问题：闭且有界还不够

上一讲定义了紧性：每个开覆盖都有有限子覆盖，并证明了紧 ⟹ 闭且有界。自然要问：反过来成立吗？

• 在 $\mathbb R^n$ 配欧氏距离时，闭且有界 ⟹ 紧，这是 $\mathbb R^n$ 的特权。
• 在一般度量空间里，不成立。

本讲先给两个反例，一个平凡、一个深刻；然后证明本讲的主定理：紧集上的任何数列都有收敛子列。这正是度量空间版的 Bolzano–Weierstrass 定理，也是引入紧性的真正目的。

为什么需要它？在 $\mathbb R$ 中，柯西收敛定理靠 BW 定理证明；而 BW 在一般度量空间失效，所以柯西收敛定理也失效。紧性把这个缺口补上。课程还强调：证明里不能假设空间完备，所以必须同时构造子列与它的极限——这正是嵌套闭球引理要做的事。

时间轴

回顾与两个反例

已知

紧性：$A$ 的每个开覆盖都有有限子覆盖。上一讲证明了两件事：

1. 紧集有界——用半径 $1$ 的球覆盖，紧性取有限子覆盖，再用三角不等式估半径。
2. 紧集闭——用 $E_n=\{y:d(y,x)>1/n\}$ 造开覆盖，紧性给出有限子覆盖后与 $x_n\to x$ 矛盾。

反例一：$(0,1]$ 不紧

取 $X=(0,1]$，距离为通常距离。

• 它作为自身的子集是闭的：整个空间总是闭的（余集为空集，空集开）。
• 它有界：包含于以 $1/2$ 为心、半径 $1$ 的球中。
• 但它不紧。取

$$U_n=\left(\frac1n,1\right],\qquad n\in\mathbb N .$$

对任意 $x\in(0,1]$，取 $n>1/x$，则 $x>1/n$，故 $x\in U_n$；这族覆盖 $X$。但任取有限个 $U_{n_1},\dots,U_{n_k}$，令 $N=\max n_i$，它们的并是 $U_N=(1/N,1]$，而 $1/(N+1)$ 不在其中。所以没有有限子覆盖。

这个反例说明：「闭且有界」在一般空间里不足以推出紧。失败的原因是空间在 $0$ 处「缺了一个点」。

反例二：连续函数空间的闭单位球不紧

取 $X=C[0,1]$，距离为

$$d(f,g)=\max_{x\in[0,1]}|f(x)-g(x)|.$$

考虑闭单位球

$$\overline{B_1(0)}=\{f\in C[0,1] : \max_{x}|f(x)|\le1\}.$$

它是闭的。 验证余集开：若 $\max|g|=M>1$，令 $s=M-1>0$。若 $d(g,h)<s$，则

$$\max|h|\ge\max|g|-\max|g-h|>M-s=1,$$

所以 $h$ 也在余集中。故余集开，闭球闭。

它显然有界——它本身就是一个球。

但它不紧。 对每个 $n\in\mathbb N$，定义 $f_n$ 如下：

$$f_n(x)= \begin{cases} 1, & 0\le x\le \dfrac{1}{n+1},\\[4pt] (n+1)(1-nx), & \dfrac{1}{n+1}\le x\le \dfrac1n,\\[4pt] 0, & \dfrac1n\le x\le 1. \end{cases}$$

图像是一个「台阶」：在前段取 $1$，随后线性下降到 $0$，之后一直是 $0$。每段都连续，且在拼接点处值相等（$x=1/(n+1)$ 时两段都得 $1$，$x=1/n$ 时都得 $0$），所以 $f_n\in C[0,1]$，并且 $\max|f_n|=1$，即 $f_n\in\overline{B_1(0)}$。

