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

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*Is it always true that $Cl_{Y}(U)= Cl_{X}(V)\cap Y$ ?

*how about this, $Cl_{X}(U)= Cl_{X}(V)$ ?

 A: $\newcommand{\cl}{\operatorname{cl}}$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 $\cl_XU=\cl_XV$.
