G is an open sphere in metric space M and A is a subset of M.
Prove that 1. $G∩Cl(A)$⊆ $Cl(G∩A)$ and 2. $Cl(G∩Cl(A))$= $Cl(G∩A)$
I am halfway done through the first part.
Suppose, $x $ is an element of $G∩Cl(A)$. Then $x$ is either in $G∩A$ => $x$ is also in $Cl(G∩A)$ or $x$ is in $G∩A^l$,
where $A^l$ is the set of limit points.
How do I show that if $x$ is in $G∩A^l$ then it'll also be in $Cl(G∩A)$? and the proof of (2) as well?
Let $x \in G\cap Cl(A)$. Let $B$ be any open ball ($\equiv $ sphere) around $x$. Then there is an open ball $D$ around $x$ contained in $G \cap B$. (Take the smaller of the radii of $G$ and $B$). Since $x \in Cl(A)$ there must be a point $y$ in $D \cap A$. It follows that $G\cap B \cap A$ is non-empty. Thus, $G \cap A$ intersects every ball $B$ around $x$. This implies that $x$ is in the closure of $G \cap A$ as required.
Second part:By the first part we have $G∩Cl(A)$⊆ $Cl(G∩A)$ and the right side is a closed set. Hence, $Cl(G∩Cl(A))$⊆ $Cl(G∩A)$. To prove the reverse inclusion note that $G\cap A \subseteq G∩Cl(A)$. Take closure on both sides to finish.
