If $U$ is a neighborhood in $D$ of $x$ and $D$ is dense, then the closure in $X$ of $U$ is a neighborhood of $x$ in $X$

Question

Let $(X, \tau )$ a topological space and $D$ a dense subset. Prove that if $x\in D$ and $U$ is a neighborhood of $x$ in $D$ (for the subspace topology) then the closure of $U$ in $X$ is a neighborhood of $x$ in $X$.

My attempt is the following:

I want to show that $\overline{U}$ contains an open neighborhood of x in X.

Let $U$ a neighborhood of $x$ in $D$. Then, exists $U'$ open in $D$ such that $U'\subseteq U$. By definition of subspace topology, exists $V$ an open set in $X$ such that $x\in V$ and $U'=V\cap D$.

I obtain that $V\cap D=U'\subseteq \overline{U'}^X=\overline{V\cap D}^X=\overline{V}^X\cap \overline{D}^X=\overline{V}^X\cap X=\overline{V}^X$.

So, $U'\subseteq U \Longrightarrow V\subseteq \overline{V}^X=\overline{U'}^X\subseteq \overline{U}^X$.

Thanks, I noticed my mistake. Do you have any hints to help me? — Sigma Algebra, Dec 08 '23 at 22:14
Yes, in some ways. But, is it true that $\overline{U}=\overline{U\cap A}$ under these hypothesis? — Sigma Algebra, Dec 08 '23 at 22:41
Yes it is, see my last comment to the answer below. So, the linked post completely answers you question, doesn't it? — Anne Bauval, Dec 09 '23 at 11:26

Anne Bauval · Accepted Answer · 2023-12-09T11:41:41.630

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Strictly speaking, your proof is correct. But your $\overline{V\cap D}^X=\overline{V}^X\cap \overline{D}^X$, or equivalently (given that $\overline{D}^X=X$) $$V\subseteq \overline{V\cap D}^X,$$ needs some argument, since $\overline{A\cap B}\subseteq\bar A\cap\bar B$ but the reverse inclusion does not hold in general.
Let us check that $V\subseteq \overline{V\cap D}^X$, i.e. that for any $v\in V$, every open neighborhood $O$ of $v$ (in $X$) meets $V\cap D$. For this, just notice that $$O\cap(V\cap D)=(O\cap V)\cap D\ne\varnothing,$$ since $O\cap V$ is a nonempty open subset of $X$ and $D$ is dense in $X$.
Note moreover that this "hard part" was sufficient. You can lighten the end of your attempt by just writing: $$U'\subseteq U\implies V\subseteq\overline{V\cap D}=\overline{U'}\subseteq\overline U,$$ and suppressing your previous line of formulas.

edited Dec 09 '23 at 11:41

answered Dec 08 '23 at 22:26

Anne Bauval

34,650

So, is it always true that if $D$ is dense and $V$ is open, $\overline{V\cap D}=\overline{V}$? – Sigma Algebra Dec 08 '23 at 22:36
Definitely: the hard part was $V\subseteq\overline{V\cap D}$, which is equivalent to $\bar V\subseteq\overline{V\cap D}$, and $\bar V\supseteq\overline{V\cap D}$ is obvious – Anne Bauval Dec 09 '23 at 11:24
Ok, so it's enough to prove that $V\subseteq \overline{V\cap D}$ because $\overline{V}$ is the smallest close set that contains $V$ and so $\overline{V}$ is necessarily contained in $\overline{V\cap D}$? – Sigma Algebra Dec 09 '23 at 17:10
Indeed! $A\subseteq\bar B\implies\bar A\subseteq\overline{\bar B}=\bar B$. – Anne Bauval Dec 09 '23 at 17:30
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Thank you very much!! – Sigma Algebra Dec 09 '23 at 20:10

If $U$ is a neighborhood in $D$ of $x$ and $D$ is dense, then the closure in $X$ of $U$ is a neighborhood of $x$ in $X$

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