Show open set $U ⊆ \overline{U ∩ A}$ where $A$ is dense in topological space $X$

Question

Let $(X, \tau)$ be a topological space. Let $U$ be an open set in $X$ and $A$ be a set in $X$ so that $\overline A = X$.

I want to show that $U ⊆ \overline{U ∩ A}$ (where the overline indicates closure).

Hmm so let $x\in U$. I could not prove $x\in \overline{U ∩ A}$ directly, so tried supposing for a contradiction that $x\notin \overline{U ∩ A}$. Then $x\in X\setminus \overline{U ∩ A}$ which is open. But I am not sure if this is the right rack~

Answer 1

Let $x \in U$. To show that $x \in \overline{U \cap A}$, we need to show that every open neighbourhood $V$ of $x$ intersects $U \cap A$: this is easy as $U \cap V$ is also an open neighbourhood of $x$ and so intersects $A$ (as $\overline{A}=X$) and so $$\emptyset \neq(U \cap V) \cap A = V \cap (U \cap A)$$

QED.

As a consequence $\overline{U} = \overline{A \cap U}$ (the left to right inclusion we now finish by taking the closure on both sides of the above; the right to left inclusion is obvious from $U \cap A \subseteq U$).

Answer 2

Following your approach note that $U\cap (X\setminus \overline {(U\cap A)})$ is an open set containing $x$. Hence it must intersect the dense set $A$. But its intersection with $A$ is empty because $X\setminus \overline {(U\cap A)}$ is contained in $X\setminus {U\cap A}$