The Opposite of Open

One of our main goals in topology is generalising analysis, so that we can apply sufficiently relevant theorems and results into new contexts. The extreme value theorem is of considerable interest.

Theorem 1. Let $f : [a, b] \to \mathbb R$ be continuous. Then there exist $c_1, c_2 \in [a, b]$ such that $f([a, b]) = [f(c_1), f(c_2)]$ .

There are many key facts about this result that we can explore, beyond some sequence-based arguments, and they boil down to the unique properties of the set $[a, b]$ . Our first observation is that

$\mathbb R \backslash [a, b] = (-\infty, a) \cup (b, \infty).$

We note that $(-\infty, a) = \bigcup_{n = 1}^{\infty}(a-n, a)$ and $(b, \infty) = \bigcup_{n = 1}^{\infty}(b, b+n)$ are open sets, so that their union $\mathbb R \backslash [a, b]$ is open too. This is our working definition of the opposite of open—a closed set.

Let $K$ be a topological space.

Definition 1. A subset $C \subseteq K$ is closed if its complement $K \backslash L$ is open. Equivalently, $C = K \backslash U$ for some open set $U \subseteq K$ .

Being the opposite of open, we would expect many properties of closed sets to resemble that of open sets. Let’s begin with some the basic topological and continuity-ish properties.

Lemma 1. The following properties hold.

$\emptyset, K$ are closed.
If $C_\alpha \subseteq K$ are closed, so is $\bigcap_{\alpha} C_\alpha$ .
If $C_1,\dots, C_n \subseteq K$ are closed, so is $\bigcup_{i=1}^n C_i$ .

Proof. We prove the second result for demonstration and leave the other properties as an exercise. Since for each $\alpha$ , there exists an open set such that $C_\alpha = K \backslash U_\alpha$ , by de Morgan’s laws,

$\displaystyle \bigcap_{\alpha} C_\alpha = \bigcap_{\alpha} (K \backslash U_\alpha) = K \backslash \bigcap_{\alpha} U_\alpha,$

where the result follows since $\bigcap_{\alpha} U_\alpha$ is open.

Lemma 2. The following properties hold.

Equip $L \subseteq K$ with the subspace topology. Then $C \subseteq L$ is closed if and only if there exists a closed set $D \subseteq K$ such that $C = L \cap D$ .
Let $L$ be a topological space and $f : K \to L$ be a function. Then $f$ is continuous if and only if for any closed set $C \subseteq L$ , $f^{-1}(C) \subseteq K$ is closed.

Proof. For the first property, use de Morgan’s laws to deduce that

$\begin{aligned} C &= L \backslash (L \cap V) \\ &= L \cap (K \backslash L \cup K \backslash V) \\ &= (L \cap K \backslash L) \cup (L \cap K \backslash V) \\ &= L \cap K \backslash V,\end{aligned}$

where $D := K \backslash V \subseteq K$ is the desired closed set if and only if $V \subseteq K$ is open. For the second property, for any subset $U \subseteq L$ , we have

$f^{-1}(L \backslash U) = f^{-1}(L) \backslash f^{-1}(U) = K \backslash f^{-1}(U),$

which is closed if and only if $f^{-1}(U)$ is open. Bookkeeping yields the desired result.

At this point we might wonder: can we list out all open and closed sets? Unlikely possible for our finite minds. But not all hope is lost.

Lemma 3. Let $L \subseteq K$ be any subset.

There exists a unique open set $\mathrm{int}(L) \subseteq K$ , called the interior of $L$ , such that for any open set $V \subseteq K$ , $V \subseteq L \Rightarrow V \subseteq \mathrm{int}(L)$ . That is, $\mathrm{int}(L)$ is the largest open set contained in $L$ .
There exists a unique closed set $\bar{L} \subseteq K$ , called the closure of $L$ , such that for any closed set $C \subseteq K$ , $L \subseteq C \Rightarrow \bar{L} \subseteq C$ . That is, $\bar{L}$ is the smallest closed set that contains $L$ .

Furthermore, $L$ is open if and only if $L = \mathrm{int}(L)$ . Similarly, $L$ is closed if and only if $L = \bar{L}$ .