关键估计：$n\ne m$ 时 $d(f_n,f_m)=1$。

不妨设 $n<m$。在点 $x=1/(n+1)$：

• $f_n(x)=1$；
• 因为 $m\ge n+1$，有 $1/(n+1)\ge 1/m$，所以 $f_m(x)=0$。

两者之差恰好是 $1$，故 $d(f_n,f_m)\ge1$；又两者取值都在 $[0,1]$ 内，$d(f_n,f_m)\le1$。所以距离恰为 $1$。

现在考虑开球族

$$\mathcal U=\{B_{1/2}(f) : f\in\overline{B_1(0)}\}.$$

它覆盖闭球（每个 $f$ 被以自身为心的球包含），且每个球都开。

没有有限子覆盖。 若 $g,h\in B_{1/2}(f)$，三角不等式给出

$$d(g,h)\le d(g,f)+d(f,h)<\frac12+\frac12=1 .$$

而任意两个不同的 $f_n,f_m$ 距离都是 $1$，所以同一个半半径球里最多只能有一个 $f_n$。有限个球最多盖住有限多个 $f_n$，其余无穷多个 $f_n$ 落在覆盖之外。因此没有有限子覆盖。$\blacksquare$

这个例子比第一个深刻得多：空间是完备的（$C[0,1]$ 配最大范数完备），闭且显然有界，却仍不紧。原因不是「缺点」，而是维数无穷——单位球里能塞进无穷多个两两距离为 $1$ 的点。

嵌套闭集引理

主定理的证明需要一个工具。

引理。 设 $A$ 紧。设 $C_1\supseteq C_2\supseteq\cdots$ 是 $A$ 中一列非空闭集（每个 $C_i$ 在 $X$ 中闭，且 $C_i\subseteq A$）。则

$$\bigcap_{i=1}^{\infty}C_i\ne\varnothing .$$

证明。 反设交集为空。由 De Morgan 律，

$$X=\left(\bigcap_i C_i\right)^c=\bigcup_i (X\setminus C_i).$$

每个 $X\setminus C_i$ 开（因 $C_i$ 闭），所以这族是 $A$ 的开覆盖。

由 $A$ 紧，存在有限子覆盖：

$$A\subseteq (X\setminus C_{i_1})\cup\cdots\cup(X\setminus C_{i_k}).$$

因为 $C_i$ 递减，$X\setminus C_i$ 递增；取 $N=\max\{i_1,\dots,i_k\}$，则有限并等于 $X\setminus C_N$。于是

$$A\subseteq X\setminus C_N, \qquad\text{即}\qquad A\cap C_N=\varnothing .$$

但 $C_N\subseteq A$ 且 $C_N\ne\varnothing$，矛盾。所以交集非空。$\square$

这个引理的直觉：紧性把「无限多个条件同时满足」压缩成「有限多个条件同时满足」，而有限的嵌套集合只要每个非空，交就非空。

推论：嵌套闭球交于一点

推论。 设 $A$ 紧，$\overline{B_{R_i}(x_i)}\cap A$ 是一列嵌套的非空闭集（每个包含下一个），且 $R_i\to0$。则它们的交集恰为一个点。

