Answer to Question #296609 in Functional Analysis for Heba

Question

Let (X,d) be a metric space and consider the subset A ⊂ X and B ⊂ X . Show that the closure satisfies the following property closure( A ∪ B) =closure(A) ∪ closure(B)

Answer 1

First of all, let us remark that the closure of a set $A$ is the smallest closed set containing $A$. Indeed, any closed set $S$ that contains $A$ should also contain its closure by its definition, and the closure is itself a closed set.

Secondly, let us remark that if $C\subseteq D$ then $\text{closure}(C) \subseteq \text{closure}(D)$. It is just a direct application of the definiton of closure (or we can use the property we just mentioned above).

Finally, let us remind that a finite union of closed sets is itself closed.

From these three properties we deduce easily the result :

$\text{closure}(A\cup B)$ is the smallest closed set containing $A\cup B$

$\text{closure}(A)\cup \text{closure}(B)$ is a closed set, it contains $A$ and B and thus $A\cup B$. Therefore, $\text{closure}(A\cup B) \subseteq \text{closure}(A)\cup \text{closure}(B)$

At the same time, $A\subseteq A\cup B \subseteq \text{closure}(A\cup B)$, so we should have $\text{closure}(A)\subseteq \text{closure}(A\cup B)$. Same argument for $B$ gives us that

$\text{closure}(A) \cup \text{closure}(B) \subseteq \text{closure}(A\cup B)$

By the double inclusion we have the result.