问题：怎样描述「边界包含在内」

上一讲定义了开集：集合里每个点都还能往外走一点。但很多常见的集合恰恰相反——它们把边界包了进来，比如闭区间 $[0,1]$。直接用「每个点都有小球」来刻画这类集合很别扭，因为边界点本身就没有小球留在集合内。

本讲用两个动作解决这个问题：

1. 把闭集定义成余集是开集，绕开直接的边界描述；
2. 证明闭集有一个非常好用的等价刻画——收敛数列的极限不逃出去。

然后再往前一步：用开集定义覆盖，用「每个开覆盖都有有限子覆盖」定义紧性。紧性看起来抽象，但它解决的问题很具体：在一般度量空间里，有界数列不一定有收敛子列；而在紧集上，这个结论会恢复。

时间轴

回顾：开集、球与集合运算

$O\subseteq X$ 开，指对每个 $x\in O$，存在 $r>0$ 使 $B_r(x)\subseteq O$，其中

$$B_r(x)=\{y : d(x,y)<r\}.$$

集合运算三件套：并 $\bigcup_\alpha U_\alpha$（存在某个 $\alpha$ 属于）、交 $\bigcap_\alpha U_\alpha$（对所有 $\alpha$ 属于）、补 $X\setminus U$（不属于 $U$）。

这些运算之所以重要，是因为开集与闭集的性质完全由它们翻译。并、交、补之间的换算律（De Morgan）会在后面反复使用。

开集的并交法则

任意并

命题。 若 $\{O_\alpha\}$ 是一族开集，则 $\bigcup_\alpha O_\alpha$ 开。

证明。 取 $x\in\bigcup_\alpha O_\alpha$，则存在 $\alpha_0$ 使 $x\in O_{\alpha_0}$。因 $O_{\alpha_0}$ 开，存在 $r>0$ 使 $B_r(x)\subseteq O_{\alpha_0}\subseteq\bigcup_\alpha O_\alpha$。$\square$

指标集多大都无所谓——可以是不可数族。

有限交

命题。 若 $O_1,\dots,O_n$ 开，则 $\bigcap_{i=1}^n O_i$ 开。

证明。 取 $x\in\bigcap_i O_i$。对每个 $i$ 选 $r_i>0$ 使 $B_{r_i}(x)\subseteq O_i$，令 $r=\min\{r_i\}>0$。有限个正数的最小值仍为正。于是 $B_r(x)\subseteq B_{r_i}(x)\subseteq O_i$ 对每个 $i$ 成立，故 $B_r(x)\subseteq\bigcap_i O_i$。$\square$

无限交的反例

在 $\mathbb R$ 中取 $O_n=(-1/n,1/n)$。每个 $O_n$ 开，但

$$\bigcap_{n=1}^{\infty}O_n=\{0\}$$

不是开集。原因就一条：$\inf_n 1/n=0$，取不到正的下界，min 那一步没有无限版本。

这个反例后面还会用到——它同时说明「开集的无限交」和「闭集的有限并」两处都必须限制在有限个。

闭集的定义

定义。 $C\subseteq X$ 是闭集，指其余集 $X\setminus C$ 是开集。

两个立刻得到的事实：$\varnothing$ 与 $X$ 都闭（它们互为余集，而两者都开），所以它们既开又闭。

引理（球的余集是开集）。 固定 $x\in X$、$r>0$，集合

$$E=\{y\in X : d(y,x)>r\}$$

是开集。

证明。 取 $y\in E$，即 $d(y,x)>r$。令

$$s=d(y,x)-r>0 .$$

对任意 $z\in B_s(y)$，用三角不等式的反向形式：

$$d(x,z)\ge d(x,y)-d(y,z)>d(x,y)-s=r .$$

于是 $z\in E$，即 $B_s(y)\subseteq E$。所以 $E$ 开。$\square$

这里和证明「球是开集」几乎一样，唯一区别是三角不等式用在了另一侧：要证距离大于某数，就得把待估的量放到不等式右边。这个技巧在后面证明紧集是闭集时还要再用一次。

由此立刻得到：闭球 $\{y : d(y,x)\le r\}$ 是闭集，因为它的余集正是上面的 $E$。

闭集的序列刻画

定理。 设 $C\subseteq X$。则 $C$ 是闭集，当且仅当：对任何满足 $x_n\in C$ 的收敛数列 $x_n\to x$，都有 $x\in C$。