证明。 非空由引理给出。若有两点 $p,q$ 同属所有集合，则对每个 $i$，

$$d(p,q)\le d(p,x_i)+d(x_i,q)\le 2R_i .$$

让 $i\to\infty$，得 $d(p,q)=0$，故 $p=q$。$\square$

半径趋零保证唯一性；紧性保证存在性。这两件事合起来，才能「无中生有」地造出一个极限点——这正是 BW 证明的关键。

主定理：紧集上的 BW 定理

定理。 设 $A$ 紧，$(x_n)$ 是 $A$ 中的数列。则 $(x_n)$ 有收敛子列，且极限属于 $A$。

证明分三步：先用有限覆盖筛出一串「装着无穷多项」的球，再用加倍把球变成嵌套闭集，最后由推论取出唯一交点作为极限。

第一步：一次覆盖

对每个 $y\in A$ 取 $B_1(y)$。这族覆盖 $A$，且都是开集。由 $A$ 紧，存在有限个 $y_1,\dots,y_k$ 使

$$A\subseteq B_1(y_1)\cup\cdots\cup B_1(y_k).$$

数列有无穷多项，却只落在有限个球里，所以至少有一个球包含无穷多项。把它记作 $B_{r_1}(y_1)$，其中 $r_1=1$。

从这一刻起，只看落在这个球里的那些项——它们构成一个子列。记它的第一项为 $x_{n_1}$。

第二步：逐层加细

设已经得到：开球 $B_{r_k}(y_k)$ 含有当前子列的无穷多项，且 $y_k\in A$。

令

$$C_k=\overline{B_{2r_k}(y_k)}\cap A .$$

$C_k$ 是紧集 $A$ 的闭子集（闭球交 $A$），所以紧——这正是上一讲「紧集的闭子集仍紧」的用处。

用半径 $r_{k+1}=r_k/4$、圆心取遍 $C_k$ 的开球覆盖 $C_k$。由紧性，有限个就够。当前子列有无穷多项落在 $C_k$ 里（它们本就在 $B_{r_k}(y_k)\subseteq C_k$ 中），所以这有限个球中至少有一个仍含无穷多项。把它记为 $B_{r_{k+1}}(y_{k+1})$，并从中选出子列的下一个指标 $n_{k+1}>n_k$。

第三步：验证嵌套

需要确认 $C_{k+1}\subseteq C_k$，否则前面的引理用不上。

因为 $B_{r_{k+1}}(y_{k+1})$ 与 $B_{r_k}(y_k)$ 都含无穷多项，它们必相交。取交点 $q$，则

$$d(y_{k+1},y_k)\le d(y_{k+1},q)+d(q,y_k)<r_{k+1}+r_k .$$

对任意 $w\in B_{2r_{k+1}}(y_{k+1})$，

$$d(w,y_k)\le d(w,y_{k+1})+d(y_{k+1},y_k)<2r_{k+1}+r_{k+1}+r_k=3r_{k+1}+r_k .$$

代入 $r_{k+1}=r_k/4$：

$$3\cdot\frac{r_k}{4}+r_k=\frac{7r_k}{4}<2r_k .$$

所以 $B_{2r_{k+1}}(y_{k+1})\subseteq B_{2r_k}(y_k)$，从而 $C_{k+1}\subseteq C_k$。

这就是「加倍」的理由。 若直接用 $B_{r_{k+1}}$，它可能有一部分伸出上一层球外，嵌套不成立；加倍到 $2r_{k+1}=r_k/2$ 后，借助三角不等式正好塞回半径 $2r_k$ 的球里。$r_k$ 每步缩小为四分之一，$2r_k$ 也趋于 $0$。

第四步：取出极限

由推论，$\bigcap_k C_k$ 恰含一个点 $p$。因为 $C_k\subseteq A$，有 $p\in A$。

对每个 $k$，$x_{n_k}\in B_{r_k}(y_k)$ 且 $p\in\overline{B_{2r_k}(y_k)}$，所以

$$d(x_{n_k},p)\le d(x_{n_k},y_k)+d(y_k,p)<r_k+2r_k=3r_k\to0 .$$

因此 $x_{n_k}\to p$。定理得证。$\square$

整个证明没有假设空间完备：极限 $p$ 是由嵌套闭球构造出来的，而不是事先假定的。这与「在 $\mathbb R$ 中先用 BW 取子列、再用柯西性说明整体收敛」的做法不同——那里完备性是白送的，这里必须自己造。

本节结论

• 紧 ⟹ 闭且有界；$\mathbb R^n$ 中反向成立，一般度量空间不成立。
• $(0,1]$ 说明空间「缺点」会导致不紧；$C[0,1]$ 的单位球说明无穷维空间即使完备、闭、有界也可能不紧。
• 紧集中嵌套非空闭集的交非空；若半径趋零，交恰为一点。
• 紧集上任何数列都有收敛子列，且极限留在紧集内——这就是度量空间版 BW 定理。
• 证明手法：有限覆盖筛选 → 加倍构造嵌套闭球 → 嵌套引理取出唯一极限。
• 该定理不依赖完备性，因此紧性可以替代完备性 + 有界性，成为分析中的常用假设。

