Z}\) with \(n_{k+1}>n_k\) and \(x_{n_{k+1}}\in A_{k+1}\text{.}\) Then \((x_{n_k})_{k=1}^\infty\) is a subsequence of \((x_n)_{n=1}^\infty\) with \(x_{n_k}\in A_k\) for all \(k\geq 1\text{.}\)

We now show \((x_{n_k})_{k=1}^\infty\) is Cauchy. Fix \(\varepsilon>0\text{.}\) Choose \(N\geq 1\) with \(\frac{1}{N}<\varepsilon\text{.}\) Fix \(k,l\in \mathbb{Z}\text{,}\) \(k,l\geq N\text{.}\) Then \(x_{n_k}\in A_k\subseteq A_n\) and \(x_{n_l}\in A_l\subseteq A_N\) (by condition (i)). So \(d(x_{n_k},x_{n_l})\leq \text{diam}(A_N)\leq \frac{1}{N}<\varepsilon\) (condition (ii)). So \((x_{n_k})_{k=1}^\infty\) is Cauchy.

(\(\impliedby\)) [wts every sequence has a Cauchy subsequence implies totally bounded]

Suppose \(X\) is not totally bounded. Then exists \(\varepsilon>0\) such that for all \(B_1,\dots,B_n\) with \(\text{diam}(B_i)<\varepsilon\) we have \(X\neq \cup_{i=1}^nB_i\text{.}\) Let \(x_i\in X\text{.}\) (note: \(X\neq \emptyset\) since \(\emptyset\) is totally bounded) Assume we've constructed \(x_1,\dots x_k\in X\) with \(d(x_i,x_j)\geq \varepsilon/2\) for all \(i,j=1,\dots,k\) with \(i\neq j\text{.}\) Since \(\text{diam}(B_{\frac{\varepsilon}{2}}(x_i))\leq \varepsilon\) for all \(i=1,\dots,k\) (triangle inequality) we have \(X\neq \cup_{i=1}^kB_{\frac{\varepsilon}{2}}(x_i)\) so exists \(x_{k+1}\in X\) with \(d(x_i,x_{k+1})\geq \frac{\varepsilon}{2}\text{.}\) So we've constructed a sequence \((x_n)_{n=1}^\infty \subseteq X\) with \(d(x_i,x_j)\geq \frac{\varepsilon}{2}\) for all \(i\neq j\) so \((x_n)_{n=1}^\infty\) has no Cauchy subsequence.

(\(\impliedby\)) holds for all metric spaces.

(\(\implies\)) If \(A\) is bounded, then every sequence \((a_n)_{n=2}^\infty \subseteq A\) is bounded, and so \((a_n)_{n=2}^\infty\) has a subsequence in \(\mathbb{C}^n\) by the Bolzano-Weierstrass Theorem (or by the Heine-Borel theorem applied to \(\overline{A}\)). So \((a_n)_{n=2}^\infty\) has a Cauchy subsequence.