Neighborhood base at the unit element in a topological group

Question

One can define the open set $U$ be such that $U$ is open if for every $x\in U$, $xV\subset U$ for some $V\in\mathcal{N}$. (2) and (3) would give the desired continuity for group multiplication and inverse operation. The following question makes me stuck:

Why necessarily the group unit $1_G\in U$ for every $U\in\mathcal{N}$?

Answer 1

It also follows from (2) and (3), … with (1):

• Let $U ∈ \mathcal N$. By (2) there is some $V ∈ \mathcal N$ with $VV ⊂ U$.
• Since $V ∈ \mathcal N$, by (3) there is some $W ∈ \mathcal N$ with $W^{-1} ⊂ V$.
• Now (1) gives you some $X ∈ \mathcal N$ with $X ⊂ W ∩ V$.

This $X$ is what you want: $$1_G ∈ X^{-1}X ⊂ W^{-1}V ⊂ VV ⊂ U.$$