理解检验

1. 为什么 $C[0,1]$ 的闭单位球里有无穷多个两两距离为 $1$ 的函数？请验证 $d(f_n,f_m)=1$。

参考答案：取台阶函数 $f_n$：在 $[0,1/(n+1)]$ 上为 $1$，在 $[1/n,1]$ 上为 $0$，中间线性连接。设 $n<m$。在 $x=1/(n+1)$ 处，$f_n(x)=1$；又因为 $m\ge n+1$，有 $x\ge1/m$，故 $f_m(x)=0$。所以两点之差为 $1$，从而 $d(f_n,f_m)\ge1$。另一方面两者取值都在 $[0,1]$，差的绝对值不超过 $1$，故 $d\le1$。合起来 $d(f_n,f_m)=1$。

2. 用两两距离为 $1$ 的这列函数，说明闭单位球不紧。

参考答案：考虑开球族 $\{B_{1/2}(f) : f\in\overline{B_1(0)}\}$，它覆盖闭球。若 $g,h\in B_{1/2}(f)$，则 $d(g,h)<1$，所以同一个球里不可能同时装下两个不同的 $f_n$。于是任何有限个开球最多装下有限多个 $f_n$，而闭球里有无穷多个 $f_n$，必有未被覆盖者。因此不存在有限子覆盖，闭球不紧。

3. 证明嵌套闭集引理：紧集中一列非空闭集递减，交非空。

参考答案：反设交集为空。由 De Morgan，$\bigcup_i(X\setminus C_i)=X$，所以 $\{X\setminus C_i\}$ 是 $A$ 的开覆盖（每个 $X\setminus C_i$ 开，因为 $C_i$ 闭）。由 $A$ 紧取有限子覆盖 $X\setminus C_{i_1},\dots,X\setminus C_{i_k}$。因 $C_i$ 递减，取 $N=\max i_j$，则这有限个并等于 $X\setminus C_N$，于是 $A\subseteq X\setminus C_N$，即 $A\cap C_N=\varnothing$。但 $C_N\subseteq A$ 且非空，矛盾。

4. 在 BW 定理的证明中，为什么必须把半径 $r_{k+1}$ 的球「加倍」成 $2r_{k+1}$？

参考答案：因为覆盖 $C_k$ 的小球圆心取在 $C_k$ 内，小球可能有一部分伸出上一层球 $B_{r_k}(y_k)$ 之外，直接用它无法保证嵌套。加倍之后，设 $w\in B_{2r_{k+1}}(y_{k+1})$，用三角不等式可估得

$$d(w,y_k)<3r_{k+1}+r_k=\frac{7r_k}{4}<2r_k,$$

于是加倍球落在 $\overline{B_{2r_k}(y_k)}$ 内，嵌套成立。同时半径 $2r_k$ 仍趋于 $0$，唯一性不受影响。

5. 为什么这个证明不能直接说「子列是柯西列，所以收敛」？

参考答案：因为「柯西 ⟹ 收敛」在一般度量空间中不成立，它需要完备性；而本定理的假设只有紧性，没有完备性。所以必须真正构造出极限点：用嵌套闭球引理在紧集内取出唯一交点 $p$，再证明子列收敛到 $p$。如果额外假设空间完备，当然可以走柯西那条路，但那会掩盖紧性的作用。

6. 举一个紧集，并说明它为什么满足定义；再举一个非紧的闭有界集。

参考答案：紧集的例子：$\mathbb R$ 中的 $[0,1]$。任给它的一个开覆盖，由 Heine–Borel 定理（或在 $\mathbb R$ 中用 BW 证明），总能取出有限子覆盖。非紧的闭有界集：$X=(0,1]$ 作为自身的子集，闭且有界，但开覆盖 $\{(1/n,1]\}$ 没有有限子覆盖；另一个例子是 $C[0,1]$ 的闭单位球。

7. 设 $A$ 紧，$(x_n)\subseteq A$ 且 $x_n\to x$。$x$ 是否一定属于 $A$？用本讲的定理说明。

参考答案：属于。因为 $A$ 紧，由 BW 定理，$(x_n)$ 有子列 $x_{n_k}\to p\in A$。但整个数列收敛到 $x$，它的任何子列也收敛到 $x$，由极限唯一性 $p=x$，所以 $x\in A$。这也顺带给出「紧集闭」的一个新证明。

