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


  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.


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 $U\cup(V\setminus A)=G\cup(V\setminus A)$.

  • $\begingroup$ 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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