2

I was reading the book Introduction to the topological groups by Taqdir Husain, he gives a proof for this theorem, but I had a problem understanding some part. The proof goes like this

It remains to show that the inversion mapping $x\to x^{-1}$ is continuous. So let $U$ be a open neighborhood of $e$. We wish to show that there exists a compact neighborhood $C$ of $e$ such that $C^{-1}\subseteq U$. Suppose this is not possible, i.e. $C^{-1}\setminus U\neq\varnothing$ for all compact neighborhood $C$ of $e$. Now define the following family $$ \mathscr{F}=\{C^{-1}\setminus U:C\text{ is a compact neighborhood of }e\} $$ In the last Lemma proves that $C^{-1}$ is compact for $C$ compact. So the family $\mathscr{F}$ is a family of compact sets. Then claims that $\mathscr{F}$ has the finite intersection property.

This last thing it's not clear to me, because I don't see why for example two elements of $\mathscr{F}$ say $A^{-1}\setminus U$ and $B^{-1}\setminus U$ must have elements in common.

The book actually says that the proof is due to Robert Ellis on A note on the continuity of the inverse. And he says that before his proof this was an open problem. I found no more proofs of this theorem, so It would be really helpful if somebody could help me.

By the way, there is another similar question in the forum but it requires the compactness Prove a compact Hausdorff space with a group structure is a topological group

Gerardo Gutierrez
  • 33
  • 4
  • Why do you think it is incorrect? – Randall Mar 25 '24 at 16:17
  • Welcome to the forum. When you ask a question you should always write what you understand about the problem and what you tried. For example, write what part of the proof you don't understand and why do you think it is wrong. Otherwise the question might get downvoted, or even closed. People are not supposed to read the whole proof and guess what part you think is wrong. – Mark Mar 25 '24 at 16:18
  • 1
    Also, please don't use images. Use MathJax when you write here. A tutorial: https://math.meta.stackexchange.com/questions/5020/mathjax-basic-tutorial-and-quick-reference – Mark Mar 25 '24 at 16:19
  • Instead of completely removing the proof you are doubting, you should edit your post to report it (mainly the part of it you don't understand). – Anne Bauval Mar 25 '24 at 22:01
  • Well, the finite intersection property is simply false in this setting. To get an example, consider an infinite group with discrete topology and $U={1}$. Are you sure you copied the proof correctly? – Moishe Kohan Mar 26 '24 at 16:23
  • Yes, I copied correctly, there Is a photo of the proof in the comment above yours – Gerardo Gutierrez Mar 26 '24 at 16:28
  • Oh, I see....... – Moishe Kohan Mar 26 '24 at 16:58

1 Answers1

1

Here is a proof that pairwise intersections are nonempty. The finite intersection property in general follows by induction.

Recall that the standing assumption is that for every compact neighborhood $C$ of $\{1\}$, $C^{-1}\setminus U$ is nonempty. Consider two compact neighborhoods $A, B$ of $1$. Take $C=A\cap B$. This is again a compact neighborhood of $1$. Then $$ (A^{-1}\setminus U)\cap (B^{-1}\setminus U)= (A^{-1} \cap B^{-1}) \setminus U = (A\cap B)^{-1} \setminus U= C^{-1}\setminus U\ne \emptyset, $$ according to the standing assumption. qed

Consider also reading the proof that every locally compact Hausdorff paratopological group is a topological group given by Alex Ravsky here, in the unnumbered Proposition in his answer.

Moishe Kohan
  • 97,719