与前后课的关系

上一讲用开覆盖定义紧性并证明它推出闭与有界；这一讲补上反向的关键缺口：一般空间里闭且有界不紧，但紧集足以保证每个数列都有收敛子列。下一讲离开度量空间，转入微分学：从差商定义导数，证明可导蕴含连续，并用同一套「插入中间项」的技巧证明和、积、商与复合的求导法则。

The problem: closed and bounded is not enough

The previous lecture defined compactness — every open cover has a finite subcover — and proved that compact implies closed and bounded. The natural question is whether the converse holds.

• In $\mathbb R^n$ with the Euclidean metric, closed and bounded implies compact; that is a privilege of $\mathbb R^n$.
• In a general metric space it fails.

This lecture gives two counterexamples, one mundane and one deep, then proves the main theorem: every sequence in a compact set has a convergent subsequence. That is the metric-space Bolzano–Weierstrass theorem, and the real purpose of compactness.

Why is it needed? In $\mathbb R$, the Cauchy convergence theorem is proved via BW; BW fails in a general metric space, so the Cauchy convergence theorem does too. Compactness fills the gap. The lecturer also stresses that the proof may not assume completeness, so both the subsequence and its limit must be constructed — which is exactly what the nested-ball lemma delivers.

Timeline

Recap and two counterexamples

What is known

Compactness: every open cover of $A$ has a finite subcover. The previous lecture proved two consequences:

1. Compact sets are bounded — cover with radius-$1$ balls, extract a finite subcover, estimate the radius by the triangle inequality.
2. Compact sets are closed — build the open cover $E_n=\{y:d(y,x)>1/n\}$, take a finite subcover, and contradict $x_n\to x$.

Counterexample 1: $(0,1]$ is not compact

Take $X=(0,1]$ with the usual distance.

• It is closed as a subset of itself: the whole space is always closed (its complement is empty, and the empty set is open).
• It is bounded: contained in the ball of radius $1$ about $1/2$.
• Yet it is not compact. Take

$$U_n=\left(\frac1n,1\right],\qquad n\in\mathbb N .$$

For any $x\in(0,1]$, choose $n>1/x$; then $x>1/n$, so $x\in U_n$, and the family covers $X$. But given finitely many $U_{n_1},\dots,U_{n_k}$, let $N=\max n_i$; their union is $U_N=(1/N,1]$, and $1/(N+1)$ lies outside it. No finite subcover exists.

The lesson: closed and bounded does not imply compact in a general space. The failure is that the space is missing the point $0$.

Counterexample 2: the closed unit ball of $C[0,1]$ is not compact

Take $X=C[0,1]$ with

$$d(f,g)=\max_{x\in[0,1]}|f(x)-g(x)|.$$

Consider the closed unit ball

$$\overline{B_1(0)}=\{f\in C[0,1] : \max_{x}|f(x)|\le1\}.$$

It is closed. Check the complement is open: if $\max|g|=M>1$, put $s=M-1>0$. If $d(g,h)<s$, then

$$\max|h|\ge\max|g|-\max|g-h|>M-s=1,$$

so $h$ also lies in the complement. Hence the complement is open and the ball is closed.

It is obviously bounded — it is a ball.

But it is not compact. For each $n\in\mathbb N$ define

$$f_n(x)= \begin{cases} 1, & 0\le x\le \dfrac{1}{n+1},\\[4pt] (n+1)(1-nx), & \dfrac{1}{n+1}\le x\le \dfrac1n,\\[4pt] 0, & \dfrac1n\le x\le 1. \end{cases}$$

The graph is a step: constant $1$ on the first piece, linear descent to $0$, then $0$. Each piece is continuous and the pieces agree at the joins ($1$ at $x=1/(n+1)$, $0$ at $x=1/n$), so $f_n\in C[0,1]$; and $\max|f_n|=1$, so $f_n\in\overline{B_1(0)}$.

The key estimate: $d(f_n,f_m)=1$ whenever $n\ne m$.

