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Does {1/n} n=1 to infinity converge? Why or why not? in a topological space

  1. Apr 1, 2012 #1
    hi,
    can someone please help me with this problem.

    Let T be the collection of all U subset R such that U is open using the usual
    metric on R.Then (R; T ) is a topological space. The topology T could also be described as
    all subsets U of R such that using the usual metric on R, R \ U is closed and
    bounded.

    Does {1/n} n=1 to infinity converge? Why or why not?

    I think it does converge...it converges to 1 for example...am I right?

    Does {n} n=1 to infinity converge? Why or why not?

    I dont think that this sequence converges in a topological space?

    thanks
     
  2. jcsd
  3. Apr 1, 2012 #2

    jgens

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    Gold Member

    Does there exist an [itex]N \in \mathbb{N}[/itex] such that the open neighborhood [itex](\frac{1}{2},\frac{3}{2})[/itex] of [itex]1[/itex] contains [itex]\frac{1}{n}[/itex] for all [itex]N<n[/itex]?

    You should say "I don't think that this sequence converges in the usual topology on [itex]\mathbb{R}[/itex]" to be more accurate. But you got this one right. Now can you prove it?
     
  4. Apr 1, 2012 #3
    thank you so much....

    It seems like I wrote everything wrong, for the second sequence that's what I meant to say, and I've already proved it.

    for the first sequence, it converges to 0 in usual topology and this is what I have so far...

    For every open U s.t. 0 is in U, there exist N such that for all n > N, 1/n is in U

    Assume 0 is in U where U is open in usual topology then there is a s> 0 s.t. (1-s, 1+s) is subset of U
    So,

    1-s < 1/n
    n < 1/( 1-s)
     
  5. Apr 1, 2012 #4

    jgens

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    Gold Member

    Given that you wrote the first part wrong, I am going to ask you to rewrite this argument carefully before I give you more help.
     
  6. Apr 1, 2012 #5

    HallsofIvy

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    Did you think to write out at least part of sequence? Taking n= 1, 2, 3, 4, 5, gives
    1, 1/2, 1/3, 1/4, 1/5. You think that is getting closer to 1?

     
  7. Apr 1, 2012 #6
    It was mistake, which I corrected later...it converges to 0 not 1.
     
  8. Apr 1, 2012 #7

    jgens

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    Gold Member

    Right. But you still have not provided a correct proof that the sequence converges to 0.
     
  9. Apr 2, 2012 #8
    agree, I"ll be working on it today...
     
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