A paratopological group with intersection of neighborhoods of a point non-closed

Question

Do you have an example of a (para)topological group $(G,\mathcal T)$ such that $\bigcap_{V\in \mathcal N_1} V$ is not closed?

$\mathcal N _1$ is the set of all neighborbood of the identity element of $G$.

Answer 1

Here is an amplification of bof's remark:

Claim: Every neighborhood $U$ of the identity contains a closed neighborhood.

Proof: The continuity of the group operations ensures that we can pick a neighborhood $V$ of the identity such that $V = V^{-1}$ and $V \cdot V \subseteq U$. We show that $\bar{V} \subseteq U$. For, $x \in \bar{V}$ means that $x W \cap V \neq \emptyset$ for all neighborhoods $W$ of the identity. In particular, there is $y$ such that $y \in x V$ and $y \in V$, or $\exists v \in V\; y = x v$ and $y \in V$. Then $x \in y V^{-1} \subseteq V \cdot V^{-1} = V \cdot V \subseteq U$, as desired.

Edit: To answer the edited question, consider the real line $\mathbb{R}$ as an additive group equipped with the topology where the open sets are $\emptyset$, $\mathbb{R}$, and intervals of the form $(r, \infty)$. It is easy to see that this is a paratopological group. However, the largest open set not containing the identity is $(0, \infty)$, and therefore the closure of $\{0\}$ is $(-\infty, 0]$. The intersection of all opens containing the identity is $[0, \infty)$; obviously this isn't closed as it doesn't contain the closure of the identity.