Prove that if $A$ is dense in $X$ and $U$ is open then $U \subseteq \overline{A \cap U}$

Question

This is Exercise 4 from Section 2.2 of Topology and Groupoids by Brown.

Exercise:

Let $X$ be a topological space and let $A \subseteq B \subseteq X$. We say that $A$ is dense in $B$ if $B \subseteq \overline A$, and $A$ is dense if $\overline A = X$. Prove that if $A$ is dense in $X$ and $U$ is open then $U \subseteq \overline{A \cap U}$.

More information:

The definition of the overline/bar notation:

Let $X$ be a topological space and let $A \subseteq X$. The closure of $A$ is the set $\overline A$ of points $x$ in $X$ such that every neighborhood of $x$ intersects $A$.

My attempt:

Consider $x \in U$. We want to show that every neighborhood of $x$ intersects $A \cap U$. Clearly every neighborhood of $x$ intersects $U$, so we just have to show that every neighborhood of $x$ intersects $A$. Since $A$ is dense in $X$, we know that $X \subseteq \overline A$, which can only mean that $\overline A = X$. Since $x \in X$, we see that every neighborhood of $x$ does intersect $A$, which concludes the proof.

My concern:

I didn't make any use of the assumption that $U$ is open, which makes me wonder if I screwed up somewhere.

I appreciate any feedback.

Answer 1

Your attempt has a gap. You've shown separately that every neighborhood of $x$ intersects $U$ and also intersects $A$. This is not the same as saying that every neighborhood of $U$ intersect $A \cap U$. More precisely, you've shown for all open $V \ni x$, $V \cap U \neq \varnothing$ and $V \cap A \neq \varnothing$, but you need to conclude $V \cap (U \cap A) \neq \varnothing$.

To fix it, use the fact that $V \cap U$ is open.

Answer 2

Joshua P. Swanson pointed out the key error in my proof, which was the conflation of the intersection with $A$ and $U$ with the intersection with the intersection $A \cap U$. However, his answer used an open set, which is not quite general enough for my purpose because I need to use any neighborhood of $x$, and open sets containing $x$ are only a subset of the neighborhoods of $x$. Below I prove the assertion from scratch.

Consider $x \in U$. We want to show that for any neighborhood $N$ of $x$, $N \cap (A \cap U) \neq \emptyset$. Using basic set theory, this is equivalent to showing that $(N \cap U) \cap A \neq \emptyset$.

Because $U$ is open, it is a neighborhood of $x$. One of the neighborhood axioms states that the intersection of two neighborhoods of $x$ is also a neighborhood of $x$. That means $N \cap U$ is a neighborhood of $x$.

Since $A$ is dense in $X$, $X \subseteq \overline A$, which means that $\overline A = X$. Since $x \in X$, we see that any neighborhood of $x$ intersects $A$. Particularly, the neighborhood $N \cap U$ intersects $A$, which means that $(N \cap U) \cap A \neq \emptyset$, which concludes the proof.