这个定理的实用价值在于：要证一个集合闭，不必去验证余集开，只要证明极限不逃出去。前者在具体例子里往往很繁琐，后者通常只有几行。

方向一：闭 ⟹ 极限封闭

设 $C$ 闭，$x_n\in C$，$x_n\to x$。反设 $x\notin C$，即 $x$ 属于开集 $X\setminus C$。于是存在 $r>0$ 使

$$B_r(x)\subseteq X\setminus C .$$

但 $x_n\to x$ 意味着充分靠后的 $x_n$ 都落在这个球里；而球整个在 $C$ 的余集中，所以这些 $x_n\notin C$，与 $x_n\in C$ 矛盾。故 $x\in C$。$\square$

方向二：极限封闭 ⟹ 闭

设 $C$ 满足「收敛列的极限仍在 $C$ 中」，要证 $X\setminus C$ 开。

取 $x\in X\setminus C$。反设不存在半径为 $r$ 的球含于余集，即对每个 $n$，球 $B_{1/n}(x)$ 都不能整个落在余集中。于是可以选出

$$x_n\in B_{1/n}(x)\cap C .$$

由 $d(x_n,x)<1/n$ 得 $x_n\to x$。按假设，极限 $x$ 应属于 $C$——但 $x$ 本来取在余集里，矛盾。所以余集开，$C$ 闭。$\square$

这两个方向都用到了同一个想法：把「不属于开集」翻译成「存在一个球，球里全在集合内」。

例子

• $[0,1]\subseteq\mathbb R$ 闭：余集 $(-\infty,0)\cup(1,\infty)$ 开。
• $[0,1)$ 既不闭也不开：余集 $(-\infty,0)\cup[1,\infty)$ 不包含 $1$ 附近的左侧点，不是开集。
• $[0,1]\cup[2,3]$ 闭。
• 单点集 $\{p\}$ 闭：余集 $\{y : d(y,p)>0\}=X\setminus\{p\}$，正是引理中 $r=0$ 的极限情形，也可以直接验证。

注意最后一条的推论：任何有限集都是闭集（有限个单点集的并）。

闭集的并交法则

定理。

• 任意多个闭集的交是闭集：若 $\{C_\alpha\}$ 闭，则 $\bigcap_\alpha C_\alpha$ 闭。
• 有限多个闭集的并是闭集：若 $C_1,\dots,C_n$ 闭，则 $\bigcup_{i=1}^n C_i$ 闭。

结论与开集完全对偶：开集是「任意并、有限交」，闭集是「任意交、有限并」。

证明。 只需用 De Morgan 律把并交互换。

$$X\setminus\bigcap_\alpha C_\alpha=\bigcup_\alpha (X\setminus C_\alpha).$$

每个 $X\setminus C_\alpha$ 开，任意多个开集的并开，所以左边是开集，即 $\bigcap_\alpha C_\alpha$ 闭。

$$X\setminus\bigcup_{i=1}^n C_i=\bigcap_{i=1}^n (X\setminus C_i).$$

每个 $X\setminus C_i$ 开，有限个开集的交开，所以 $\bigcup_i C_i$ 闭。$\square$

对应地，无限多个闭集的并可能不是闭集。例如

$$\bigcup_{n=1}^{\infty}\left[\frac1n,1\right]=(0,1],$$

而 $(0,1]$ 不闭。这和「开集无限交失效」是同一个现象的两种说法。

覆盖与紧性

覆盖

设 $A\subseteq X$，$\{U_\alpha\}$ 是一族子集。若

$$A\subseteq\bigcup_\alpha U_\alpha,$$

称 $\{U_\alpha\}$ 是 $A$ 的一个覆盖。

• 子覆盖：从原族中取一部分，仍然覆盖 $A$。
• 开覆盖：每个 $U_\alpha$ 都是开集。
• 有限子覆盖：子覆盖只含有限个集合。

覆盖本身没有任何要求——集合可以是开的、闭的、什么都不是。

例子。 在 $\mathbb R$ 中取 $U_n=(-n,n)$，$n\in\mathbb N$。这族集合覆盖整个 $\mathbb R$，且每个都开，所以是开覆盖。但它没有有限子覆盖：任取有限个，最大的那个是 $(-N,N)$，盖不住 $N+1$。

紧性

定义。 $A\subseteq X$ 是紧集，指 $A$ 的每个开覆盖都有有限子覆盖。

量词结构值得慢慢读一遍：

$$\forall\ \text{开覆盖}\ \{U_\alpha\}\ \exists\ \text{有限子族}\ \{U_{\alpha_1},\dots,U_{\alpha_k}\}\ \text{仍覆盖 } A .$$

