Let $(G, .)$ be a group. Further assume that $G$ is endowed with some topology such that the map $G × G → G$ given by $(x, y) → xy$ and the map $G → G$ given by $x → x ^{−1}$ are continuous. In this case $(G, .)$ is said to be a topological group wrt the topology on it. Let $e$ be the identity element of $G$.
Let $H_0$ denote intersection of all neighbourhoods of identity. Prove that it is closure of ${e}$.
Attempt :
- $e \in H_0$
- If $x \in H_0=(\cap U_\alpha$ where $U_\alpha$ are neighbourhoods of identity), then $x \in U_\alpha, \forall \alpha$ , so $x \in \bar e$ , $\bar e =$ closure of identity ${e}$, so $H_0 \subset \bar e$.
- Take $y \in \bar e \setminus H_0$. Then $y \in \bar e$ but $\exists U_\beta$ such that $y \notin U_\beta$. I'm done if I contradict this isn't it? Now , since $y \in \bar e$, any nbhd of $y$ say $U_y \cap \{e\} \neq \emptyset$, so, $\{e\} \in U_y \cap U_\beta$. So $\exists$ $V \subset U_y \cap U_\beta$ open which is a nbhd of $e$.
I don't know if I am doing it correct and how to proceed.
Since you asked for a different approach with respect to the one given here, I'll try to give you an answer, but maybe it would be appreciated if you edit your question in order to recognize you already read the previous answer and it was not the answer you were looking for.
Now let, as you did, $\left\{U_\alpha\mid \alpha\in I\right\}$ be the family of neighborhoods of $e$. Our approach will be the following: first we will show that $\bigcap_{\alpha\in I}U_\alpha$ is closed and secondly we will show that every closed $C$ in $G$ which contains $e$ must contain $\bigcap_{\alpha\in I}U_\alpha$ as well.
For the first claim, if $x\notin \bigcap_{\alpha\in I}U_\alpha$ then there exists $\beta\in I$ such that $x\notin U_\beta$. Since $x\mapsto x^{-1}$ is an homeomorphism, $U_\beta^{-1}=\{h^{-1}\mid h\in U_\beta\}$ is still an open neighborhood of $e$ and so $xU_\beta^{-1}$ is an open neighborhood of $x$. We claim that $xU_\beta^{-1}\cap U_\beta=\emptyset$. Assume by contradiction that $y\in xU_\beta^{-1}\cap U_\beta$. Then there exists $z\in U_\beta$ such that $y=xz^{-1}$, which implies that $x=yz\in U_\beta$; a contradiction. Therefore $G\setminus \bigcap_{\alpha\in I}U_\alpha$ is open and hence $\bigcap_{\alpha\in I}U_\alpha$ is closed.
Let us prove the second claim now. Assume that $e\in C$ for some closed subset $C$ of $G$. Pick a generic $x\in\bigcap_{\alpha\in I}U_\alpha$. The family of open neighborhoods of $x$ is of the form $\{xU_\alpha\mid \alpha\in I\}$. Since $x\mapsto x^{-1}$ is an homeomorphism, we can also write this family as $\{xU_\alpha^{-1}\mid \alpha\in I\}$. For every $\alpha\in I$, we have that $e=x\cdot x^{-1}\in xU_\alpha^{-1}$. Thus for every open neighborhood $xU_\alpha^{-1}$ of $x$, $e\in C\cap xU_\alpha^{-1}$ and so $x\in\overline{C}=C$. In conclusion, $\bigcap_{\alpha\in I}U_\alpha\subseteq C$ as claimed.