Open subsets of the closure I want to prove that every open subset of a topological subpace is an open subset of its closure.
Let $Y$ be a topological space and $X$ a subspace of $Y$. If $U$ is an open subset of $X$, we have $U=U\cap \overline X$, thus U is an open subset of $\overline X$ also.
Am I right?
Thanks in advance
 A: Yes, this is correct. More generally, if $U$ is an open subset of $X$, and $A$ is any subset of $X$ containing $U$, then $U\cap A=U$ is an open subset of $A$.
Added: By ‘this’ I meant the assertion that an open set is open in its closure; I was called away and didn’t look closely enough at the argument, which seems to go with the different (and false) proposition that if $U$ is an open subset of $X$, where $X\subseteq Y$, then $U$ is open in $\operatorname{cl}_YX$. A counterexample to this is to take $Y=\Bbb R$, $X=\Bbb Q$, and $U=\Bbb Q\cap(0,1)$. Then $U$ is relatively open in $\Bbb Q$, but $U$ is not an open set in $\operatorname{cl}_{\Bbb R}\Bbb Q=\Bbb R$.
A: No, this is incorrect. An open subset $O$ of $X$ is an intersection $X\cap O'$ with $O'$ an open subset of $Y$. In general $O$ will not, in general, be open in $\overline{X}$, for there is no reason that there exists an open subset $O'\subset Y$ with both 
$$\begin{cases}
O=X\cap O' \\
\qquad \text{and}\\
O=\overline{X}\cap O'
\end{cases}$$
For instance, consider $X=\Bbb Q\subset \Bbb R=Y$. Then $O=\Bbb Q_{+}^*=(0,\infty)\cap\Bbb Q$, the positive rational numbers, is open in $X$, and not in $Y$.