Assume $n<m$. At $x=1/(n+1)$:

• $f_n(x)=1$;
• since $m\ge n+1$, we have $1/(n+1)\ge1/m$, so $f_m(x)=0$.

The difference is exactly $1$, so $d(f_n,f_m)\ge1$; both functions take values in $[0,1]$, so $d\le1$. Hence the distance is exactly $1$.

Now consider the family of open balls

$$\mathcal U=\{B_{1/2}(f) : f\in\overline{B_1(0)}\}.$$

It covers the ball (each $f$ lies in the ball centred at itself) and every member is open.

No finite subcover exists. If $g,h\in B_{1/2}(f)$, the triangle inequality gives

$$d(g,h)\le d(g,f)+d(f,h)<\frac12+\frac12=1 .$$

Since distinct $f_n,f_m$ are at distance $1$, any half-radius ball contains at most one $f_n$. Finitely many balls therefore contain only finitely many $f_n$, leaving infinitely many uncovered. So no finite subcover exists. $\blacksquare$

This example is far deeper than the first: the space is complete (with the sup norm), closed and plainly bounded, yet not compact. The cause is not a missing point but infinite dimension — the unit ball can hold infinitely many points pairwise at distance $1$.

The nested closed set lemma

The main proof needs one tool.

Lemma. Let $A$ be compact. Let $C_1\supseteq C_2\supseteq\cdots$ be a sequence of nonempty closed sets contained in $A$ (each closed in $X$). Then

$$\bigcap_{i=1}^{\infty}C_i\ne\varnothing .$$

Proof. Suppose the intersection is empty. By De Morgan,

$$X=\left(\bigcap_i C_i\right)^c=\bigcup_i (X\setminus C_i).$$

Each $X\setminus C_i$ is open (since $C_i$ is closed), so this is an open cover of $A$.

By compactness of $A$ there is a finite subcover:

$$A\subseteq (X\setminus C_{i_1})\cup\cdots\cup(X\setminus C_{i_k}).$$

Because the $C_i$ decrease, the complements $X\setminus C_i$ increase; with $N=\max\{i_1,\dots,i_k\}$ the finite union equals $X\setminus C_N$. Hence

$$A\subseteq X\setminus C_N, \qquad\text{i.e.}\qquad A\cap C_N=\varnothing .$$

But $C_N\subseteq A$ and $C_N\ne\varnothing$, a contradiction. Therefore the intersection is nonempty. $\square$

Intuition: compactness compresses "infinitely many conditions hold simultaneously" into "finitely many conditions hold simultaneously", and finitely many nested nonempty sets intersect.

Corollary: nested closed balls meet in one point

Corollary. Let $A$ be compact and let $\overline{B_{R_i}(x_i)}\cap A$ be a nested sequence of nonempty closed sets (each containing the next) with $R_i\to0$. Then their intersection is exactly one point.

Proof. Nonemptiness comes from the lemma. If $p,q$ both lie in every set, then for each $i$,

$$d(p,q)\le d(p,x_i)+d(x_i,q)\le 2R_i .$$

Letting $i\to\infty$ gives $d(p,q)=0$, so $p=q$. $\square$

Radii tending to zero give uniqueness; compactness gives existence. Together they conjure a limit point out of nothing — exactly what the BW proof needs.

Main theorem: Bolzano–Weierstrass on compact sets

Theorem. Let $A$ be compact and let $(x_n)$ be a sequence in $A$. Then $(x_n)$ has a convergent subsequence, and its limit lies in $A$.

The proof has three moves: use finite covers to single out a chain of balls each holding infinitely many terms, double them to obtain nested closed sets, then use the corollary to extract the unique intersection point as the limit.

Step 1: one covering

For each $y\in A$ take $B_1(y)$. These cover $A$ and are open. By compactness, finitely many $y_1,\dots,y_k$ satisfy

$$A\subseteq B_1(y_1)\cup\cdots\cup B_1(y_k).$$

The sequence has infinitely many terms but lands in finitely many balls, so at least one ball contains infinitely many terms. Denote it $B_{r_1}(y_1)$ with $r_1=1$.

