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Proof: A point is a limit point of S is a limt of a sequence

  1. Nov 14, 2015 #1
    Hi guys,

    I attempted to prove this theorem, but just wanted to see if it a valid proof.

    Thanks!

    1. The problem statement, all variables and given/known data


    Prove that x is an accumulation point of a set S iff there exists a sequence ( s n ) of points in S \ {x} that converges to x

    2. Relevant equations

    N * ( x; ε ) is the x - deleted ε - neighbourhood of x

    3. The attempt at a solution

    Suppose that x is an accumulation point of a set S. Then N * ( x; ε ) ⋀ S ≠ / Ø

    Now consider any y such that y ∈ N * ( x; ε ) ⋀ S. Since N * ( x; ε ) ⋀ S ⊆ S \ {x} ,we know that S \ {x} ≠ Ø .

    Since y ∈ N * ( x; ε ) , 0 < | y - x | < ε , which means | y - x | < ε for any y. Now consider any sequence of y ∈ N * ( x; ε ) and call this ( s n ).

    For every n > N, where N can be any number of the index, |( s n ) - x|< ε which means ( s n ) converges to x.

    Conversely, suppose that there exists a sequence ( s n ) of points in S \ {x} that converges to x .

    Then for ∀ ε >0, ∃ N such that ∀ n>N, it is the case that |( s n )-x|< ε . But this is the ε - neighbourhood of x, so that ( s n ) consists of all y such that y ∈ N * ( x; ε ) ⋀ S . It follows then that N * ( x; ε ) ⋀ S ≠ Ø , and so x is an accumulation point
     
  2. jcsd
  3. Nov 14, 2015 #2

    HallsofIvy

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    No, this is not valid because the "[itex]N(x,\epsilon)[/itex]" you are talking about is for a specific [itex]\epsilon[/itex]. There is no reason to think that the [itex]s_n[/itex] get close to x. Instead, look at all the [itex]N(x, \epsilon)[/itex] with [itex]\epsilon= 1/n[/itex], n= 1, 2, 3, ...
     
  4. Nov 14, 2015 #3

    fresh_42

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    How is accumulation point defined here? What is S? Which metric do you use, if at all? In which space is the question located? ℝ? How can you conclude that ##N^*(x;ε) ≠ ∅##?
     
  5. Nov 14, 2015 #4

    HallsofIvy

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    I believe zigzagdoom said, or at least implied, the definition of "accumulation" point: "p is an accumulation point of set S if and only if every neighborhood of x contains at least one point, other than x itself, of S- equivalently, that every "deleted" neighbor hood of x (a neighborhood of x with x "deleted" (removed from the set). A "neighborhood" of x in a metric space is [itex]N(x, \epsilon)= \{ p| d(x, p)< \epsilon\}[/itex]. A "deleted neighborhood" is [itex]N*(x, \epsilon)= \{p| 0< d(p, x)< \epsilon\}[/itex]. This is true for all metric spaces so it is not necessary nor desirable to specify a metric or the space.[/B]
     
    Last edited: Nov 14, 2015
  6. Nov 14, 2015 #5

    fresh_42

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    Yep. But Hausdorff would do. Without being metric. I remember a textbook in physics where a complete space has been defined by: "... when for every x there is a sequence converging to x."
    However, whether at school or in science: you can never be too young to get used to state problems in a proper way. That often leads to a much clearer understanding and may help to reveal the solution by itself. Plus it avoids almost always mistakes. (personal opinion)
     
    Last edited: Nov 14, 2015
  7. Nov 14, 2015 #6
    Thanks I will rework and see if I can incorporate this
     
  8. Nov 14, 2015 #7
    Thanks,

    In response to your questions: x is an accumulation point if N(x;ε) ∧ S ≠ ∅. S is not defined, but I have taken it to mean some metric space (the metric has not been specified). The space of the question has not been specified. I have assumed x is an accumulation point, so that N(x;ε) ∧ S ≠ ∅(by definition).
     
  9. Nov 14, 2015 #8

    mfb

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    For ever real ε>0. This is an important part of the definition.
     
