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I want to prove that

($\ast$) If $G$ is a Hausdorff topological group, then for every neighbourhood $U$ of $e$, there exists a symmetric neighbourhood $V$ of $e$ ($V^{-1}=V$), s.t $V*V \subset U$

I had seen a lot related proofs regarding preimage of the map $(x,y)\longmapsto x^{-1}y$ for $G \times G \longrightarrow G$ or $U\times U \longrightarrow G$ but I can't understand them.

For example:

  1. http://www.math.wm.edu/~vinroot/PadicGroups/topgroups.pdf (Prop 1.1 on p2)

  2. In topological groups. Is every neighborhood of $e$ supset of a square of a symmetric neighborhood of $e$?

  3. Is a discrete subgroup of a Hausdorff group closed?

My problem is they all take the premimage $\overline U$ of $U$ for $(x,y)\longmapsto x^{-1}y$, but how then the open set $W$ is obtained so that $xy^{-1}\in{U}\forall x,y\in W$?

I mean $\overline U$ is of the form $\bigcup (x,y)\subset G\times G$, and taking projections $\pi_1(\overline U),\pi_2(\overline U)$ doesn't work, so I am stuck.

So how to proof ($\ast$)? My goal is to prove it, hence every other methods without using preimage are also welcome.

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CYC
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2 Answers2

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Let $\mu$ denote the multiplication. By continuity, $\mu^{-1}(U)$ is a neighbourhood of $(e,e)$, that means there are [open, if desired] neighbourhoods $W_1,W_2$ of $e$ with $W_1\times W_2 \subset \mu^{-1}(U)$. Now let $W := W_1 \cap W_2$. As an intersection of two neighbourhoods of $e$, it is again a neighbourhood of $e$. Next, let $V := W \cap W^{-1}$. Then $V$ is a symmetric [open, if $W_1,W_2$ are open] neighbourhood of $e$ with $\mu(V\times V) \subset \mu(W_1\times W_2) \subset U$.

Daniel Fischer
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  • How to obtain $W_1$ and $W_2$? – CYC Aug 14 '15 at 13:14
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    By definition of the product topology. The product topology has a basis consisting of "open rectangles", that is, sets of the form $A\times B$ with $A,B$ open. So every open set $O$ containing some point $(a,b)$ contains an open rectangle with $(a,b) \in A\times B \subset O$. – Daniel Fischer Aug 14 '15 at 13:16
  • Why is $W^{-1}$ open? Or is there some other reason why $V$ is open? – SihOASHoihd Dec 24 '18 at 02:58
  • @SihOASHoihd It's only open if $W$ is open. It is in any case a neighbourhood of $e$, because the inversion $x\mapsto x^{-1}$ is continuous. As an involution, it is therefore a homeomorphism. And $V$ also isn't necessarily open, but it certainly is a neighbourhood of $e$. If we want an open $V$, we pick $W_1, W_2$ at the start as open neighbourhoods of $e$, then also $W, W^{-1}$ and consequently $V$ are open. – Daniel Fischer Oct 02 '19 at 23:41
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  1. If $U$ is an open neighborhood of $e$, by the continuity of the multiplication, there are two open neighborhoods $V_1, V_2$ of $e$ such that $V_1\cdot V_2\subseteq U$. Let $V := V_1\cap V_2$, then $V\cdot V\subseteq V_1\cdot V_2\subseteq U$.

  2. Similarly, if $U$ is an open neighborhood of $e$, then there is an open neighborhood $V$ of $e$ such that $V\cdot V^{-1} \subseteq U$.

  3. We also observe that if $V$ is an open neighborhood of $e$, then $W := V\cdot V^{-1}$ is a symmetric open neighborhood of $e$. (This is a hint from James Munkres's book, page 146)

Back to the main problem.

By 1, there is an open neighborhood $W$ of $e$ such that $W\cdot W \subseteq U$.

By 2, there is an open neighborhood $Z$ of $e$ such that $Z\cdot Z^{-1} \subseteq W$.

Let $V := Z\cdot Z^{-1}$, then by 3, $V$ is a symmetric open neighborhood of $e$. We have $$V\cdot V \subseteq W\cdot W \subseteq U$$

HBK
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