Is a nonempty intersection of a collection of closed sets closed?

Question

My intuition is yes, but how would you prove it? Would the nonempty intersection of a collection of closed sets always be closed?

Answer 1

We use the following proposition*:

If $\{ U_i : i \in \mathcal{A}\}$ is any collection of open subsets of $M$, then $\displaystyle\bigcup_{i \in \mathcal{A}} U_i$ is open.

We will also use one of De Morgan's Laws:

Let $M$ be a set, $\{B_i : i \in \mathcal{A}\}$ a collection of subsets of $M$. Then $\displaystyle\bigcup_{i \in \mathcal{A}}(M \setminus B_i) = M \setminus \displaystyle\bigcap_{i \in \mathcal{A}} B_i$

The trick to dealing with closed sets is to take their complement so that we instead deal with open sets.

We want to show that:

If $\{ F_i : i \in \mathcal{A}\}$ is any collection of closed subsets of $M$, then $\displaystyle\bigcap_{i \in \mathcal{A}} F_i$ is closed.

Since $F_i$ is closed, we have that $M \setminus F_i$ is open.

So $\displaystyle\bigcup_{i \in \mathcal{A}}(M \setminus F_i)$ is open by the proposition.

By De Morgan's Laws, $\displaystyle\bigcup_{i \in \mathcal{A}}(M \setminus F_i) = M \setminus \Bigg(\displaystyle\bigcap_{i \in \mathcal{A}}F_i\Bigg)$.

Hence, by definition $\displaystyle\bigcap_{i \in \mathcal{A}}F_i$ is closed.

*EDIT: I've just realised that this is tagged general-topology in a recent edit, in which case the proposition is true by definition and isn't actually a proposition at all. However, for a metric space we can actually prove it.

Answer 2

This is correct and easy to prove.

The proof is from ProofWiki:

Theorem

Let $T = \left({S, \tau}\right)$ be a topological space. Then the intersection of an arbitrary number of closed sets of $T$ (either finitely or infinitely many) is itself closed.

Proof

Let $I$ be an indexing set (either finite or infinite).

Let $\displaystyle \bigcap_{i \mathop \in I} V_i$ be the intersection of a indexed family of closed sets of $T$ indexed by $I$.

Then from De Morgan's laws: Difference with Intersection:

$\displaystyle S \setminus \bigcap_{i \mathop \in I} V_i = \bigcup_{i \mathop \in I} \left({S \setminus V_i}\right)$

By definition of closed set, each of $S \setminus V_i$ are by definition open in $T$.

We have that $\displaystyle \bigcup_{i \mathop \in I} \left({S \setminus V_i}\right)$ is the union of a indexed family of open sets of $T$ indexed by $I$.

Therefore, by definition of a topology, $\displaystyle \bigcup_{i \mathop \in I} \left({S \setminus V_i}\right) = S \setminus \bigcap_{i \mathop \in I} V_i$ is likewise open in $T$.

Then by definition of closed set, $\displaystyle \bigcap_{i \mathop \in I} V_i$ is closed in $T$. $\blacksquare$