注意「有限」出现在结论里：无论别人给出多么庞大、多么刁钻的开覆盖，你总能从中挑出有限个完成任务。这正是紧性强大的地方——它把无限问题压缩成有限问题。

为什么需要紧性

回顾主线：在 $\mathbb R$ 中，柯西收敛定理靠 Bolzano–Weierstrass 定理证明，而 BW 说的是「有界数列有收敛子列」。

在一般度量空间里，BW 不成立（上一讲的 $X=(0,1]$、$x_n=1/n$ 就是反例）。于是柯西收敛定理也随之失效。

紧性正是修复这件事的条件：在紧集上，任何数列都有收敛子列。这就是下一讲要证的「度量空间版 BW 定理」，也是引入紧性的主要动机。

紧集必有界且闭

定理。 若 $A$ 紧，则 $A$ 有界且闭。

有界性

对每个 $y\in A$ 取半径 $1$ 的开球 $B_1(y)$。这族球显然覆盖 $A$（每个点被自己为心的球包含），且都是开集。

由紧性，存在有限个 $y_1,\dots,y_k\in A$ 使

$$A\subseteq\bigcup_{i=1}^{k}B_1(y_i).$$

令

$$r=1+\max_{1\le i\le k}d(y_1,y_i).$$

对任意 $x\in A$，存在某个 $i$ 使 $x\in B_1(y_i)$，即 $d(x,y_i)<1$。三角不等式给出

$$d(x,y_1)\le d(x,y_i)+d(y_i,y_1)<1+\max_{i}d(y_1,y_i)=r .$$

所以 $A\subseteq B_r(y_1)$，有界。$\square$

闭性

用闭集的序列刻画：取 $x_n\in A$、$x_n\to x$，要证 $x\in A$。

反设 $x\notin A$。对每个 $n$ 定义

$$E_n=\left\{y\in X : d(y,x)>\frac1n\right\}.$$

前面已证每个 $E_n$ 是开集。并且 $\bigcup_n E_n=X\setminus\{x\}$：若 $y\ne x$，取 $n>1/d(y,x)$ 即可。因为 $x\notin A$，所以

$$A\subseteq X\setminus\{x\}=\bigcup_{n=1}^{\infty}E_n$$

是一个开覆盖。

由 $A$ 紧，存在有限子覆盖。注意 $E_n$ 随 $n$ 增大而增大（$1/n$ 变小，条件变松），所以有限个 $E_{n_1},\dots,E_{n_k}$ 中最大指标那个 $E_N$ 包含其余所有，从而

$$A\subseteq E_N,\qquad\text{即}\qquad \forall y\in A,\ d(y,x)>\frac1N .$$

但 $x_n\in A$ 且 $x_n\to x$，充分靠后时 $d(x_n,x)<1/N$，矛盾。所以 $x\in A$，$A$ 闭。$\square$

这个证明的结构值得记住：为了证明一个点不在闭包里，就用「距离大于 $1/n$」造一族递增的开集把它隔离出去。下一讲证明「闭子集仍紧」时会再次出现类似的构造。

反向不成立

在 $\mathbb R^n$ 中（配欧氏距离），闭且有界 ⟹ 紧。这是下一讲会讨论的定理，也是 $\mathbb R^n$ 特别好用的原因。

但在一般度量空间里这条不成立。上一讲的例子 $X=(0,1]$：作为自身子集它既闭又有界，却不紧——开覆盖 $\{(1/n,1]\}$ 没有有限子覆盖。

紧集的闭子集仍紧

定理。 设 $A$ 紧，$C\subseteq A$ 且 $C$ 在 $X$ 中闭。则 $C$ 紧。

证明。 设 $\{O_\alpha\}$ 是 $C$ 的任意开覆盖。因为 $C$ 闭，$X\setminus C$ 开。于是

$$\{O_\alpha\}\cup\{X\setminus C\}$$

是 $A$ 的开覆盖：$A$ 中不属于 $C$ 的点被 $X\setminus C$ 盖住，属于 $C$ 的点被某个 $O_\alpha$ 盖住。

由 $A$ 紧，取有限子覆盖。若其中有 $X\setminus C$，把它去掉——它对覆盖 $C$ 毫无贡献（$C$ 与 $X\setminus C$ 不相交）。剩下的有限个 $O_\alpha$ 仍然覆盖 $C$。所以 $C$ 的任意开覆盖都有有限子覆盖，$C$ 紧。$\square$

