Defn: $\textbf{A}$ is a dense subset of $\mathbb{R^n}$ if $\mathbb{R^n} \subset \overline{A}$
where $\overline{A}= A$ closure
Defn: The closure of $A$, is the set $\overline{A}$ consisting of all limit points of $A$.
1)Show that if $\textbf{U}$ is open in $\mathbb{R^n}$, then $A\cap U$ is dense in $\textbf{U}$.
Attempt:
1) We are given that $\mathbb{R^n} \subset \overline{A}$ and $\textbf{U}$ is open in $\mathbb{R^n}$.
This means that $U \subset \mathbb{R^n} \subset \overline{A}$. Therefore $\overline{A}$ contains all of the limit points of $\textbf{U}$. So if we take the intersection of $\overline{A}$ and $U$, we will obtain all of the limit points of $\overline{A\cup U}$ which is what we were trying to show.
Is this the right thought process?
Note: I feel I maybe should've used the sequences of the limit points in some fashion, but I could not deduce any set of steps to lead me to a conclusion.