Show that if $(X,*,\cal T)$ is a t. g. then $\text{cl}Y_1*\text{cl}Y_2=\text{cl}(Y_1*Y_2)$ and $(\text{cl}Y)^{-1}=\text{cl}Y^{-1}$ where $Y_1,Y_2,Y⊆X$

Question

Definition

A topological group is a group $(X,*)$ equipped with a topology $\cal T$ with resepct the functions $$ p:X\times X\ni (x_1,x_2)\longrightarrow x_1*x_2\in X\quad\text{and}\quad s:X\ni x\longrightarrow x^{-1}\in X $$ are continuous.

Now for any $Y_1,Y_2,Y\in\mathcal P(X)$ we put $$ Y_1*Y_2:=\{x\in X:x=y_1*y_2\text{ }\text{where }y_i\in Y_i\text{ for }i=1,2\}\quad\text{and}\quad Y^{-1}:=\{x\in X:x=y^{-1}\text{ for }y\in Y\} $$ so that we let to prove that the identities $$ \operatorname{cl}(Y_1*Y_2)=\operatorname{cl}Y_1*\operatorname{cl}Y_2\quad\text{and}\quad\operatorname{cl}Y^{-1}=(\operatorname{cl}Y)^{-1} $$ holds for any $Y_1,Y_2,Y\in\mathcal P(X)$. So if $A_{x_1*x_2}$ is an open neighborhood of $x_1*x_2\in\operatorname{cl}Y_1*\operatorname{cl}Y_2$ then $p^{-1}[A_{x_1*x_2}]$ is an open neighborhood of $(x_1,x_2)$ in $X\times X$ but it is a well know result that $$ \operatorname{cl}(Y_1\times Y_2)=\operatorname{cl}Y_1\times\operatorname{cl}Y_2 $$ so that if $(x_1,x_2)\in\operatorname{cl}Y_1\times\operatorname{cl}Y_2$ then $$ p^{-1}[A_{x_1*x_2}]\cap (Y_1\times Y_2)\neq\emptyset $$ and so by the inclusion $$ p\Big[p^{-1}[A_{x_1*x_2}]\cap(Y_1\times Y_2)\Big]\subseteq p\big[p^{-1}[A_{x_1*x_2}]\big]\cap p[Y_1\times Y_2]\subseteq A_{x_1*x_2}\cap(Y_1*Y_2) $$ we conclude that $$ A_{x_1*x_2}\cap(Y_1*Y_2)\neq\emptyset $$ which implies $x_1*x_2\in\operatorname{cl}(Y_1*Y_2)$ and so finally $$ \operatorname{cl}Y_1*\operatorname{cl}Y_2\subseteq\operatorname{cl}(Y_1*Y_2) $$ Another possible way to infer the last inclusion is the follow: so by continuity of $p$ we observe that the inclusion $$ Y_1\times Y_2\subseteq p^{-1}[Y_1*Y_2]\subseteq\operatorname{cl}p^{-1}[Y_1*Y_2]\subseteq p^{-1}\big[\operatorname{cl}(Y_1*Y_2)\big] $$ holds but by continuity of $p$ the set $p^{-1}\big[\operatorname{cl}(Y_1*Y_2)\big]$ is closed so that effectively the inclusion $$ \operatorname{cl}Y_1\times\operatorname{cl}Y_2=\operatorname{cl}(Y_1\times Y_2)\subseteq p^{-1}\big[\operatorname{cl}(Y_1*Y_2)\big] $$ holds and thus finally we conclude that $$ \operatorname{cl}Y_1*\operatorname{cl}Y_2=p[\operatorname{cl}Y_1\times\operatorname{cl}Y_2]\subseteq p\Big[p^{-1}\big[\operatorname{cl}(Y_1*Y_2)\big]\Big]\subseteq\operatorname{cl}(Y_1*Y_2) $$ as we desidred.

So as you can see I am not able to prove that the inclusion $$ \operatorname{cl}(Y_1*Y_2)\subseteq\operatorname{cl}Y_1*\operatorname{cl}Y_2 $$ holds so that I ask to prove it. Moreover is the inclusion $$ \operatorname{cl}Y_1*\operatorname{cl}Y_2\subseteq\operatorname{cl}(Y_1*Y_2) $$ well proved? Finally how prove that the identity $$ \operatorname{cl}Y^{-1}=(\operatorname{cl}Y)^{-1} $$ holds? So could someone help me, please?

Answer 1

Both proofs of $\newcommand{\cl}{\operatorname{cl}}$ $$ \cl (Y_1) * \cl (Y_2) \subseteq \cl(Y_1 * Y_2) \tag1 $$ are correct. Here is another one:

Exercise: If $f \colon T \to S$ is continuous map between topological spaces, then for any $A \subseteq T$ one has that $$ f[\cl(A)] \subseteq \cl(f[A]). \tag2 $$ (In fact, the converse also holds.)

Thus, $(1)$ follows from this with $f = p$ and $A = Y_1 \times Y_2$.

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And, unless I missing something obvious, the other inclusion does not holds in general:

It is clear that $(\Bbb R,+)$ is a topological group with the usual topology, that $\Bbb Z$ and $\sqrt2\Bbb Z$ are closed, and $\Bbb Z+\sqrt2\Bbb Z$ is dense. Hence, $$ \cl(\Bbb Z+\sqrt2\Bbb Z) = \Bbb R \neq \Bbb Z+\sqrt2\Bbb Z = \cl(\Bbb Z)+\cl(\sqrt2\Bbb Z). $$

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Finally, since $s$ is its own inverse, it is a homeomorphism; so, the above exercise tells us that $$ s[\cl(Y)] = \cl(s[Y]). $$ (If $f \colon S \to T$ is a homeomorphism, then by the exercise we have that $f^{-1}[\cl(f[A])] \subseteq \cl(f^{-1}[f[A]]) = \cl(A)$, and applying $f[\_]$ we obtain the equality in $(2)$.)