这个证明很短，但用了两个关键事实：$C$ 闭（所以余集开，可以加进覆盖）与 $A$ 紧（所以有限子覆盖存在）。少了任何一个都不行。

本节结论

• 闭集 = 余集开；$\varnothing$ 与 $X$ 既开又闭。
• 闭集 ⟺ 收敛列的极限留在集合内；这个刻画是证明闭性的主力工具。
• 开集：任意并、有限交；闭集：任意交、有限并。无限情形两边都会失效。
• 覆盖不要求开；开覆盖 + 紧性才产生有限子覆盖。
• 紧集必有界且闭；$\mathbb R^n$ 中反向成立，一般度量空间不成立。
• 紧集的闭子集仍紧。

理解检验

1. 为什么不能把闭集直接定义成「每个点都有一个球不完全在集合外」？

参考答案：这种说法是开集定义的否定，刻画的是「不是开集」，而不是「闭」。例如 $[0,1)$ 不是开集，但它也不是闭集；而 $\mathbb R$ 中 $(0,1]$ 的余集 $(-\infty,0]\cup(1,\infty)$ 不是开集，说明 $(0,1]$ 不闭。「闭」的正确含义是余集开，它表达的是「补集里的点都有回旋余地」，与「集合内每个点都有球」是两件不同的事。

2. 证明：闭球的余集 $E=\{y:d(y,x)>r\}$ 是开集。

参考答案：取 $y\in E$，令 $s=d(y,x)-r>0$。对任意 $z\in B_s(y)$，由反向三角不等式

$$d(x,z)\ge d(x,y)-d(y,z)>d(x,y)-s=r,$$

故 $z\in E$，即 $B_s(y)\subseteq E$。由于 $y$ 任意，$E$ 开。关键在于要用「距离的差」而不是「距离的和」。

3. 证明：若 $C$ 闭且 $x_n\in C$、$x_n\to x$，则 $x\in C$。

参考答案：反设 $x\notin C$，则 $x$ 在开集 $X\setminus C$ 中，存在 $r>0$ 使 $B_r(x)\subseteq X\setminus C$。由 $x_n\to x$，存在 $N$ 使 $n\ge N$ 时 $x_n\in B_r(x)$，于是 $x_n\notin C$，与 $x_n\in C$ 矛盾。所以 $x\in C$。

4. 举出一族闭集，其并不是闭集，并说明它与「开集无限交失效」的关系。

参考答案：取 $C_n=[1/n,1]$，每个闭，但 $\bigcup_n C_n=(0,1]$，不闭（$0$ 是极限点却不在集合中）。它的余集是 $(-\infty,0)\cup(1,\infty)$，不包含 $0$ 附近的点，不是开集。取余集后，这族闭集的并就变成一族开集 $(-1/n,1/n)$ 的交，交集为 $\{0\}$ 不闭——两件事通过 De Morgan 律完全对应。

5. 用紧性证明：紧集有界。（写出半径 $r$ 的取法）

参考答案：对每个 $y\in A$，取开球 $B_1(y)$，这族球覆盖 $A$。紧性给出有限子覆盖 $B_1(y_1),\dots,B_1(y_k)$。令 $r=1+\max_i d(y_1,y_i)$。任取 $x\in A$，有某个 $i$ 使 $d(x,y_i)<1$，于是

$$d(x,y_1)\le d(x,y_i)+d(y_i,y_1)<1+\max_i d(y_1,y_i)=r,$$

所以 $A\subseteq B_r(y_1)$。

6. 为什么「紧集的闭子集仍紧」的证明里，必须把 $X\setminus C$ 加进覆盖？

参考答案：因为 $C$ 的开覆盖 $\{O_\alpha\}$ 只覆盖 $C$，未必覆盖 $A$ 中位于 $C$ 之外的点。要给 $A$ 造一个开覆盖，必须补上 $A\setminus C$；而 $C$ 闭保证 $X\setminus C$ 开，可以合法地加入开覆盖。取到有限子覆盖后，$X\setminus C$ 对覆盖 $C$ 没有贡献，可以删去，剩下的有限个 $O_\alpha$ 就覆盖了 $C$。

7. 在 $\mathbb R$ 中，$[0,1]$ 紧而 $(0,1]$ 不紧。给出后者的一个没有有限子覆盖的开覆盖。

参考答案：取 $U_n=(1/n,1]$，$n\in\mathbb N$。每个 $U_n$ 是 $(0,1]$ 中的开集（相对于 $(0,1]$）。对任意 $x\in(0,1]$，取 $n>1/x$，则 $1/n<x$，故 $x\in U_n$；所以这族覆盖 $(0,1]$。任取有限个 $U_{n_1},\dots,U_{n_k}$，令 $N=\max n_i$，则这些集合的并是 $U_N=(1/N,1]$，而 $1/(N+1)\in(0,1]$ 不在其中。所以没有有限子覆盖。

