Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

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

share|improve this question
5  
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
1  
Is this homework? Where does it come from? –  Old John Jul 16 '12 at 21:25
2  
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
2  
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
add comment

1 Answer

up vote 1 down vote accepted

Following Leonid's suggestion:

(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.

share|improve this answer
    
I would like to ask whether the Hausdorff axiom is used. I may have missed it. –  Frank Jul 18 '12 at 1:14
    
@Mohammed: Ah, good point, I intended to comment on that and then forgot. No, I don't think I used it. –  joriki Jul 18 '12 at 5:37
add comment

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.