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Analysis Limit Point Proof

  1. Oct 18, 2009 #1
    1. The problem statement, all variables and given/known data
    Let x [tex]\in[/tex] R, and let A [tex]\subset[/tex] R. Let (an) be a sequence with an [tex]\in[/tex] A and (an) [tex]\neq[/tex] x for all n [tex]\in[/tex] N, and assume
    that x = lim (an.) Prove that x is a limit point of A.


    2. Relevant equations



    3. The attempt at a solution
    Suppose that x=lim(an). Meaning there exists a N [tex]\in[/tex] N so that for n[tex]\geq[/tex]N:
    |an-x|<[tex]\epsilon[/tex]

    Which would make the [tex]\epsilon[/tex]- neighborhood of (an)= (an- [tex]\epsilon[/tex], an+ [tex]\epsilon[/tex])

    After this I don't know what to do. I need to show that every epsilon neighborhood of V(x) intersects A in some point other than x.

    Any tips would be great!
     
  2. jcsd
  3. Oct 18, 2009 #2
    Careful, your e-neighborhood is off. Remember |a_n - x| < e means x - e < a_n < x + e or a_n is in (x - e, x + e). Now you know that a_n =/= x for each n, so what can you conclude.

    Of course this proposition is an if and only if, but this direction is probably clearer since "all but a finitely many points" certainly implies "at least one".
     
  4. Oct 18, 2009 #3
    alright so if the epsilon neighborhood is (x-e,x+e) and (an) is unequal to x for all n in N would that mean that for some N in N when n>=N that (an) is contained in the epsilon neighborhood. Which would mean that x is the limit point of (an)?
     
  5. Oct 18, 2009 #4
    oh wait, nevermind, i think I got it. I was trying to make it more complicated than it really was. I am finding this whole open sets, closed sets and limit points idea kind of confusing.
     
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