Suppose $(G,\mathcal T)$ is a paratopological group and $a,b\in G$ and every neighborhood of $a$ contains $b$. Can we say every neighborhood of $b$ contains $a$?
clearly every closed neighborhood of $b$ contains $a$. So I wonder if any open neighborhood of $1$ contains a closed neighborhood of $1$.
If the answer to the above question is negative I add another assumption:
How if we have a neighborhood $U$ of $1$, with $\overline{U^{-1}}$ compact?
How if we have a neighborhood $U$ of $1$, with $\overline{U^{-1}}$ compact?
It seems the following.
The answer is positive, because in this case $G$ is a (locally compact) topological group. So it has a base at the unit consisting of symmetric $(U=U^{-1})$ neighborhoods.
Now we prove the first claim. A family $\{xU:x\in \overline{U^{-1}}\}$ is an open cover of the set $\overline{U^{-1}}$. Since the set $\overline{U^{-1}}$ is compact, this cover has a finite subcover. That is there exists a finite subset $F$ of the set $\overline{U^{-1}}$ such that $\overline{U^{-1}}\subset FU$. Then $U\subset U^{-1}F^{-1}$. Since a non-empty open set is not a finite union of nowhere dense sets, a set $U^{-1}$ has a non-empty interior. Since it has a compact closure, $G$ is a locally compact paratopological group. So it is a topological group by my old result (see Proposition below).
clearly every closed neighborhood of $b$ contains $a$. So I wonder if any open neighborhood of $1$ contains a closed neighborhood of $1$.
This is another question, which has a negative answer. Consider the following counterexample. Define on the group $\Bbb Z$ of integers a topology $\mathcal T$ as follows. Let $p>1$ be a natural number. For each positive integer $n$ put $U_n=p^n(\Bbb N\cap\{0\})$. Put a family $\{m+U_n: m\in\Bbb Z, m\in\Bbb N\}$ as a base of the topology $\mathcal T$. It is easy to check that $(\Bbb Z,\mathcal T)$ is a Hausdorff paratopological group, but $\overline{U_n}=p^n\Bbb Z\not \subset U_1$ for each $n$.
To prove our main result we shall need four lemmas.
Lemma 1. Let $G$ be a paratopological group, $K\subset G$ be a compact, $C\subset G$ be a closed set and $K\cap C=\varnothing$. Then there exists a neighborhood $U$ of the unit such that $UK\cap C=\varnothing$.
Proof. For every point $x\in K$ there exists a neighborhood $V(x)$ of the unit such that $V(x)x\cap C=\varnothing$. Let $U(x)$ be a neighborhood of the unit such that $U(x)^2\subset V(x)$. There exists a finite set $F$ such that $K\subset\bigcup\{U(x)x:x\in F\}$. Put $U=\bigcap\{U(x):x\in F\}$. Then $$UK\cap C\subset U\bigcup\{U(x)x:x\in F\}\cap C \subset \bigcup\{U(x)^2x:x\in F\}\cap C=\varnothing.$$ $\square$
Let $H$ be a subgroup of a paratopological group $(G,\tau)$. Define a topology $\tau/H$ on the space of left cosets $G/H$ of the group $G$ in the following way. A set $U$ is open in $G/H$ if and only if $\pi^{-1}(U)$ is open in $G$, where $\pi:G\to G/H$ is the quotient map, $\pi(x)=xH$.
Lemma 2. If $H$ is a compact subgroup of a paratopological group $G$ then the quotient map $\pi:G\to G/H$ is closed.
Proof. Let $F$ be a closed subset of the group $G$. Let $\tilde x\in G/H\setminus \pi(F)$. Consider an arbitrary point $x\in\pi^{-1}(\tilde x)$. Then $xH\cap F=\varnothing$. By Lemma 1 there exists an open neighborhood $U$ of the unit such that $UxH\cap F=\varnothing$. Then $\tilde x\in\pi(Ux)$ and $\pi(Ux)\cap\pi(F)=\varnothing$. Thus the map $\pi$ is closed. $\square$
Lemma 3. Let $G$ be a paratopological group and $H$ be a normal subgroup of the group $G$. If $H$ and $G/H$ are topological groups then $G$ is a topological group.
Proof. Let $U$ be an arbitrary neighborhood of the unit. There exist neighborhoods $V,W$ of the unit such that $V\subset U$, $(V^{-1})^2\cap H\subset U$ and $W\subset V$, $W^{-1}\subset VH$. If $x\in W^{-1}$ then there exist elements $v\in V,h\in H$ such that $x=vh$. Then $h=v^{-1}x\in V^{-1}W^{-1}\cap H\subset U$. Therefore $x\in VU\subset U^2$. Hence $G$ is a topological group.$\square$
Lemma 4. A compact paratopological group is a topological group.
