Question #296609

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)


1
Expert's answer
2022-02-13T17:58:31-0500

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

Secondly, let us remark that if CDC\subseteq D then closure(C)closure(D)\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 :

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

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

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

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

By the double inclusion we have the result.


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