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I understand that a topological group is a group $G$ endowed with a topology $\tau$ on $G$ such that addition and inverse are continuous on $\tau$.

Now, the definition of continuity is that for all $U\in\tau$, $f^{-1}(U)\in\tau$. But in this case $+^{-1}(U)$ is a subset of $G\times G$ because the domain of $+$ is $G\times G$, and we endowed $G$ with a topology, not $G\times G$. So, where am I messing up?

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    $\begingroup$ $G\times G$ has the product topology $\endgroup$ Feb 14, 2019 at 17:55
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    $\begingroup$ I love this question. $\endgroup$ Feb 14, 2019 at 18:11
  • $\begingroup$ @BenBlum-Smith why? $\endgroup$ Feb 15, 2019 at 3:47
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    $\begingroup$ @AlexKruckman - I recognize an earnest desire to understand. $\endgroup$ Feb 16, 2019 at 4:55
  • $\begingroup$ @BenBlum-Smith That's a good reason! $\endgroup$ Feb 16, 2019 at 5:22

1 Answer 1

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The topology of $G\times G$ is the product topology, which is the topology of the unions of sets of the form $A\times B$, with $A,B\in\tau$.

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    $\begingroup$ And in brief, continuity of $+$ in general unfolds to: whenever $x+y \in U \in \tau$, there exist $V,W$ with $x \in V \in \tau, y \in W \in \tau$, and $V+W = \{ x' + y' \mid x'\in V, y' \in W \} \subseteq U$. $\endgroup$ Feb 14, 2019 at 19:00
  • $\begingroup$ @DanielSchepler Now I understand why some authors require the topology to be Hausdorff. $\endgroup$
    – Garmekain
    Feb 16, 2019 at 12:12

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