An example of a group with a topology












6












$begingroup$


Do you know an example of a group with a topology satisfying both the following two conditions




  1. the product is separately continuous but not jointly continuous

  2. the inversion map is continuous.






abstract-algebra general-topology examples-counterexamples topological-groups







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  • $begingroup$
    One important example of a group with a topology is a Lie Group. With that in mind, you might find this post to be interesting.
    $endgroup$
    – Omnomnomnom
    Sep 16 '18 at 18:20
















6












$begingroup$


Do you know an example of a group with a topology satisfying both the following two conditions




  1. the product is separately continuous but not jointly continuous

  2. the inversion map is continuous.






abstract-algebra general-topology examples-counterexamples topological-groups







share|cite|improve this question











$endgroup$












  • $begingroup$
    One important example of a group with a topology is a Lie Group. With that in mind, you might find this post to be interesting.
    $endgroup$
    – Omnomnomnom
    Sep 16 '18 at 18:20














6












6








6


3



$begingroup$


Do you know an example of a group with a topology satisfying both the following two conditions




  1. the product is separately continuous but not jointly continuous

  2. the inversion map is continuous.






abstract-algebra general-topology examples-counterexamples topological-groups







share|cite|improve this question











$endgroup$




Do you know an example of a group with a topology satisfying both the following two conditions




  1. the product is separately continuous but not jointly continuous

  2. the inversion map is continuous.






abstract-algebra general-topology examples-counterexamples topological-groups




abstract-algebra general-topology examples-counterexamples topological-groups






share|cite|improve this question















share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








edited Sep 16 '18 at 18:00









José Carlos Santos

175k24134243




175k24134243










asked Sep 16 '18 at 17:48









W4cc0W4cc0

1,90621227




1,90621227












  • $begingroup$
    One important example of a group with a topology is a Lie Group. With that in mind, you might find this post to be interesting.
    $endgroup$
    – Omnomnomnom
    Sep 16 '18 at 18:20


















  • $begingroup$
    One important example of a group with a topology is a Lie Group. With that in mind, you might find this post to be interesting.
    $endgroup$
    – Omnomnomnom
    Sep 16 '18 at 18:20
















$begingroup$
One important example of a group with a topology is a Lie Group. With that in mind, you might find this post to be interesting.
$endgroup$
– Omnomnomnom
Sep 16 '18 at 18:20




$begingroup$
One important example of a group with a topology is a Lie Group. With that in mind, you might find this post to be interesting.
$endgroup$
– Omnomnomnom
Sep 16 '18 at 18:20










2 Answers
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Take, for instance, $(mathbb{Q},+)$, endowed with the Zariski topology (that is, a non-empty set $A$ is open if and only if $A^complement$ is finite). Then the inversion ($xmapsto-x$) is clearly continuous and addition is clearly separately continuous. But it is not jointly continuous since, for instance ${(x,y)inmathbb{Q}^2,|,x+y=0}$ is not a closed set.






share|cite|improve this answer











$endgroup$





















    6












    $begingroup$

    Let $G$ be any infinite group and give it the cofinite topology. Then the product is separately continuous as is inversion, since any bijection $Gto G$ is continuous. But the product is not jointly continuous, since ${1}$ is closed but its preimage is not. (Or, you can just cite the fact that any $T_0$ topological group is Hausdorff, so $G$ cannot be a topological group.)






    share|cite|improve this answer









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      2 Answers
      2






      active

      oldest

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      2 Answers
      2






      active

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      active

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      active

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      10












      $begingroup$

      Take, for instance, $(mathbb{Q},+)$, endowed with the Zariski topology (that is, a non-empty set $A$ is open if and only if $A^complement$ is finite). Then the inversion ($xmapsto-x$) is clearly continuous and addition is clearly separately continuous. But it is not jointly continuous since, for instance ${(x,y)inmathbb{Q}^2,|,x+y=0}$ is not a closed set.






      share|cite|improve this answer











      $endgroup$


















        10












        $begingroup$

        Take, for instance, $(mathbb{Q},+)$, endowed with the Zariski topology (that is, a non-empty set $A$ is open if and only if $A^complement$ is finite). Then the inversion ($xmapsto-x$) is clearly continuous and addition is clearly separately continuous. But it is not jointly continuous since, for instance ${(x,y)inmathbb{Q}^2,|,x+y=0}$ is not a closed set.






        share|cite|improve this answer











        $endgroup$
















          10












          10








          10





          $begingroup$

          Take, for instance, $(mathbb{Q},+)$, endowed with the Zariski topology (that is, a non-empty set $A$ is open if and only if $A^complement$ is finite). Then the inversion ($xmapsto-x$) is clearly continuous and addition is clearly separately continuous. But it is not jointly continuous since, for instance ${(x,y)inmathbb{Q}^2,|,x+y=0}$ is not a closed set.






          share|cite|improve this answer











          $endgroup$



          Take, for instance, $(mathbb{Q},+)$, endowed with the Zariski topology (that is, a non-empty set $A$ is open if and only if $A^complement$ is finite). Then the inversion ($xmapsto-x$) is clearly continuous and addition is clearly separately continuous. But it is not jointly continuous since, for instance ${(x,y)inmathbb{Q}^2,|,x+y=0}$ is not a closed set.







          share|cite|improve this answer














          share|cite|improve this answer



          share|cite|improve this answer








          edited Dec 29 '18 at 11:29

























          answered Sep 16 '18 at 17:56









          José Carlos SantosJosé Carlos Santos

          175k24134243




          175k24134243























              6












              $begingroup$

              Let $G$ be any infinite group and give it the cofinite topology. Then the product is separately continuous as is inversion, since any bijection $Gto G$ is continuous. But the product is not jointly continuous, since ${1}$ is closed but its preimage is not. (Or, you can just cite the fact that any $T_0$ topological group is Hausdorff, so $G$ cannot be a topological group.)






              share|cite|improve this answer









              $endgroup$


















                6












                $begingroup$

                Let $G$ be any infinite group and give it the cofinite topology. Then the product is separately continuous as is inversion, since any bijection $Gto G$ is continuous. But the product is not jointly continuous, since ${1}$ is closed but its preimage is not. (Or, you can just cite the fact that any $T_0$ topological group is Hausdorff, so $G$ cannot be a topological group.)






                share|cite|improve this answer









                $endgroup$
















                  6












                  6








                  6





                  $begingroup$

                  Let $G$ be any infinite group and give it the cofinite topology. Then the product is separately continuous as is inversion, since any bijection $Gto G$ is continuous. But the product is not jointly continuous, since ${1}$ is closed but its preimage is not. (Or, you can just cite the fact that any $T_0$ topological group is Hausdorff, so $G$ cannot be a topological group.)






                  share|cite|improve this answer









                  $endgroup$



                  Let $G$ be any infinite group and give it the cofinite topology. Then the product is separately continuous as is inversion, since any bijection $Gto G$ is continuous. But the product is not jointly continuous, since ${1}$ is closed but its preimage is not. (Or, you can just cite the fact that any $T_0$ topological group is Hausdorff, so $G$ cannot be a topological group.)







                  share|cite|improve this answer












                  share|cite|improve this answer



                  share|cite|improve this answer










                  answered Sep 16 '18 at 17:56









                  Eric WofseyEric Wofsey

                  193k14221352




                  193k14221352






























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