5
$\begingroup$

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 $V\cap (X-U)W=\emptyset$?

$\endgroup$
3
$\begingroup$

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 $V\cap (X - U)W = \emptyset$.

| cite | improve this answer | |
$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.