与前后课的关系

上一讲建立了开集与并交法则；这一讲把它们对偶到闭集，并引入覆盖与紧性。紧性的意义暂时只是「一个定义加两个性质」，下一讲才会兑现：用嵌套闭集引理证明紧集上的 Bolzano–Weierstrass 定理，从而说明紧性正是让「有界 ⟹ 有收敛子列」重新成立的条件。这一讲末尾的「闭子集仍紧」会在那里作为关键工具出现。

The problem: describing "the boundary is included"

The previous lecture defined open sets: every point can move a little without leaving. Many familiar sets do the opposite — they contain their boundary, such as the closed interval $[0,1]$. Describing such sets by "every point has a small ball" is awkward, because boundary points have no ball that stays inside.

This lecture solves the problem in two moves:

1. Define a closed set as one whose complement is open, bypassing any direct description of the boundary.
2. Prove a much more usable characterisation: limits of convergent sequences do not escape.

Then one step further: use open sets to define covers, and "every open cover has a finite subcover" to define compactness. Compactness looks abstract, but it solves something concrete: in a general metric space a bounded sequence need not have a convergent subsequence; on a compact set it does.

Timeline

Recap: open sets, balls and set operations

$O\subseteq X$ is open if for every $x\in O$ there is $r>0$ with $B_r(x)\subseteq O$, where

$$B_r(x)=\{y : d(x,y)<r\}.$$

The three set operations: union $\bigcup_\alpha U_\alpha$ (in at least one), intersection $\bigcap_\alpha U_\alpha$ (in all), complement $X\setminus U$ (not in $U$).

They matter because the properties of open and closed sets are stated entirely through them. The conversion rules between union, intersection and complement (De Morgan) will be used repeatedly.

Union and intersection rules for open sets

Arbitrary unions

Proposition. If $\{O_\alpha\}$ is a family of open sets, then $\bigcup_\alpha O_\alpha$ is open.

Proof. Take $x\in\bigcup_\alpha O_\alpha$; then $x\in O_{\alpha_0}$ for some $\alpha_0$. Since $O_{\alpha_0}$ is open, some $r>0$ has $B_r(x)\subseteq O_{\alpha_0}\subseteq\bigcup_\alpha O_\alpha$. $\square$

The index set may be arbitrarily large — even uncountable.

Finite intersections

Proposition. If $O_1,\dots,O_n$ are open, then $\bigcap_{i=1}^n O_i$ is open.

Proof. Take $x\in\bigcap_i O_i$. For each $i$ choose $r_i>0$ with $B_{r_i}(x)\subseteq O_i$ and set $r=\min\{r_i\}>0$. The minimum of finitely many positive numbers is positive. Then $B_r(x)\subseteq B_{r_i}(x)\subseteq O_i$ for every $i$, so $B_r(x)\subseteq\bigcap_i O_i$. $\square$

The infinite-intersection counterexample

In $\mathbb R$ take $O_n=(-1/n,1/n)$. Each is open, yet

$$\bigcap_{n=1}^{\infty}O_n=\{0\}$$

is not open. The reason is sharp: $\inf_n 1/n=0$, there is no positive lower bound, and the min step has no infinite analogue.

The counterexample returns later — it also shows why "infinite intersection of open sets" and "finite union of closed sets" both need the finiteness restriction.

Definition of a closed set

Definition. $C\subseteq X$ is closed if its complement $X\setminus C$ is open.

Two immediate facts: $\varnothing$ and $X$ are both closed (they are each other's complements and both are open), so they are both open and closed.

Lemma (the complement of a ball is open). Fix $x\in X$ and $r>0$. The set

$$E=\{y\in X : d(y,x)>r\}$$

is open.

Proof. Take $y\in E$, so $d(y,x)>r$. Put

$$s=d(y,x)-r>0 .$$

For any $z\in B_s(y)$, the reverse triangle inequality gives

$$d(x,z)\ge d(x,y)-d(y,z)>d(x,y)-s=r .$$

Hence $z\in E$, i.e. $B_s(y)\subseteq E$, and $E$ is open. $\square$

This is almost the proof that a ball is open, with the triangle inequality used on the other side: to prove a distance is greater than something, the quantity being estimated must sit on the right of the inequality. The same trick returns when proving compact implies closed.

It follows at once that the closed ball $\{y : d(y,x)\le r\}$ is closed, since its complement is exactly $E$.

