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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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    @BrianM.Scott Yes. I got confused because I am now studying topological vector spaces.2012-09-19

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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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    @Eli: You’re right: I really am having trouble with typos today. Thanks for catching that. I’m glad that it finally made sense.2012-09-20