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Let M be a metric space. Let G be an open subset of M and A be any subset of M. It is to be shown that G intersection cl(A) is contained in cl(G intersection A). Initially I thought that every interior point is a limit point but that was shown to be wrong. Now I don’t know how to attack this problem. Any solution will be appreciated.

General Grievous
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1 Answers1

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Let $x_0 \in G \cap cl(A)$. Then there is a sequence $(x_n)$ in $A$ with $x_n \to x_0$.

Since $G$ is open, there is $N \in \mathbb N$ such that $x_n \in G$ for all $n>N$.

Then we have $x_n \in G \cap A$ for all $n>N$. This gives $x_0 = \lim x_n \in cl(G \cap A)$ .

Fred
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