The sequential characterisation of closed sets

Theorem. Let $C\subseteq X$. Then $C$ is closed if and only if: whenever $x_n\in C$ and $x_n\to x$, the limit satisfies $x\in C$.

The practical value: to show a set is closed you need not check that its complement is open; it suffices to show limits do not escape. The former is often tedious in examples; the latter is usually a few lines.

Direction 1: closed ⟹ limits stay inside

Let $C$ be closed, $x_n\in C$, $x_n\to x$. Suppose $x\notin C$, so $x$ lies in the open set $X\setminus C$. Then some $r>0$ satisfies

$$B_r(x)\subseteq X\setminus C .$$

But $x_n\to x$ means all sufficiently late terms lie in this ball, which sits entirely in the complement of $C$ — so those $x_n\notin C$, contradicting $x_n\in C$. Hence $x\in C$. $\square$

Direction 2: limits stay inside ⟹ closed

Assume $C$ has the property that limits of convergent sequences in $C$ stay in $C$; show $X\setminus C$ is open.

Take $x\in X\setminus C$. Suppose no ball around $x$ lies in the complement, i.e. for every $n$ the ball $B_{1/n}(x)$ fails to be contained in the complement. Then we may choose

$$x_n\in B_{1/n}(x)\cap C .$$

Since $d(x_n,x)<1/n$, we get $x_n\to x$. By hypothesis the limit belongs to $C$ — but $x$ was chosen in the complement, a contradiction. So the complement is open and $C$ is closed. $\square$

Both directions rest on one translation: "not in the open set" becomes "there is a ball entirely inside the set".

Examples

• $[0,1]\subseteq\mathbb R$ is closed: its complement $(-\infty,0)\cup(1,\infty)$ is open.
• $[0,1)$ is neither open nor closed: its complement $(-\infty,0)\cup[1,\infty)$ omits points just left of $1$, so it is not open.
• $[0,1]\cup[2,3]$ is closed.
• A singleton $\{p\}$ is closed: its complement $\{y : d(y,p)>0\}$ is the $r=0$ limiting case of the lemma, or can be checked directly.

A consequence of the last point: every finite set is closed (a finite union of singletons).

Union and intersection rules for closed sets

Theorem.

• The intersection of arbitrarily many closed sets is closed: if $\{C_\alpha\}$ is closed, so is $\bigcap_\alpha C_\alpha$.
• The union of finitely many closed sets is closed: if $C_1,\dots,C_n$ are closed, so is $\bigcup_{i=1}^n C_i$.

The statements are exactly dual to those for open sets: open sets allow arbitrary unions and finite intersections; closed sets allow arbitrary intersections and finite unions.

Proof. Apply De Morgan to exchange union and intersection.

$$X\setminus\bigcap_\alpha C_\alpha=\bigcup_\alpha (X\setminus C_\alpha).$$

Each $X\setminus C_\alpha$ is open, and arbitrary unions of open sets are open, so the left side is open; hence $\bigcap_\alpha C_\alpha$ is closed.

$$X\setminus\bigcup_{i=1}^n C_i=\bigcap_{i=1}^n (X\setminus C_i).$$

Each $X\setminus C_i$ is open, and finite intersections of open sets are open, so $\bigcup_i C_i$ is closed. $\square$

Correspondingly, the union of infinitely many closed sets need not be closed, e.g.

$$\bigcup_{n=1}^{\infty}\left[\frac1n,1\right]=(0,1],$$

which is not closed. This is the same phenomenon as the failure of infinite intersections of open sets, read through De Morgan.

Covers and compactness

Covers

Let $A\subseteq X$ and let $\{U_\alpha\}$ be a family of subsets. If

$$A\subseteq\bigcup_\alpha U_\alpha,$$

the family is a cover of $A$.

• A subcover is a subfamily that still covers $A$.
• An open cover is a cover whose members are all open.
• A finite subcover is a subcover consisting of finitely many sets.

A cover imposes no condition on the sets themselves — they may be open, closed, or neither.

Example. In $\mathbb R$ take $U_n=(-n,n)$, $n\in\mathbb N$. The family covers $\mathbb R$ and every member is open, so it is an open cover. It has no finite subcover: among finitely many, the largest is some $(-N,N)$, which misses $N+1$.

Compactness

Definition. $A\subseteq X$ is compact if every open cover of $A$ has a finite subcover.

