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Discrete groups

  1. Feb 17, 2008 #1
    1. The problem statement, all variables and given/known data
    Prove: a topological group is discrete if the singleton containing the identity is an open set.

    The statement is in here http://en.wikipedia.org/wiki/Discrete_group



    3. The attempt at a solution
    Is that because if you multiply the identity with any element in the group, you get a new element with nothing surrounding it because it's like you can also multiply the area around the identity to the new position. In other words mapping open sets to open set?

    f is cts => open sets are mapped to open sest in a topological group.
     
    Last edited: Feb 17, 2008
  2. jcsd
  3. Feb 17, 2008 #2

    StatusX

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    It's not true that continuous functions map open sets to open sets. For example, the map f(x)=x^2 maps the open set (-1,1) to the non-open set [0,1). Rather, the inverse image of an open set under a continuous map is an open set. So look at the preimage of the identity under certain maps, say, f_g:G->G, where f_g(h)=gh.
     
  4. Feb 17, 2008 #3
    I realised that after posting. I may have remembered under some circumstances, open sets are mapped to open sets. What is this circumstance?
     
  5. Feb 17, 2008 #4

    morphism

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    Maps that do that are called 'open maps'. Just like continuous maps don't have to be open, open maps don't have to be continuous either.
     
  6. Feb 17, 2008 #5
    Under what circumstances are open maps continous and vice versa?
     
  7. Feb 17, 2008 #6
    But do all f_g have to be continous maps? It isn't implied from the axioms for a topological group.
     
  8. Feb 18, 2008 #7

    morphism

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    Multiplication maps are always continuous. This follows from the fact that multiplication is jointly continuous in a topological group.
     
  9. Feb 18, 2008 #8
    RIght, I found a proof of it using product topology and component mapping, pi which is open.
     
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