From now on, look only at the terms in this ball — they form a subsequence. Call its first term $x_{n_1}$.

Step 2: refining

Suppose we have an open ball $B_{r_k}(y_k)$ containing infinitely many terms of the current subsequence, with $y_k\in A$.

Set

$$C_k=\overline{B_{2r_k}(y_k)}\cap A .$$

$C_k$ is a closed subset of the compact set $A$ (a closed ball intersected with $A$), hence compact — this is exactly where the previous lecture's "closed subsets of compact sets are compact" is used.

Cover $C_k$ with open balls of radius $r_{k+1}=r_k/4$ centred at points of $C_k$. By compactness finitely many suffice. The current subsequence has infinitely many terms in $C_k$ (they already lie in $B_{r_k}(y_k)\subseteq C_k$), so one of these finitely many balls still contains infinitely many terms. Call it $B_{r_{k+1}}(y_{k+1})$, and choose the next index $n_{k+1}>n_k$ from it.

Step 3: verifying nestedness

We need $C_{k+1}\subseteq C_k$, otherwise the lemma does not apply.

Both $B_{r_{k+1}}(y_{k+1})$ and $B_{r_k}(y_k)$ contain infinitely many terms, so they intersect. Pick $q$ in the intersection; then

$$d(y_{k+1},y_k)\le d(y_{k+1},q)+d(q,y_k)<r_{k+1}+r_k .$$

For any $w\in B_{2r_{k+1}}(y_{k+1})$,

$$d(w,y_k)\le d(w,y_{k+1})+d(y_{k+1},y_k)<2r_{k+1}+r_{k+1}+r_k=3r_{k+1}+r_k .$$

Substituting $r_{k+1}=r_k/4$:

$$3\cdot\frac{r_k}{4}+r_k=\frac{7r_k}{4}<2r_k .$$

So $B_{2r_{k+1}}(y_{k+1})\subseteq B_{2r_k}(y_k)$, hence $C_{k+1}\subseteq C_k$.

That is why the ball is doubled. Using $B_{r_{k+1}}$ directly, part of it could stick outside the previous ball and nestedness would fail; after doubling to $2r_{k+1}=r_k/2$, the triangle inequality fits it inside the ball of radius $2r_k$. Since $r_k$ quarters at each step, $2r_k\to0$.

Step 4: extracting the limit

By the corollary, $\bigcap_k C_k$ consists of exactly one point $p$. Since $C_k\subseteq A$, we have $p\in A$.

For each $k$, $x_{n_k}\in B_{r_k}(y_k)$ and $p\in\overline{B_{2r_k}(y_k)}$, so

$$d(x_{n_k},p)\le d(x_{n_k},y_k)+d(y_k,p)<r_k+2r_k=3r_k\to0 .$$

Hence $x_{n_k}\to p$, completing the proof. $\square$

Nothing assumes completeness: the limit $p$ is constructed from nested closed balls rather than presumed. This differs from the $\mathbb R$ argument, where completeness is free and one first extracts a subsequence by BW before showing the whole Cauchy sequence converges.

What this lecture establishes

• Compact implies closed and bounded; the converse holds in $\mathbb R^n$ but not in general metric spaces.
• $(0,1]$ shows that a missing point breaks compactness; the unit ball of $C[0,1]$ shows that in infinite dimensions even a complete, closed, bounded set may fail to be compact.
• Nested nonempty closed sets in a compact set intersect; if their radii tend to zero, the intersection is a single point.
• Every sequence in a compact set has a convergent subsequence with limit in the set — the metric-space Bolzano–Weierstrass theorem.
• The technique: finite cover to filter → doubling to build nested closed balls → nested lemma to extract the unique limit.
• The theorem does not need completeness, so compactness can replace "complete plus bounded" as the working hypothesis in analysis.

Comprehension check

1. Why does the closed unit ball of $C[0,1]$ contain infinitely many functions pairwise at distance $1$? Verify $d(f_n,f_m)=1$.

