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Homework Help: Limit points

  1. Dec 6, 2007 #1
    1. The problem statement, all variables and given/known data
    I have seen two definitions of limit points. Are they the same:

    1)x is a limit point of a set A in X iff each nbhd of x contains a point of A other than x

    2) x is a limit point of A if it lies in the closure of A - {x}

    2. Relevant equations

    3. The attempt at a solution
  2. jcsd
  3. Dec 6, 2007 #2
    No, not quite. The second is not a definition of a limit point.

    A point x in a metric space is said to be a limit point if every neighborhood of x contains at least one element in the metric space not equal to x.

    Consider A', the set of all limit points of A. The closure of A is A U A'.
  4. Dec 6, 2007 #3
    Actually, I just realized that those two definitions are the same for the following reason:

    x is in the closure of A-{x}
    x is in every closed set containing A-{x}
    there does not exist a nbhd U_x of s.t. [tex] U_x \cap (A-{x}) = \emptyset [/tex]
  5. Dec 6, 2007 #4
    Am I right?
  6. Dec 6, 2007 #5
    If you're right, you'll be able to prove it.

    You're just making everything exponentially more complicated than it actually is. True mathematicians aim for simplicity.
  7. Dec 6, 2007 #6
    What are you talking about? I am asking if my proof in the third post makes sense.
  8. Dec 6, 2007 #7
    Someone, please, am I right?
  9. Dec 6, 2007 #8


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    It looks OK, but you're somewhat over-complicating it. Try to use the fact that y is in the closure of B iff every nbhd of y intersects B.
  10. Dec 6, 2007 #9
    Then the equivalence of those two definitions is immediate, isn't it? I really do not understand why two people have said I am overcomplicating this...
  11. Dec 7, 2007 #10
    I was saying if it makes sense, it has a proof. So if you can't come up with a solid proof, there's a higher chance it doesn't make sense.

    But see, in your second "definition" you talk about the closure of a set A. But the closure is defined as the set A unioned with the set of all of A's limit points. And then you still need a definition for a limit point. So it comes down to the fact that you want to use the term in the definition, which just further complicates things. That's what I mean.
  12. Dec 7, 2007 #11
    The closure of A is defined as the intersection of all closed sets containing A. It is equivalently A union A'. Both of these definitions are commonly found in the literature and indeed, they are the seem. I think that you are complicating things and that you should make sure you know more about the topic before you say that something is or is not a definition.
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