Answer to Question #296609 in Functional Analysis for Heba

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 "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.


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