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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

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  • $\begingroup$ I don't understand : Consider $X=\Bbb Q\subset \Bbb R=Y$. Then $U=\Bbb Q_{+}^*$, the positive rational numbers, is open in $X$, and not in $Y$... $\endgroup$ – Olivier Bégassat Nov 20 '13 at 4:25
  • $\begingroup$ Make sure you pay close attention to which set you're taking the closure of. I agree with Brian that the theorem is referring to the closure of $U$ in $X$. You would need to rewrite your proof to show that $U = U \cap \overline{U}$ (where $\overline{U}$ is the closure of $U$ in $X$) so $U$ is open in $\overline{U}$. $\endgroup$ – manthanomen Nov 20 '13 at 5:15
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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 $\operatorname{cl}_{\Bbb R}\Bbb Q=\Bbb R$.

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  • $\begingroup$ Does this apply to OP's question? Although $U$ is an open subset of $X$ and $U$ is contained in $\overline{X}$, $\overline{X}$ is only a subset of $X$ if $X$ is closed, so we can't conclude that $U$ is open in $\overline{X}$. $\endgroup$ – manthanomen Nov 20 '13 at 4:48
  • $\begingroup$ @manthanomen: Yes, it applies. The theorem in the original question is that if $U$ is open in $X$, then $U$ is open in $\operatorname{cl}_XU$; I’ve simply replaced $\operatorname{cl}_XU$ by an arbitrary subset of $X$ containing $U$. $\endgroup$ – Brian M. Scott Nov 20 '13 at 4:52
  • $\begingroup$ The statement's language is ambiguous, but from the proof attempt, I think the theorem refers to the closure of $X$ in $Y$, not the closure of $U$ in $X$. $\endgroup$ – manthanomen Nov 20 '13 at 5:01
  • $\begingroup$ @manthanomen: I think it likelier that the proof attempt is based on a misunderstanding of the theorem. $\endgroup$ – Brian M. Scott Nov 20 '13 at 5:03
  • $\begingroup$ Yes, that's likelier, I misunderstood and thought you were saying that OP's proof attempt was correct. $\endgroup$ – manthanomen Nov 20 '13 at 5:12
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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 $Y$.

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