Read the quantifiers slowly:

$$\forall\ \text{open cover}\ \{U_\alpha\}\ \exists\ \text{finite subfamily}\ \{U_{\alpha_1},\dots,U_{\alpha_k}\}\ \text{still covering } A .$$

"Finite" appears in the conclusion: however large or awkward the open cover someone hands you, you can always extract finitely many sets that finish the job. That is the power of compactness — it compresses an infinite problem into a finite one.

Why compactness is needed

Recall the main line: in $\mathbb R$, the Cauchy convergence theorem is proved via Bolzano–Weierstrass, which says every bounded sequence has a convergent subsequence.

In a general metric space BW fails (the previous lecture's $X=(0,1]$, $x_n=1/n$). The Cauchy convergence theorem then fails too.

Compactness is exactly the hypothesis that repairs this: on a compact set every sequence has a convergent subsequence. That is the metric-space BW theorem proved next lecture, and the main motivation for the definition.

Compact sets are bounded and closed

Theorem. If $A$ is compact, then $A$ is bounded and closed.

Boundedness

For each $y\in A$ take the open ball $B_1(y)$. These balls cover $A$ (each point lies in its own ball) and are open.

By compactness there are finitely many $y_1,\dots,y_k\in A$ with

$$A\subseteq\bigcup_{i=1}^{k}B_1(y_i).$$

Let

$$r=1+\max_{1\le i\le k}d(y_1,y_i).$$

For any $x\in A$, some $i$ satisfies $x\in B_1(y_i)$, i.e. $d(x,y_i)<1$. The triangle inequality gives

$$d(x,y_1)\le d(x,y_i)+d(y_i,y_1)<1+\max_{i}d(y_1,y_i)=r .$$

So $A\subseteq B_r(y_1)$ and $A$ is bounded. $\square$

Closedness

Use the sequential characterisation: take $x_n\in A$ with $x_n\to x$ and show $x\in A$.

Suppose $x\notin A$. For each $n$ define

$$E_n=\left\{y\in X : d(y,x)>\frac1n\right\}.$$

Each $E_n$ is open by the lemma above, and $\bigcup_n E_n=X\setminus\{x\}$: if $y\ne x$, choose $n>1/d(y,x)$. Since $x\notin A$,

$$A\subseteq X\setminus\{x\}=\bigcup_{n=1}^{\infty}E_n$$

is an open cover.

By compactness there is a finite subcover. The sets $E_n$ increase with $n$ (as $1/n$ shrinks the condition loosens), so among finitely many $E_{n_1},\dots,E_{n_k}$ the one with the largest index, say $E_N$, contains all the others. Hence

$$A\subseteq E_N,\qquad\text{i.e.}\qquad \forall y\in A,\ d(y,x)>\frac1N .$$

But $x_n\in A$ and $x_n\to x$, so eventually $d(x_n,x)<1/N$ — a contradiction. Therefore $x\in A$ and $A$ is closed. $\square$

The structure is worth remembering: to prove a point lies outside a closed hull, construct an increasing family of open sets "distance greater than $1/n$" that isolates it. A similar construction appears next lecture when proving that closed subsets of compact sets are compact.

The converse fails

In $\mathbb R^n$ with the Euclidean metric, closed and bounded implies compact. That is the theorem discussed next lecture, and the reason $\mathbb R^n$ is so convenient.

In a general metric space it fails. The previous lecture's $X=(0,1]$: as a subset of itself it is closed and bounded, yet not compact — the open cover $\{(1/n,1]\}$ has no finite subcover.

Closed subsets of compact sets are compact

Theorem. Let $A$ be compact and let $C\subseteq A$ be closed in $X$. Then $C$ is compact.

Proof. Let $\{O_\alpha\}$ be an arbitrary open cover of $C$. Since $C$ is closed, $X\setminus C$ is open. Hence

$$\{O_\alpha\}\cup\{X\setminus C\}$$

is an open cover of $A$: points of $A$ outside $C$ are covered by $X\setminus C$, points of $C$ by some $O_\alpha$.

By compactness of $A$, extract a finite subcover. If $X\setminus C$ is among the chosen sets, discard it — it contributes nothing to covering $C$ (it is disjoint from $C$). The remaining finitely many $O_\alpha$ still cover $C$. Hence every open cover of $C$ has a finite subcover, so $C$ is compact. $\square$

Short, but it uses two essential facts: $C$ closed (so the complement is open and may join the cover) and $A$ compact (so a finite subcover exists). Neither can be dropped.

