While I was trying to solve an Exercize about the Vaught trasforms (Descriptive Set theory), I conjectured the following:
Conjecture: Let $G$ be a topological group (even Polish whenever necessary) and $H\subseteq G$ a subset. Then for every $g\in G$ $$(1) (H)^°g=(Hg)^°,$$ $$(2) \overline{H}g=\overline{Hg},$$ $$(3) (H)^cg=(Hg)^c,$$ where $(.)^°$ is the interior operator, $\overline{(.)}$ is the closure operator and $(.)^c$ is the complementation.
Question: is the above conjecture true?
Attempt: $(1)$ $(\subseteq)$ is obvious; for the converse $(\supseteq)$, notice that $$x\in (Hg)^°\iff \text{$\exists U(x)$ nhbd s.t. $U(x)\subseteq Hg$},$$ $$y\in (H)^°\iff \text{$\exists U(y)$ nhbd s.t. $U(y)\subseteq H$}$$ If $x\in (Hg)^°$, then $x=hg$ and $U(h)g$ is an open nhbd of $x$ contained in $Hg$ and hence $U(h)$ is an open nhbd of $h$ in $H$.
Is this reasoning right? If it is, then in a similar way $(2)$ can be shown.
However, $(3)$ is more difficult for me and I'm looking for a proof (or a counterexample, whenever $(3)$ does not hold).
Thank you in advance for your help.
(1) does hold, for the ($\supseteq$)direction, just notice $U(x)g^{-1}\subseteq H$ is an open nbhd of $h$ if $x=hg$.
(2) is similar.
(3) is false. Let $G=H=\Bbb Z$ and $g=2$. Then $(H)^cg=\emptyset$, but $(Hg)^c=\text{the set of all odd numbers}$.