Answer to Question #200195 in Real Analysis for Rajkumar

Question

Prove that the union of two closed sets is a closed set. Give an example to show that

union of an infinite number of closed sets need not be a closed set.

Answer 1

Solution:

Let $(X, \mathcal{T})$ be a topological space.

By definition, for any subset $A \subset X$ we have

A is open $\Longleftrightarrow A \in \mathcal{T}$ and A is closed $\Longleftrightarrow C(A)$ is open.

a)$\emptyset$ is closed: Because $\emptyset=C(X)$ and $X \in \mathcal{T}$

b) X is closed: Because $X=C(\emptyset)$ and $\emptyset \in \mathcal{T}$

c) A finite union of closed sets is closed: Let $A_{1}, \ldots, A_{N}$ be closed subsets in X. Then $C\left(A_{1}\right), \ldots, C\left(A_{N}\right)$ are open by definition. Furthermore, we have

$C\left(\bigcup_{n=1}^{N} A_{n}\right)=\bigcap_{n=1}^{N} C\left(A_{n}\right)$

from set theory, and this is open. So its complement

$\bigcup_{n=1}^{N} A_{n}$ is closed by definition, as desired.

d) An arbitrary intersection of closed sets is closed: Let $\left\{A_{\alpha}\right\}_{\alpha \in I}$ be a collection of closed subsets in X. Then $C\left(A_{\alpha}\right)$ is open for every \alpha \in I by definition. Furthermore, we have

$C\left(\bigcap_{\alpha \in I} A_{\alpha}\right)=\bigcup_{\alpha \in I} C\left(A_{\alpha}\right)$

from set theory, and this is open. So its complement

$\bigcap_{\alpha \in I} A_{a}$

is closed by definition, as desired.

From all these parts, we showed that the union of two closed sets is a closed set.

Example:

Define $A_{n}=\left[\frac{1}{n}, 1-\frac{1}{n}\right]$ for n>1.

Then, obviously each $A_{n}$ is closed, but $\bigcup A_{n}=(0,1)$

which is open and not closed.