  10. Nov 14, 2015 #9

    fresh_42

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    Sorry I read ∧ as AND. And one reason I was asking is, I looked it up to remember its exact definition and to avoid mistakes. I found your problem above as such in the case S is any sequence. This left me alone with the question what your definition is. If I may try to understand it better: x is an accumulation point of S iff for every ε > 0 there can be found an open neighborhood N(x;ε) of x such that (S ∩ N(x;ε)) \ {x} ≠ ∅.
    One has to be precise here because the gap between the definition and what has to be shown is rather close. It's almost the same.
     
  11. Nov 14, 2015 #10
    .
    Thanks again.

    The precise definition that I have in the book is:

    "x is an accumulation point of S if every deleted neighbourhood of x contains a point of S. That is, for every ε > 0, N * ( x;ε ) ≠ ∅."

    N * ( x;ε ) ≠ ∅ is the deleted neighbourhood of x.
     
  12. Nov 14, 2015 #11

    fresh_42

    Staff: Mentor

    I've corrected my definition:
    ##x## is an accumulation point of ##S## if and only if for every ##ε > 0## there can be found an open neighborhood ##N(x;ε)## of ##x## such that ##S ∩ N^*(x;ε) ≠ ∅##
    Ok, now you can apply Hallsoflvy's advice in post #2 and create your convergent sequence.
     
  13. Nov 14, 2015 #12
    Second attempt at proof

    Suppose that x is an accumulation point of a set S. Then for every ε > 0, it is the case that N * ( x; ε ) ⋀ S ≠ Ø.

    Now consider any y such that y ∈ N * ( x; ε ) ⋀ S. Since N * ( x; ε ) ⋀ S ⊆ S \ {x} ,we know that S \ {x} ≠ Ø.

    Now let a sequence of a number of y be denoted by ( s n ). Construct the sequence by letting ϵ = 1/n, where n= 1, 2, 3, .... and picking each point within the sequence so that N * ( x;ϵ ) contains each successive member y of (sn).

    Then we have that for every ε we select, ∃N such that for all n > N, |(sn) - x| < ε. This means ( s n ) converges to x.

    Conversely, suppose that there exists a sequence ( s n ) of points in S \ {x} that converges to x .

    Then for ∀ ε >0, ∃ N such that for all n>N, it is the case that | ( s n ) - x | < ε . But this is the ε - neighbourhood of x, so that ( sn ) consists of all y such that y ∈ N * ( x; ε ) ⋀ S . It follows then that N * ( x; ε ) ⋀ S ≠ Ø , and so x is an accumulation point.
     
  14. Nov 14, 2015 #13

    fresh_42

    Staff: Mentor

    correct

    You don't consider any ##y##, you choose such an ##y_ε## out of ##N^* ( x; ε ) ∩ S## which you know exists. Out of these ##y_ε## you construct a convergent sequence using Hallsofly's narrowing down ##ε##.

    ##s_n## in N^*(x;ε) ∩ S.
    > 0 if
    N = next number above 1/ε
    1/n < 1/N
    correct. A little bit bumpy, but ok. Your ##s_n## are already in N^*(x;ε) ∩ S making it non empty and that's all you wanted to show. Forget about the y.
     
  15. Nov 14, 2015 #14
    Thanks a lot for this - guidance has been very useful in helping me think about the problem.
     
  16. Nov 14, 2015 #15

    fresh_42

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    Never mind. But you see the two definitions aren't far apart. That's why you have to be precise. One can easily mess things up.
     
  17. Nov 14, 2015 #16

    fresh_42

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    Just out of curiosity. In the discrete metric ##d(x,y) = 0## for ##x = y## and ##d(x,y) = 1 ## for ##x ≠ y## there won't be any accumulation points, right? Ok, that does not contradict what you said but it would change the proof dramatically.
     
  18. Nov 14, 2015 #17

    mfb

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    There are no accumulation points in that metric, right. There are also no series in S\{x} converging to x, for any x in S.
     
  19. Nov 14, 2015 #18

    fresh_42

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    ... which makes the proof rather short because all elements of the empty set have blue eyes.
     
  20. Nov 14, 2015 #19

    mfb

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    This special case would be easier to show than the general case, sure.

    @zigzagdoom: Don't forget the other direction.
     
  21. Nov 14, 2015 #20

    fresh_42

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    But he had both directions. Did I overlook something?
     
    Last edited: Nov 14, 2015
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