Filter of all neighborhoods of an element

Question

Recently, I was reading a booklet about topological groups and I had difficulty with the definition of filter of all neighborhoods of an element.

How I do proof that, given a topological group $G$ and $g\in G$ the set $V(g)$ of all neighborhoods of $g$ is a filter?

In fact,

1. $V(g)\neq\emptyset$ because $G\in\tau_G$ and therefore is open.
2. If $V_1,V_2\in V(g)$ then $V_1\cap V_2\in V(g)$ because $V_1$ and $V_2$ are open.
3. $\emptyset\notin V(g)$?
4. If $V_1\in V(g)$ and $V_1\subset V_2$ then $V_2\in V(g)$?

Answer 1

Topological groups aren't special in this regard. If you have a topological space $(X,\tau)$, then the set $\mathcal{V}(x)$ of neighborhoods of $x\in X$ is a filter.

Note that a neighborhood is not necessarily open: a subset $V\subseteq X$ is a neighborhood of $x$ if (and only if) $V$ cointains an open set $V'$ such that $x\in V'$.

Now $\mathcal{V}(x)$ is a filter because:

1. $X\in \mathcal{V}(x)$, so $\mathcal{V}(x)\ne\emptyset$
2. for every $V\in\mathcal{V}(x)$, $x\in V$, so $\emptyset\notin\mathcal{V}(x)$
3. if $V_1,V_2\in\mathcal{V}(x)$, then $V_1\supseteq V_1'$ and $V_2\supseteq V_2'$ where $V_1'$ and $V_2'$ are open sets containing $x$; then $V_1\cap V_2\supseteq V_1'\cap V_2'$ and $V_1'\cap V_2'$ is an open set containing $x$; thus $V_1\cap V_2\in\mathcal{V}(x)$
4. if $V\in\mathcal{V}(x)$ and $U\supseteq V$, then $U\supseteq V'$ where $V'\subseteq V$ is an open set containing $x$, so $U\in\mathcal{V}(x)$

What's special in topological groups is that if $g\in G$, where $G$ is a topological group, then $$ \mathcal{V}(g)=\{gV:V\in\mathcal{V}(e)\}=\{Vg:V\in\mathcal{V}(e)\} $$ where $e$ is the neutral element. This follows from the fact that the maps $\lambda_g\colon G\to G$, $\lambda_g(x)=gx$ and $\rho_g\colon G\to G$, $\rho_g(x)=xg$ are homeomorphisms.