Reference answer: Take the step functions $f_n$: equal to $1$ on $[0,1/(n+1)]$, $0$ on $[1/n,1]$, linear in between. Let $n<m$. At $x=1/(n+1)$, $f_n(x)=1$; and since $m\ge n+1$ we have $x\ge1/m$, so $f_m(x)=0$. The difference is $1$, hence $d(f_n,f_m)\ge1$. Both functions take values in $[0,1]$, so the difference never exceeds $1$ and $d\le1$. Therefore $d(f_n,f_m)=1$.

2. Use this family to show the closed unit ball is not compact.

Reference answer: Consider the open balls $\{B_{1/2}(f) : f\in\overline{B_1(0)}\}$; they cover the ball. If $g,h\in B_{1/2}(f)$, then $d(g,h)<1$, so one ball cannot contain two distinct $f_n$. Hence finitely many balls contain only finitely many $f_n$, while the ball contains infinitely many, so some are uncovered. No finite subcover exists and the ball is not compact.

3. Prove the nested closed set lemma: a decreasing sequence of nonempty closed sets in a compact set has nonempty intersection.

Reference answer: Suppose the intersection is empty. By De Morgan, $\bigcup_i(X\setminus C_i)=X$, so $\{X\setminus C_i\}$ is an open cover of $A$ (each $X\setminus C_i$ is open because $C_i$ is closed). Compactness gives a finite subcover $X\setminus C_{i_1},\dots,X\setminus C_{i_k}$. Since the $C_i$ decrease, with $N=\max i_j$ the finite union equals $X\setminus C_N$, so $A\subseteq X\setminus C_N$, i.e. $A\cap C_N=\varnothing$. But $C_N\subseteq A$ and is nonempty, a contradiction.

4. In the BW proof, why must the ball of radius $r_{k+1}$ be doubled to $2r_{k+1}$?

Reference answer: The small balls covering $C_k$ are centred inside $C_k$ and may stick partly outside the previous ball $B_{r_k}(y_k)$, so they cannot be used directly to get nestedness. After doubling, for $w\in B_{2r_{k+1}}(y_{k+1})$ the triangle inequality gives

$$d(w,y_k)<3r_{k+1}+r_k=\frac{7r_k}{4}<2r_k,$$

so the doubled ball lies inside $\overline{B_{2r_k}(y_k)}$ and nestedness holds. The radii $2r_k$ still tend to $0$, so uniqueness is preserved.

5. Why can the proof not simply say "the subsequence is Cauchy, hence convergent"?

Reference answer: Because "Cauchy implies convergent" fails in a general metric space; it requires completeness, while the hypothesis here is only compactness. The limit must genuinely be constructed: the nested-ball lemma produces a unique intersection point $p$ inside the compact set, and the subsequence is then shown to converge to $p$. Assuming completeness would work but would hide the role of compactness.

6. Give a compact set and explain why it satisfies the definition; then give a closed, bounded, non-compact set.

Reference answer: A compact example is $[0,1]\subseteq\mathbb R$: every open cover admits a finite subcover, by Heine–Borel (or via BW on $\mathbb R$). A closed, bounded, non-compact example is $X=(0,1]$ as a subset of itself: the open cover $\{(1/n,1]\}$ has no finite subcover. Another is the closed unit ball of $C[0,1]$.

7. Let $A$ be compact, $(x_n)\subseteq A$ and $x_n\to x$. Must $x\in A$? Use this lecture's theorem.

Reference answer: Yes. Since $A$ is compact, BW gives a subsequence $x_{n_k}\to p\in A$. But the whole sequence converges to $x$, so every subsequence converges to $x$; uniqueness of limits gives $p=x$, hence $x\in A$. This also yields a new proof that compact sets are closed.

Relation to neighbouring lectures

The previous lecture defined compactness by open covers and showed it implies closed and bounded; this one fills the crucial gap in the converse — closed and bounded is not enough in a general space, but compactness guarantees a convergent subsequence for every sequence. The next lecture leaves metric spaces for differentiation: the derivative is defined through difference quotients, differentiability is shown to imply continuity, and the same "insert an intermediate term" trick proves the sum, product, quotient and chain rules.