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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 $G$.

Minimus Heximus
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    Every topological group is regular, meaning that every neighborhood of a point $p$ contains a closed neighborhood of $p$. Therefore, the intersection of all neighborhoods of a point is the same as the intersection of the closed neighborhoods, and so it is closed. (As a matter of fact, topological groups are completely regular.) – bof May 29 '14 at 10:37
  • This question is too elementary for Math Overflow. It would be more appropriate for Math Stack Exchange. – bof May 29 '14 at 10:41
  • @bof If it was not migrated i could edit my mistake. I meant a paratopological group. – Minimus Heximus May 29 '14 at 17:04
  • @bof: If you can give me an example of a paratopological group with that condition let me know. Otherwise I may need to ask a new question in mathoverflow. – Minimus Heximus May 29 '14 at 17:08

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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.

user43208
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  • My question is now adjusted to this answer! – Minimus Heximus May 30 '14 at 12:32
  • I even guess (and I'm almost sure) that $\bigcap_{V\in \mathcal N_1} V$ is closed iff the paratopological group $(G,\mathcal T)$ is topological. – Minimus Heximus May 30 '14 at 12:37
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    @user138171 Isn't the Sorgenfrey line a counterexample? (This is the additive group $\mathbb{R}$, with topology given by a base consisting of half-open intervals $[a, b)$ with $a < b$.) – user43208 May 30 '14 at 13:10
  • yep, I thought it follows from a theorem which say if $\bigcap_{V\in \mathcal N_1} V$ is closed and $\mathcal F$ is a filter on $G$ with $\mathcal F \to a$ then $$\mathcal F^{-1}\to a^{-1}$$ provided $\mathcal F ^{-1}$ is convergent. I forgot the last condition. – Minimus Heximus May 30 '14 at 13:32