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Let $A$ be a subset of a topological space $X$ and let $B$ be a subset of $A$. I need to prove that the $\mathrm{Cl}(A)-\mathrm{Cl}(B)$ is contained in $\mathrm{Cl}(A-B)$. $\mathrm{Cl}$ denotes the closure of a set.

I first let $x$ be in $\mathrm{Cl}(A)-\mathrm{Cl}(B)$, then $x$ is in $\mathrm{Cl}(A)-B$.

After that step I'm not sure how to proceed. Any help would be deeply appreciated.

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Since $x$ is not in $\text{Cl}(B)$, some open neighborhood $U$ of $x$ does not intersect $B$. But $x \in \text{Cl}(A)$, so $U$ intersects $A$. Why does this give you what you want?

Edit: to finish the proof, let $V$ be any open neighborhood of $x$, so we want to show that $V$ intersects $A - B$. Since $U \cap V$ is an open neighborhood of $x$ and $x \in \text{Cl}(A)$ there exists $y \in A \cap U \cap V$. In particular $y \in U \subset X - B$, so actually $y \in (A - B) \cap V$ as desired.

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If U intersects A then U intersects A-B? –  james Dec 15 '11 at 18:49
    
@james $U$ is fixed, here. Given an arbitrary open set $V$ containing $x$, what can you say about $V$ (looking at $V\cap U$)? –  David Mitra Dec 15 '11 at 19:05
    
Im not sure, I want to say that x is in the closure of V. –  james Dec 15 '11 at 19:27
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