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Let $A \subset X$ be closed and $U \subset A$ open in $A$. Let $V$ be any open set in $X$ with $U \subset V$.

Prove $U \cup (V \setminus A)$ is open in $X$.

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Please don't phrase your queries in the imperative mode (as if you are giving orders or assignments for the group). If this is homework, please tag it with the [homework] tag. In any case, you should say what you have tried or why you are having trouble with it. –  Arturo Magidin Dec 23 '10 at 15:44
    
@Sivaram I actually think that your reply was completely appropriate. If I may ask, what was the reason for deleting it? –  Alex B. Dec 23 '10 at 16:00
    
@Alex: Can't speak for Sivaram, but his original hint (before his edit, when it was more detailed) assumed that $U$ was open, not "open in $A$". –  Arturo Magidin Dec 23 '10 at 16:02
    
@Arturo It is true that it didn't mention subspace topology, but just because a hint doesn't give away everything, it doesn't mean that it should be deleted. That's why I was curious. To be honest, in view of your request that the poster demonstrate his progress, which hasn't happened yet, I would rather support lighter hints. –  Alex B. Dec 23 '10 at 16:09
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@Sivaram You might want to google "subspace topology". Any subspace of a topological space inherits a topology from the ambient space. Its definition is essentially given in Arturo's answer. –  Alex B. Dec 24 '10 at 2:15

1 Answer 1

HINT: Since $U$ is open in $A$, there is an open set $\mathscr{U}$ of $X$ such that $U = A\cap\mathscr{U}$. Consider the open sets $\mathscr{U}\cap V$ and $V\setminus A$.

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