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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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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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