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Let $X$ be a topological space and $B$ is a Borel set of this space, i.e. $B\in\mathcal{B}(X)$ where $\mathcal{B}(X)$ is the smallest $\sigma$-algebra which contains all open subsets of $X$.

Let $B\subset A$ where $A$ is compact. Is it true that $A' = (A\setminus B) \cup\partial B$ is a compact set?

If it is not true in general, does it hold if $X$ is also metrizable (separable)?

Finally, is a condition that $B$ is a Borel set is crucial?

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1 Answer 1

up vote 4 down vote accepted

Unwinding the definition, one can see that $A' = A \backslash B^\circ$, where $B^\circ$ is the interior of $B$. A compact set minus an open set is always compact; this is an easy exercise. (Contrary to a previous edit, it isn't necessary to assume that $A$ is closed or that $X$ is Hausdorff.) No other assumptions on $A,B,X$ are needed.

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Do you use here that $B^\circ = B\setminus \partial B$? –  Ilya Jun 8 '11 at 12:48
    
@Gortaur: Yes, the interior of a set is the set itself minus its boundary. –  Asaf Karagila Jun 8 '11 at 12:54

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