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Prove that a sequence converges in this topological space iff

  1. Dec 8, 2013 #1
    1. The problem statement, all variables and given/known data

    Consider (R,C). Prove that a sequence converges in this topological space iff it is bounded below
    define ##C = ## ##\left \{ (a,\infty)|a\in R \right \} \bigcup \left \{ \oslash , R \right \}##

    2. Relevant equations



    3. The attempt at a solution

    So I am not very clear on how to wrap my head around what (R,C) actually means. I know that R is the underlying set and that C is the topology. So does that mean that, in this case, that all open sets in the underlying set, R, are of this form (a,∞)? And does that mean any open ball around a point in R is of that same form? Please help
     
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  3. Dec 8, 2013 #2

    Dick

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    Yes, that's what it means.
     
  4. Dec 8, 2013 #3
    OK so if we let ##\left \{ X_n \right \} \rightarrow x##, it follows that for each neighborhood ##V## of x, ##\exists N_0 \in \mathbb{N}## s.t ##X_n \in V## for ##n > N_0##.

    Since ##V## is a neighborhood of x it contains an open set, call it ##O## that contains x.

    Here I get stuck. Does this mean that ##\left \{ X_n \right \}## is in ##O## for ##n > N_0## and ##O## is bounded below by x?
     
    Last edited: Dec 8, 2013
  5. Dec 8, 2013 #4

    Dick

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    You are kind of taking it the wrong way, you don't even know ##X_n## has a limit yet. Suppose ##X_n## is bounded below by a number ##B##. What kinds of points do you think might be candidates for the limit? You know what the neighborhoods look like. Try to think in terms of pictures rather than writing a bunch of symbols down first.
     
  6. Dec 8, 2013 #5
    arent you suppose to assume a sequence converges for the => direction?
     
  7. Dec 8, 2013 #6

    Dick

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    Yes, you are. I was thinking of the opposite direction. So ok, what are all the neighborhoods that can contain x?
     
  8. Dec 8, 2013 #7
    neighborhoods that can contain x,
    ##(a,\infty)## where a<x?
     
  9. Dec 8, 2013 #8

    Dick

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    Exactly. So ##(x-1,\infty)## contains all but a finite number of the ##X_n##. Does that show ##X_n## is bounded?
     
  10. Dec 8, 2013 #9
    well we are suppose to show that ##X_n## is bounded below. I suppose it would be bounded below by x-1 if it is increasing, and bounded above and below if it is decreasing? The neighborhood going towards infinity is confusing me. I guess if it is decreasing it can't start at infinity it has to start at a point, and it would stay in ##(x-1,\infty)## and be bounded above by x-1
     
  11. Dec 8, 2013 #10

    Dick

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    You are probably over thinking this. There is some N such that ##(x-1,\infty)## contains all ##X_n## with n>N. So those points are all bounded below by x-1. The points you haven't bounded yet is the finite collection ##X_1, X_2, ... X_n##. Are those bounded below (not necessarily by x-1)?
     
  12. Dec 8, 2013 #11
    oh i see. So take that finite collection and take the minimum, then that finite collection will be bounded below by that min?
     
  13. Dec 8, 2013 #12

    Dick

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    Sure, so the whole sequence is bounded below by the minimum of that min and x-1. Now try the other direction. It's actually even easier.
     
  14. Dec 8, 2013 #13
    ok, so we assume that there is a sequence bounded bellow by some number. Call it ##\delta##. Thus ##X_n## > ##\delta## ##\forall n##. Let ε > 0. ##N_\delta## = ##\left ( \delta - ε, \infty \right )## is the set of all neighborhoods of ##\delta##. Notice that ##X_n \in N_\delta## ##\forall n## since ##X_n## > ##\delta - ε## ##\forall n##. Thus ##\left \{ X_n \right \} \rightarrow \delta##,
     
  15. Dec 8, 2013 #14

    Dick

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    You've got the concept fine. Your notation is a little lacking. Having a lower bound δ means ##X_n \ge \delta##. Don't you mean to define ##N_\epsilon=\left ( \delta - ε, \infty \right )##? Try and rewrite that a bit. And what's maybe interesting about this topology is that ##\left \{ X_n \right \} \rightarrow \delta-1## is also true.
     
    Last edited: Dec 8, 2013
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