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Let $Y$ is a dense subspace of topological space $X$ and $U \mathop \subset \limits^{open} Y$. My purpose is to show that $Cl_{Y}(U)= Cl_{X}(V)\cap Y$. and i have two Question;

  1. Is it always true that $Cl_{Y}(U)= Cl_{X}(V)\cap Y$ ?
  2. how about this, $Cl_{X}(U)= Cl_{X}(V)$ ?
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What is the set $V$...? –  icurays1 Jan 30 '13 at 4:24
    
@icurays1 : Open set in $X$ s.t. $U=V\cap Y$ –  TXC Jan 30 '13 at 4:25

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$\newcommand{\cl}{\operatorname{cl}}$The answer to both questions is yes.

Question 1. Is it always true that $\cl_YU=Y\cap\cl_XV$?

Clearly $U\subseteq Y\cap\cl_XV$, and $Y\cap\cl_XY$ is closed in $Y$, so $\cl_YU\subseteq Y\cap\cl_XV$. Now suppose that $y\in Y\cap\cl_XV$, and let $G$ be an open nbhd of $y$ in $Y$. There is an open $W$ in $X$ such that $G=Y\cap W$, and since $y\in\cl_XV$, clearly $W\cap V\ne\varnothing$. $Y$ is dense in $X$, so $Y\cap W\cap V\ne\varnothing$. But $Y\cap W\cap V=(Y\cap W)\cap(Y\cap V)=G\cap U$, so $G\cap U\ne\varnothing$, $y\in\cl_YU$, and $Y\cap\cl_XV\subseteq\cl_YU$. Thus, it is always true that $\cl_YU=Y\cap\cl_XV$.

Question 2. Is it always true that $\cl_XU=\cl_XV$?

Clearly $\cl_XU\subseteq\cl_XV$, so it suffices to show that $\cl_XV\subseteq\cl_XU$. Suppose that $x\in\cl_XV$, and let $G$ be any open nbhd of $x$ in $X$; of course $G\cap V\ne\varnothing$. $G\cap V$ is therefore a non-empty open set in $X$, and $Y$ is dense in $X$, so $G\cap U=G\cap V\cap Y\ne\varnothing$, $x\in\cl_XU$, and $\cl_XU=\cl_XV$.

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In general and without dense condition, is it true that $Cl_{Y} U= Cl_{X} U \cap Y$? –  TXC Jan 30 '13 at 5:16
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@TXC: Yes. If $y\in Y\cap\cl_XU$, and $V$ is any open nbhd of $y$ in $Y$, let $W$ be open in $X$ with $V=Y\cap W$. Then $W\cap U\ne\varnothing$, and since $U\subseteq Y$, $W\cap U=V\cap U$. Thus, $V\cap U\ne\varnothing$, and $y\in\cl_YU$. –  Brian M. Scott Jan 30 '13 at 5:34

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