Suppose that $G \subset A\subset \mathbb{R}^n$ and $A$ is closed. How would I show that $\bar{G}\subset A?$ ($\bar{G}$ means the closure of G.)
My first guess was to use the proposition the that the closure of a set is closed (which I proved). So, $G\subset A \implies \bar{G}\subset \bar{A} \implies \bar{G}\subset A.$ But this proof seems incomplete.
Any ideas?
Use the facts that for any subsets of metric or topological spaces, $X \subseteq Y \implies \overline{X} \subseteq \overline{Y}$ and $A = \overline{A}$ if $A$ is closed. You should prove these if you haven't.
You could also approach the problem more directly and take a limit point $x$ of $G$, show that it must be a limit point of $A$, and thus since $A$ is closed, $x \in A$ which then would imply $\overline{G} \subseteq A$.