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Let $X$ be a topological space with no special properties. My book claims the following in an exercise:

Suppose:

  1. $A \subset X$ is closed.

  2. $U \subset A$ is open (with respect to the induced topology on $A$).

  3. $V \subset X$ is open with $U \subset V$.

Show: $U \cup (V - A)$ is open in $X$.

I can see why this would be true in the presence of some separation axioms, but the book asks me to prove it in general.

Many thanks for your help.

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Revised to match revised problem statement.

HINT: Since $U$ is open in $A$, there is a $W$ open in $X$ such that $U=W\cap A$. Let $G=W\cap V$, and show that $G\cap A=U$. Then show that $U\cup(V\setminus A)=G\cup(V\setminus A)$.

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  • $\begingroup$ I'm very sorry but I forgot one of the hypotheses in my question. I have added it now (to point 3.). $\endgroup$ – Frank Apr 12 '15 at 16:29
  • $\begingroup$ @Frank: Ah, okay; now the statement is indeed generally true. $\endgroup$ – Brian M. Scott Apr 12 '15 at 16:34

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