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In the definition of topological groups we impose both $(x,y)\to xy$ and $x\to x^{-1}$ to be continuous.

However, I cannot find an example where the first condition holds but the second fails.

Is the second one redundant?

Thanks!

  • 0
    Off-topic, but slightly related. If $X$ is a group and a complete metric space, and the map $(x,y) \mapsto xy$ is separately continuous, then in fact $X$ is a topological group. That is: if follows that $(x,y) \mapsto xy$ is jointly continuous and that $x \mapsto x^{-1}$ is continuous.2012-01-02

2 Answers 2

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Take $\mathbb{Z}$ with the usual group operation and topology given by the open sets $(n, \infty), n \in \mathbb{Z}$ (together with the empty set and the entire space). The group operation is continuous since the preimage of $(n, \infty)$ is a union of the open sets $(a, \infty) \times (b, \infty)$ where $a + b = n$, but inversion is not since the preimage of $(n, \infty)$ is $(-\infty, -n)$ which is not open.

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    Hi, Qiaochu! Do you have any good books in mind where I can pick up basics about topological groups? I think I am fine with both topology and groups but need some background in 'topological groups'. Thanks!2012-01-02
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Sorgenfrey Line is another example of paratopological group. See: Topological group and related structure, Book by Arhangel'skii and Tkachenko Page 13 example 1.2.1.

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    The proof follows from the fact that sets of the form $[a,b)$, where $a, b \in \mathbb{R}$, form the bases for the sorgenfrey line. Then inverse image of the open set $[1,2)$ under the map $x \rightarrow -x$ is $(-2,-1]$ which is not open in sorgenfrey line. So, inverse function is not continuous.2012-01-04