What this lecture establishes

• A closed set is one whose complement is open; $\varnothing$ and $X$ are both open and closed.
• Closed ⟺ limits of convergent sequences stay inside; this is the main tool for proving closedness.
• Open sets: arbitrary unions, finite intersections. Closed sets: arbitrary intersections, finite unions. Infinite cases fail on both sides.
• A cover need not be open; only an open cover plus compactness yields a finite subcover.
• Compact sets are bounded and closed; in $\mathbb R^n$ the converse holds, in general metric spaces it does not.
• Closed subsets of compact sets are compact.

Comprehension check

1. Why can a closed set not simply be defined as "every point has a ball that is not entirely outside the set"?

Reference answer: That phrasing is the negation of the open-set condition and characterises "not open", not "closed". For instance $[0,1)$ is not open, but it is not closed either; and in $\mathbb R$ the complement of $(0,1]$ is $(-\infty,0]\cup(1,\infty)$, which is not open, so $(0,1]$ is not closed. The correct meaning of "closed" is the complement is open, which says points in the complement have room to move — a different statement from "every point of the set has a ball inside".

2. Prove that the complement of a closed ball, $E=\{y:d(y,x)>r\}$, is open.

Reference answer: Take $y\in E$ and set $s=d(y,x)-r>0$. For any $z\in B_s(y)$, the reverse triangle inequality gives

$$d(x,z)\ge d(x,y)-d(y,z)>d(x,y)-s=r,$$

so $z\in E$ and $B_s(y)\subseteq E$. Since $y$ was arbitrary, $E$ is open. The key is to use a difference of distances, not a sum.

3. Prove: if $C$ is closed, $x_n\in C$ and $x_n\to x$, then $x\in C$.

Reference answer: Suppose $x\notin C$. Then $x$ lies in the open set $X\setminus C$, so some $r>0$ has $B_r(x)\subseteq X\setminus C$. Since $x_n\to x$, there is $N$ with $x_n\in B_r(x)$ for $n\ge N$, hence $x_n\notin C$, contradicting $x_n\in C$. Therefore $x\in C$.

4. Give a family of closed sets whose union is not closed, and relate it to the failure of infinite intersections of open sets.

Reference answer: Take $C_n=[1/n,1]$; each is closed, but $\bigcup_n C_n=(0,1]$, which is not closed ($0$ is a limit point outside the set). Its complement $(-\infty,0)\cup(1,\infty)$ omits points near $0$, so it is not open. Taking complements turns this union into the intersection of the open sets $(-1/n,1/n)$, whose intersection is $\{0\}$ and is not open — the two facts correspond exactly through De Morgan.

5. Use compactness to prove that a compact set is bounded (show how $r$ is chosen).

Reference answer: For each $y\in A$, the open balls $B_1(y)$ cover $A$. Compactness gives a finite subcover $B_1(y_1),\dots,B_1(y_k)$. Let $r=1+\max_i d(y_1,y_i)$. For any $x\in A$ some $i$ satisfies $d(x,y_i)<1$, so

$$d(x,y_1)\le d(x,y_i)+d(y_i,y_1)<1+\max_i d(y_1,y_i)=r,$$

hence $A\subseteq B_r(y_1)$.

6. In the proof that a closed subset of a compact set is compact, why must $X\setminus C$ be added to the cover?

Reference answer: The open cover $\{O_\alpha\}$ covers only $C$ and need not cover points of $A$ outside $C$. To construct an open cover of $A$, those points must be added; closedness of $C$ guarantees $X\setminus C$ is open, so it may legally join the cover. After extracting a finite subcover, $X\setminus C$ contributes nothing to covering $C$ and can be discarded, leaving finitely many $O_\alpha$ that still cover $C$.

7. In $\mathbb R$, $[0,1]$ is compact but $(0,1]$ is not. Give an open cover of $(0,1]$ with no finite subcover.

Reference answer: Take $U_n=(1/n,1]$ for $n\in\mathbb N$. Each $U_n$ is open in $(0,1]$ (relative to that space). For any $x\in(0,1]$, choose $n>1/x$; then $1/n<x$, so $x\in U_n$, and the family covers $(0,1]$. Given finitely many $U_{n_1},\dots,U_{n_k}$, let $N=\max n_i$; their union is $U_N=(1/N,1]$, and $1/(N+1)\in(0,1]$ lies outside it. So no finite subcover exists.

Relation to neighbouring lectures

The previous lecture built open sets and their union/intersection rules; this one dualises them to closed sets and introduces covers and compactness. For now compactness is "a definition plus two properties"; the next lecture cashes it in by proving Bolzano–Weierstrass on compact sets, showing that compactness is exactly the condition that restores "bounded implies a convergent subsequence". The closing result here, that closed subsets of compact sets are compact, appears there as a key tool.