# Subspace of Hausdorff space

Let $X$ be a Hausdorff space and $A\subset X$.how to show that the following statements are equivalent:

(I) For every $x\in A$, $x$ has a neighborhood $U\subset X$ such that $A\cap U$ is closed in $U$.

(II) $A$ can be written as the intersection of a closed set and an open set.

(III) $\bar{A}-A$ is closed

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What have you tried? What are your thoughts on the problem? What are the difficulties you are encountering? –  M Turgeon Jul 16 '12 at 20:59
Is this homework? Where does it come from? –  Old John Jul 16 '12 at 21:25
I suggest proving equivalence in the order 3->2->1->3. The first two steps aren't that hard. NB for readers: neighborhoods are not assumed to be open in this problem. –  user31373 Jul 17 '12 at 3:49
Mohammed, it seems that quite a number of your questions have drawn comments on the lack of context and requests to provide the source of the problem -- I suggest that you take this to heart and start providing this information of your own accord. It's more fun to work on a problem when you know why someone is interested in it, and how difficult it's likely to be, and how reliable it is that it's well-posed. Also some people like to know whether it's homework, and the way you've been posting problems looks quite similar to people posting homework problems, though it seems you weren't. –  joriki Jul 17 '12 at 7:50

(III) $\Rightarrow$ (II): $\bar A$ is closed and $X-(\bar A-A)$ is open, and their intersection is $\bar A\cap(X-(\bar A-A))=\bar A\cap (A\cup(X-\bar A))=A$.
(II) $\Rightarrow$ (I): Let $A=C\cap U$ with $C$ closed and $U$ open. Then $U$ is a neighbourhood of every $x\in A$, and $A\cap U=C\cap U\cap U=C\cap U$; thus $U-(A\cap U)=U-(C\cap U)=U-C=U\cap(X-C)$, which is open, so $A\cap U$ is closed in $U$.
(I) $\Rightarrow$ (III): If $U$ is a neighbourhood of $x\in A$ such that $A\cap U$ is closed in $U$, then $U$ is disjoint from $\bar A-A$; otherwise it would contain a point outside of $A$ each of whose neighbourhoods contains a point of $A$, so $U-(A\cap U)$ couldn't be open in $U$. Thus the union of all the open neighbourhoods contained in all these neighbourhoods for all $x\in A$ is disjoint from $\bar A-A$. It is open and contains $A$. If we add the complement of $\bar A$, which is open, we get the complement of $\bar A-A$, and as the union of two open sets this is open; hence $\bar A-A$ is closed.
@Joriki For $(I)\rightarrow (II)$, you are proving $\bar{A}^c$ and $A$ are both open. So $\bar{A}-A$ is closed? Did I miss anything? –  claire Jul 11 '14 at 15:18