Let $G$ be a topological group acting continuously on a topological space $X$. Let $A$ be a $G$ - invariant subspace of $X$. Then is it true that $\bar A$ is also $G$ - invariant? (where $\bar A$ is the closure of $A$ in $X$)
My attempt -
Let $g\in G$ and $x\in \bar A$. To prove that $g\cdot x \in \bar A$
Since $x\in \bar A$, there exists a net of points $\{a_\alpha\}\subseteq A$ such that $a_\alpha\longrightarrow x$ . Since $G$ acts continuously, $g\cdot a_\alpha\longrightarrow g\cdot x$ . Then since $A$ is $G$ - invariant, $\{g\cdot a_\alpha\}$ is a net of points in $A$ converging to $g\cdot x$ $\Longrightarrow g\cdot x\in \bar A$.
Can someone tell me if I have done this correctly?
Thank you.
If $f$ is a continuos function $f (\bar {A})\subseteq \bar {f (A)}$. If $L_g$ is the left moltiplication for g , set $f=L_g$.