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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!

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A paratopological group is required to satisfy only your first condition. See, e.g.: books.google.com/… –  David Mitra Jan 2 '12 at 11:21
    
@DavidMitra thanks David! But Qiaochu gave a simple and direct answer. Enjoy it! –  Hui Yu Jan 2 '12 at 11:23
    
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. –  GEdgar Jan 2 '12 at 16:09

2 Answers 2

up vote 17 down vote accepted

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! –  Hui Yu Jan 2 '12 at 16:50

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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If the example is not too long, maybe you could write it in your answer in order to make it more self-contained. –  Davide Giraudo Jan 2 '12 at 16:45
    
note that the operation here is addition, that $x \rightarrow -x$ is not continuous is quite clear. –  Henno Brandsma Jan 2 '12 at 18:06
    
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. –  Abcd J Jan 4 '12 at 15:37

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