Prove: $\overline{U}=\overline{A\cap U}$

Question

Let $X$ be a topology space, $U\subseteq X$ is open set in $X$ and $A\subseteq X$ is dense in $X$

Prove: $\overline{U}=\overline{A\cap U}$

$\subseteq:$

Let $u\in U\subset \overline{U}$ be a non empty open set now $\overline{A}=X$ so $u\in\overline{A}\cap \overline{U}=\overline{A\cap U}$

It is correct? how should I attack $\supseteq$?

Answer 1

Since $A\cap U\subseteq U$, you can immediately derive that $\overline{A\cap U}\subseteq\overline{U}$, because it's a general fact that doesn't depend on $A$ being dense or $U$ being open.

The hypotheses are necessary to go the other way around (which is the part you attempted, but with no success, I'm afraid).

Let $x\in\overline{U}$. We need to show that $x\in\overline{A\cap U}$. Let $V$ be an open neighborhood of $x$. Then $U\cap V$ is an open set so, by density, it contains a point $a\in A$. Since $a\in U\cap V$, we have $a\in A\cap U$. Thus $a\in V\cap(A\cap U)$.

Answer 2

If $u \in \overline {U}$ and $V$ is any open set containing $u$ then $U \cap V$ is an open set containing $u$. Since $A$ is dense there some point $x$ in $A \cap (U \cap V)$. Thus every neighborhood of $u$ intersects $A \cap U$ which implies $u \in \overline {A\cap U}$.

Answer 3

$\textbf{Definition}$. We say that a set $A\subset X$ is dense if $\overline{A}=X$.

We're going to prove that $A$ is dense iff each non-empty open set $U\subset X$ contains at least one point of $A$.

$\boxed{\Rightarrow}$ Let $U\subset X$ be an open set, that is to say, is a neighbourhood of each of its points $x\in U\subset X=\overline{A}$, because of $A$ is dense. Furthermore, $x\in \overline{A}$, so each neighbourhood of $x$ meets $A$ and then $U\cap A \neq \emptyset$.

$\boxed{\Leftarrow}$ Let $x\in X$ an arbitrary point and take a neighbourhood $V_x$ of $x$. By definition, there is an open set $U_x$ such that $x\in U_x \subset V_x$. As $U_x$ is open, contains by hypothesis at least one point of $A$, so $V_x\cap A\neq \emptyset$. However, that's the definition of an adherent point.

Now, we're going to prove the question. Since $A\cap U\subseteq U$, it's clear that $\overline{A\cap U}\subseteq\overline{U}$. On the other hand, let $x\in\overline{U}$ and $V_x$ be an open neighborhood of $x$. Then, $U\cap V_x$ is an open set because is the intersection of two open sets. As we proved before, it contains a point $a\in A$, that is to say, $a\in U\cap V_x$. Thus, $$a\in (U\cap V)\cap A= (U\cap A)\cap V_x \Rightarrow (U\cap A)\cap V_x \neq \emptyset$$ and that's the definition of the adherent points of $A\cap U$.

Answer 4

First prove the very useful lemma, if U is open, then U $\cap$$ \overline A$ $\subseteq$ $\overline {U \cap A}.$

For dense A, open U, U = U $\cap \overline A$ $\subseteq$ $\overline {U \cap A}$ $\subseteq$ $\overline U.$

$\overline U$ = $\overline {U \cap A}$ follows by taking the closure of the above sets.