Proof. For any $x \in L$ , call $x$ an interior point of $L$ if it has at least one (not necessarily unique) neighbourhood $V_x$ contained in $L$ (that is, $V_x \subseteq L$ ). Define $\mathrm{int}(L)$ to be the set of interior points of $L$ . Then

$\displaystyle \mathrm{int}(L) = \bigcup_{x \in \mathrm{int}(L)} \{x\} \subseteq \bigcup_{x \in \mathrm{int}(L)} V_x \subseteq \bigcup_{x \in \mathrm{int}(L)} L = L$

is open as a union of open sets. Furthermore, for any open set $V \subseteq K$ , if $V \subseteq L$ , then each point in $V$ is an interior point of $L$ . Hence, $V \subseteq \mathrm{int}(L)$ . For the closed set properties, define $\bar{L} := K \backslash \mathrm{int}(K \backslash L)$ and perform relevant bookkeeping.

As a fun result, we can define the boundary of a set. This is essentially what we mean by the perimeter of a two-dimensional geometric shape.

Definition 2. The boundary and exterior of a set $L \subseteq K$ are the sets defined by

$\partial L := \bar L \backslash L,\quad \mathrm{ext}(L) := K \backslash \bar L.$

Clearly, $\mathrm{ext}(L)$ is open and

$K = \mathrm{int}(L) \sqcup \partial L \sqcup \mathrm{ext}(L),$

so that $\partial L$ is closed.

Rather conveniently, we can even characterise closed sets using convergence, assuming that we are working in a first-countable space.

Lemma 3. Fix $L \subseteq K$ . Then $x \in \bar L$ if and only if for any neighbourhood $V$ of $x$ , $L \cap V \neq \emptyset$ .

Proof. We will prove by contrapositive in both directions. For the direction $(\Rightarrow)$ , suppose $x \notin \bar L$ . Then $x \in K \backslash \bar{L}$ . Since the latter is open, there exists a neighbourhood $V$ of $x$ such that

$L \cap V \subseteq L \cap (K \backslash \bar{L}) \subseteq L \cap (K \backslash L) = \emptyset.$

For the direction $(\Leftarrow)$ , suppose there exists a neighbourhood $V$ of $x$ such that $L \cap V = \emptyset$ . Then $V \subseteq K \backslash L$ . We note that $K \backslash V$ is a closed set containing $L$ , and thus $K \backslash V \supseteq \bar{L}$ . However, $x \notin K \backslash V$ , so that $x \notin \bar{L}$ , as required.

Theorem 1. Fix $L \subseteq K$ . Fix $x \in K$ .

If there exists an $L$ -sequence $x_n \to x$ , then $x \in \bar L$ .
Suppose $K$ is first-countable. If $x \in \bar L$ , then there exists an $L$ -sequence $x_n \to x$ .

Proof. For the first property, fix any neighbourhood $V$ of $x$ . Since $x_n \to x$ , there exists $N \in \mathbb N$ such that $x_n \in V$ for $n \geq N$ . In particular, $x_N \in L \cap V$ . Thus, $L \cap V \neq \emptyset$ . By Lemma 3, $x \in \bar L$ .

For the second property, fix $x \in \bar L$ . Since $K$ is first-countable, there exists a countable collection $\{B_n : n \in \mathbb N\}$ of basis elements containing $x$ . Define the open sets

$\displaystyle V_n := \bigcap_{k=1}^n B_k \supseteq \{x\}.$

Since $x \in \bar L$ , by Lemma 3, each $L \cap V_n \neq \emptyset$ . Define the sequence $x_n \in L \cap V_n$ . We claim that $x_n \to x$ . Fix any neighbourhood $V$ of $x$ . Then there exists $N \in \mathbb N$ such that $B_N \subseteq V$ . In particular, for $n \geq N$ ,

$x_n \in L \cap V_n \subseteq V_n \subseteq B_n \subseteq V,$

so that $x_n \to x$ , as required.

In particular, the set $[a, b]$ is usually discussed in the context of the normed space $(\mathbb R, | \cdot |)$ , which is a metric space, and thus first-countable, so that Theorem 1 applies.

However, closed sets by themselves are nowhere nearly as interesting as compact sets. Observe that $[a, b]$ is not only closed, but bounded—in that for any $x \in [a, b]$ , there exists $M \geq 0$ such that $|x| \leq M$ . Generally, the notion of $\leq$ doesn’t make sense in topology. But we will soon see that the closeness and boundedness together can be generalised into the notion of compact sets, which is, really, the key ingredient to generalising the extreme value theorem.