I found a proof that any topological group is regular here, but I got lost in the last part. The whole argument goes like this:

Consider the map $f:G \times G \to G$ defined by $f(a,b)=ab^{-1}$. This map is always continuous in a topological group. Now take $x \in U$, where $U$ is an open set in $G$. Then $f^{-1}(U)$ contains $(x,e)$, so we have $(x,e)\in V \times W \subseteq f^{-1}(U)$ for some open subsets $V$ and $W$ such that $x\in V$ and $e\in W$.

Hence, $x\in V$. Furthermore, $V\cap (X-U)W=\emptyset$, since any element in the intersection corresponds to $a\in V$, $b\in W$, such that $ab^{-1}\notin U$, which is a contradiction.

Since $(X-U)W$ is an open set containing $X-U$, $X$ is regular.

The boldface argument is what I don't understand. How can I see that $V\cap (X-U)W=\emptyset$?


1 Answer 1


If $y\in V\cap (X - U)W$, then $y\in V$ and $y = st$ for some $s\notin U$ and $t\in W$. Then $(y,t)\in V\times W$, so $f(y,t) \in U$, i.e., $yt^{-1}\in U$. On the other hand, $s = yt^{-1}\in U$. This contradiction shows $V\cap (X - U)W = \emptyset$.


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