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Accumulation point of [0,1]

  1. Jun 23, 2011 #1
    are 0,1 accumulation points in (0,1) ? how about in [0,1] ?
    if 0 and 1 are accumulation points in [0,1] interval what is the open subset they are in ?
    i need explanation about this.....pls...
     
  2. jcsd
  3. Jun 23, 2011 #2

    micromass

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    Hi mbarby! :smile:

    An accumulation point of a set S is a point x such that every neighborhood of x contains a point of S (that is not x). Right?

    Well, take (0,1), then every neighborhood around 1 contains a point of (0,1), thus 1 is an accumulation point of (0,1). And so is 0.
    Likewise, 0 are 1 are accumulation points of [0,1].
     
  4. Jun 23, 2011 #3
    what really confuses me here is that:
    lets assume 1 is accumulation point in (0,1)
    then shouldnt it be contained in an interval like (1-e, 1+e) (e=epsilon)
    but we dont have 1+e since it exceeds interval border..
    where am i wrong now ?
     
  5. Jun 23, 2011 #4

    micromass

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    You are considering (0,1) as a subset of the space [itex]\mathbb{R}[/itex], or are you considering (0,1) as a subset of itself?

    Anyway, 1 is an accumulation point because every interval (1-e,1+e) around 1 contains a point of (0,1). I don't see what 1+e exceeding the interval boundary has to do with this?
     
  6. Jun 23, 2011 #5
    sorry for the questions , i am not a math guy. but topology is one of the topics i want to learn. so somethings are as clear to me as it is to you guys.
    (0,1) is the space's itself. so we cant take an interval of (1-e, 1+e) without exceeding 1 by +e.
    but what i get from your reaction is that we can take the (1-e, 1+e) interval even if it exceeds the boundary, is that right ?
    if so why is that possible ? or isnt there a rule against it ? etc.
    thx for the quick explanations by the way.
     
  7. Jun 23, 2011 #6

    micromass

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    Ah, I think I get it. But if your space is (0,1), then 1 doesn't belong to this space, thus cannot be an accumulation point. In [0,1], 1 is an accumulation point of (0,1). Indeed, the set (1-e,1] is open in [0,1] and contains points from (0,1).
     
  8. Jun 23, 2011 #7
    thx a lot this explains a great to me.
    is it true , then , if i say any interval having the border of the interval is open, or sth similar to that ?
     
  9. Jun 23, 2011 #8
    That's the problem right there. 1 is an accumulation point OF (0,1). But it's wrong to say that 1 is IN (0,1). I believe your use of "in" rather than "of" is causing you to mis-think about this.
     
  10. Jun 23, 2011 #9

    micromass

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    Well, yes, something similar. Things like ]a,1] and [0,a[ are open in [0,1]. But things like [0.5,1] isn't...
     
  11. Jun 23, 2011 #10
    thx all guys,
    that eased my mind. i was tearing myself apart to understand where i was making the mistakes :/ ...
     
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