IV.1 · Open & Closed Sets — Problem 1 #

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Problem. In a metric space $(S,\rho)$, prove that $S$ and $\varnothing$ are closed, that a finite union of closed sets is closed, and that an arbitrary intersection of closed sets is closed.

Hint #

A set is closed exactly when its complement is open, so translate everything into a statement about complements and apply DeMorgan’s laws:

$$ \Big(\bigcap_{i \in I} A_i\Big)^c = \bigcup_{i \in I} A_i^c, \qquad \Big(\bigcup_{i \in I} A_i\Big)^c = \bigcap_{i \in I} A_i^c. $$

Then use the known theorems that arbitrary unions of open sets are open and finite intersections of open sets are open.

Solution #

To prove the first two parts, note that $S$ and $\varnothing$ are complements of $\varnothing$ and $S$, respectively. Since the latter are both open by a theorem from class, both of the former must be closed.

For the other two parts, we will need DeMorgan’s laws. Recall that

$\left(\bigcap_{i \in I} A_i\right)^c = \bigcup_{i \in I} A_i^c$, and
$\left(\bigcup_{i \in I} A_i\right)^c = \bigcap_{i \in I} A_i^c$.

Thus for part (d), we use the first equality to see that the complement of $\bigcap_{i \in I} A_i$ is a union of complements of closed sets. That is, it is a union of open sets, which is open by a theorem from class. Hence $\bigcap_{i \in I} A_i$ is closed since its complement is open.

Finally, for part (c), we use the second equality to see that the complement of $\bigcup_{i \in I} A_i$ is an intersection of complements of closed sets. That is, it is an intersection of open sets, and in fact, an intersection of a finite number of such. By a theorem from class, this intersection must be open. Hence $\bigcup_{i \in I} A_i$ is closed since its complement is open.

Long solution #

Strategy. The word “closed” is defined through the word “open”: a set is closed exactly when its complement is open. So the whole problem is a translation exercise. Each statement about closed sets becomes, after taking complements, a statement about open sets — and the statements we land on (the whole space and the empty set are open, arbitrary unions of open sets are open, finite intersections of open sets are open) were all proved in class. The dictionary between the two languages is the pair of DeMorgan laws, which swap unions with intersections when you take complements. Notice the pleasant duality: for open sets it is unions that are unrestricted and intersections that must be finite; after complementing, closed sets inherit the mirror image — unrestricted intersections, finite unions.

Definition / known facts. A set $A \subseteq S$ is closed iff its complement $A^c = S \setminus A$ is open. From class we take as known:

$S$ and $\varnothing$ are open.

The union of any collection of open sets is open.

The intersection of finitely many open sets is open.

Step 0: DeMorgan’s laws. For any index set $I$ (finite or infinite) and any sets $A_i \subseteq S$,

$$ \Big(\bigcap_{i \in I} A_i\Big)^c = \bigcup_{i \in I} A_i^c, \qquad \Big(\bigcup_{i \in I} A_i\Big)^c = \bigcap_{i \in I} A_i^c. $$

These are pure set theory, and it is worth seeing why they hold, since the whole proof rests on them. Take the first one and chase an arbitrary element $x \in S$:

$$ x \in \Big(\bigcap_{i \in I} A_i\Big)^c \;\iff\; x \notin \bigcap_{i \in I} A_i \;\iff\; \text{not } (\forall i \in I:\ x \in A_i) \;\iff\; \exists i \in I:\ x \notin A_i \;\iff\; x \in \bigcup_{i \in I} A_i^c . $$

The key move is the middle one: negating “for all $i$, $x \in A_i$” gives “for some $i$, $x \notin A_i$.” Since the two sets have exactly the same elements, they are equal. The second law is proved the same way (or by applying the first law to the sets $A_i^c$ and taking complements of both sides, using $(B^c)^c = B$).

(a) $S$ is closed.

By the definition in the box, $S$ is closed iff its complement $S^c$ is open. Every point of $S$ is a point of $S$, so nothing is left over: $S^c = S \setminus S = \varnothing$. And $\varnothing$ is open by fact 1. Hence $S$ is closed.

(b) $\varnothing$ is closed.

Likewise $\varnothing$ is closed iff $\varnothing^c$ is open. Removing nothing from $S$ leaves everything: $\varnothing^c = S \setminus \varnothing = S$. And $S$ is open by fact 1. Hence $\varnothing$ is closed.

(So $S$ and $\varnothing$ are each simultaneously open and closed. That is not a contradiction — “closed” does not mean “not open.” The two words are not opposites: a set can be both, either, or neither.)

(c) A finite union of closed sets is closed.

Let $I$ be a finite index set, say $I = \{1, 2, \dots, n\}$, and suppose each $A_i$ is closed. Unwinding the definition: each complement $A_i^c$ is open. We want to show $\bigcup_{i \in I} A_i$ is closed, i.e. that its complement is open. By the second DeMorgan law,

$$ \Big(\bigcup_{i \in I} A_i\Big)^c = \bigcap_{i \in I} A_i^c = A_1^c \cap A_2^c \cap \cdots \cap A_n^c . $$

The right-hand side is an intersection of open sets, and — this is the crucial point — an intersection of only finitely many of them, because $I$ is finite. By fact 3 it is therefore open. So the complement of $\bigcup_{i \in I} A_i$ is open, which by definition means $\bigcup_{i \in I} A_i$ is closed.

Infinitely many closed intervals whose union is the half-open interval (0,1], which is not closed

Why (c) needs finiteness: each closed interval $A_n = [1/n, 1]$ contains its left endpoint, but the left endpoints $1/n$ march down to $0$, and $0$ itself is never reached. The union is $(0,1]$, which fails to contain its limit point $0$.

Remark (finiteness is essential in (c)). In $\mathbb{R}$ with the usual metric, each interval $A_n = [\,1/n,\, 1\,]$ is closed, yet

$$ \bigcup_{n \geq 1} A_n = (0, 1], $$

which is not closed: the point $0$ is a limit point of $(0,1]$ (every ball $(-r, r)$ around $0$ contains points $1/n \in (0,1]$), but $0 \notin (0,1]$. To see where the proof of (c) would break for infinite $I$: the complement of the union is still $\bigcap_n A_n^c$ by DeMorgan, but now it is an intersection of infinitely many open sets, and fact 3 does not apply. Indeed $\bigcap_n A_n^c = (0,1]^c = (-\infty, 0\,] \cup (1, \infty)$ is not open — no ball around $0$ stays inside it.

(d) An arbitrary intersection of closed sets is closed.

Now let $I$ be any index set — finite, countable, or uncountable — and suppose each $A_i$ is closed, so each $A_i^c$ is open. By the first DeMorgan law,

$$ \Big(\bigcap_{i \in I} A_i\Big)^c = \bigcup_{i \in I} A_i^c . $$

The right-hand side is a union of open sets. Fact 2 places no restriction at all on how many open sets may be unioned, so this union is open regardless of the size of $I$. Therefore the complement of $\bigcap_{i \in I} A_i$ is open, which means $\bigcap_{i \in I} A_i$ is closed.

This completes all four parts. $\blacksquare$