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I have this problem:

let $Y$ be a closed subspace in $X$. If $A\subset X$ and $H$ is an open neighborhood of $Y\cap A$ in $Y$. Prove that $A\cap (\overline{Y\setminus H}) = \emptyset$. ($\overline{Y\setminus H}$) is the closure of $Y \setminus H$ in $X$.

I don´t see why it works, neither how to prove it, I hope someone can help me! Thank you

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Can you give some background information about how you run into this problem? Also, what do you mean by 'Y be a closed subspace of X'? Subspace in the linear sense or just a subset with topology induced from $X$? – Hui Yu Sep 19 '12 at 5:50
@Hui Yu: This is clearly just topology. – Brian M. Scott Sep 19 '12 at 5:51
@BrianM.Scott Yes. I got confused because I am now studying topological vector spaces. – Hui Yu Sep 19 '12 at 6:16

If $H$ is an open set in $Y$, there must be an open set $U$ in $X$ such that $H=U\cap Y$. We want to show that $A\cap\operatorname{cl}_X(Y\setminus H)=\varnothing$.

Now $Y\setminus H=Y\setminus(U\cap Y)=Y\setminus U$. (You may have to think a little about that last step. $Y\setminus(U\cap Y)=\{y\in Y:y\notin U\cap Y\}$, and $Y\setminus U=\{y\in Y:y\notin U\}$; do you see why those must be equal?)

$Y$ is closed in $X$, and $U$ is open in $X$, so $Y\setminus U$ is closed in $X$. (Why?) Thus, $Y\setminus H$ is closed in $X$, and therefore $\operatorname{cl}_X(Y\setminus H)=Y\setminus H$. Thus, $A\cap\operatorname{cl}_X(Y\setminus H)=A\cap(Y\setminus H)$, and we want to show that this is empty. But $H\supseteq A\cap Y$, so ... ?

Here’s a diagram that you may find helpful: vertical shading is $H$, horizontal shading is $A\cap Y$. $Y\setminus H=Y\setminus U$ is the region between the outline of $Y$ and the shaded set $H$.

enter image description here

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Thank you! but i have some questions I dont see why H⊇A, because we only know that H⊇Y∩A. Also dont know why if Y∖U is closed then Y∖H, is it beacause U⊇H??? – Ali Sep 20 '12 at 4:41
@Eli: That wasn’t a very good picture: I shouldn’t have drawn it as if $H\supseteq A$. I’ll replace it with a better one. However, in my argument I never claimed that $H\supseteq A$, so it doesn’t affect the argument. For your other question, recall that I began by proving that $Y\setminus U=Y\setminus H$, so if one is closed, the other must be: they’re the same set. – Brian M. Scott Sep 20 '12 at 4:58
Oh i didn´t paid enough attention. Sorry. – Ali Sep 20 '12 at 5:03
You wrote that H⊇A... I think this is true but only in Y so when we take A∩(Y∖H) that´s empty? is that a good argument or am I missing something. Sorry for all the doubts but topology is hard for me – Ali Sep 20 '12 at 5:09
@Eli: Oh, I didn’t even see that when I reread it moment ago. That was a typo, which I’ve now fixed. And yes, you have the right idea now: $A\cap(Y\setminus H)=\{x\in A:x\in Y\setminus H\}=\{x\in A:x\in Y\text{ and }x\notin H\}=\{x\in A\cap Y:x\notin H\}=(A\cap Y)\setminus H=\varnothing$. – Brian M. Scott Sep 20 '12 at 5:17

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