Proof. Let $G$ be a compact paratopological group. At first we consider the case when $G$ is $T_1$. Let $C$ be a closed subset of the group $G$. Then $C^{-1}=\{x\in G:xC\ni e\}$. Let $x\not\in C^{-1}$. Then for every point $y\in C$ there exist open neighborhoods $Ox(y)\ni x$, $Oy\ni y$ such that $Ox(y)Oy\not\ni e$. Choose a finite subcover $\{Oy:y\in Y\}$ of the set $C$ and put $Ox=\bigcap\{Ox(y):y\in Y\}$. Then $OxC\not\ni e$ and therefore the set $C^{-1}$ is closed. Therefore the map $i:G\to G$, $i:x\mapsto x^{-1}$ is continuous.
Now consider the general case. Let $\mathcal B$ be a base at the unit $e$ of the group $G$. Put $B=\bigcap\mathcal B$. Since the multiplication of the group $G$ is continuous at the unit, for each neighborhood $U\in\mathcal B$ there exists a neighborhood $V\in\mathcal B$ such that $VV\subset U$. This implies that $B$ is a semigroup. Put $B'=\overline{\{e\}}=\{x\in G:\forall U\in\mathcal B$ $xU\ni e\}=\{x\in G:xB\ni e\}= B^{-1}$. Hence $B'$ is a closed semigroup of the group $G$. Since $B'$ is a compact semigroup then there exists a minimal closed right ideal $H\subset B'$. Choose an arbitrary element $x\in H$. Then $x^2H=H\ni x$ hence $x^{-1}\in H$ and $e\in H$. Thus $H=B'$ and $xB'=B'$ for every element $x\in B'$. Therefore $B'$ is a group and $B'=B$. Since $B=\bigcap\mathcal B$ then $g^{-1}Bg\subset B$ for every $g\in G$. Hence $B$ is a normal subgroup of the group $G$. The group $B$ is topological, because $B$ is endowed with the trivial topology. Since $G/B$ is $T_1$ compact topological group then Lemma 3 implies that $G$ is a topological group. $\square$
Proposition. A locally compact paratopological group is a topological group.
Proof. Let $G$ be a locally compact paratopological group. Let $\mathcal B$ be a base at the unit of the group $G$. Put $B=\bigcap\{\overline U:U\in\mathcal B\}$. Then $B$ is compact closed subset of $G$. Since the multiplication of the group $G$ is continuous at the unit, for each neighborhood $U\in\mathcal B$ there exists a neighborhood $V\in\mathcal B$ such that $VV\subset U$. This implies that $B$ is a semigroup. Also for each neighborhood $U\in\mathcal B$ and each element $g\in G$ there exists a neighborhood $V\in\mathcal B$ such that $VV\subset U$. This implies that $g^{-1}Bg\subset B$ for every $g\in G$. Since $B$ is a compact semigroup then there exists a minimal closed right ideal $H\subset B$. Choose an arbitrary element $x\in H$. Then $x^2H=H\ni x$ hence $x^{-1}\in H$ and $e\in H$. Thus $H=B$ and $B$ is a normal subgroup of the group $G$. Lemma 4 implies that $B$ is a topological group. Let $\pi:G\to G/B$ be the quotient map. Since $B$ is compact then, by Lemma 2, $\pi$ is a closed map and therefore $G/B$ is locally compact. Let $x,y\in G$ and $xB\not=yB$. Then $x\bigcap\{\overline U:U\in\mathcal B\}\cap y\bigcap\{\overline U:U\in\mathcal B\}=\varnothing$ and therefore there exists a neighborhood $U\in\mathcal B$ such that $x\overline {U^2}\cap y\overline {U^2}=\varnothing$. Then $xUB\cap yUB=\varnothing$ hence $\pi(xU)\cap\pi(yU)=\varnothing$ and thus $G/B$ is a Hausdorff locally compact topological group. Hence by Ellis Theorem [Ell] $G/B$ is a topological group. Lemma 3 implies that $G$ is a topological group. $\square$
References
[Ell] Robert Ellis, Locally compact transformation groups, Duke Math J. 27, (1957), 119-125.
The answer is negative. Let $\tau=\{[a, +\infty): a\in \Bbb R\}$. Then $(\Bbb R, \tau)$ is a paratopological group under addition. However, $0$ and $1$ witness the fact.
