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Set of integers is closed.

  1. Dec 11, 2011 #1
    When considered as a subset of [itex]\mathbb{R}^2[/itex], [itex]\mathbb{Z}[/itex] is a closed set.

    We will show, by definition, that [itex]\mathbb{Z} \subset \mathbb{R}^2[/itex] is closed.
    That is, we need to show that, if [itex]n[/itex] is a limit point of [itex]\mathbb{Z}[/itex], then [itex]n \in \mathbb{Z}[/itex].

    I think this becomes vacuously true, since our hypothesis is false, i.e. because [itex]\mathbb{Z}[/itex] has no limit points. Is this true, or am I just being silly?

    Thank you!

    Edit: (I know this can be proved, again by definition, by showing that [itex]\mathbb{R}^2 - \mathbb{Z}[/itex] is open.)
  2. jcsd
  3. Dec 12, 2011 #2


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    Any Cauchy sequence must ultimately be constant.
  4. Dec 12, 2011 #3
    I'm a little confused how that is relevant to a direct proof of showing that the set of all integers is closed.

    Edit: Actually, if the statement is vacuously true, then it actually shows nothing at all, since then we could conclude nothing about the set of all integers, correct? Serves me right for trying to do math while overtired.
    Last edited: Dec 12, 2011
  5. Dec 12, 2011 #4
    I think lavinia is using the following proposition:
    Working in the reals, given a set A and a point x, x is a limit point of A if and only if there is a nonconstant sequence xn contained in A that converges to x.
    No, I think what you said at first is fine. A set is closed iff it contains all its limit points. If it has no limit points, then it is certainly closed. You might want to rigorously prove that [itex] \mathbb{Z} [/itex] has no limit points, but it's pretty straightforward.
  6. Dec 12, 2011 #5
    Oh I see. At this point in the textbook, he hasn't said anything other than the definition of a sequence itself.

    That's exactly what I was thinking as well.

    I do think I finally see where I was confused now. By vacuous truth, I showed that the set was closed, but that did nothing to show that the set was not also open. So it still remains for me to show that [itex]\mathbb{Z}[/itex] is not open, correct? (I am considering/mimicking how to show that the empty set is open, which just so happens to be closed as well.)

    Edit: What I just said makes zero sense. Disregard.
    Last edited: Dec 12, 2011
  7. Dec 12, 2011 #6
    Why do you want to show that [itex] \mathbb{Z} [/itex] is not open? Is that just another part of the question? I ask only because open and closed are not antonyms like they are in their normal English usage. Just because a set is closed doesn't mean it isn't open, too (although this is the case with [itex] \mathbb{Z}[/itex]). I hope this isn't confusing you!
    Last edited: Dec 12, 2011
  8. Dec 12, 2011 #7
    I just realized while walking from the library that nothing I typed made any sense whatsoever, since I made no hypothesis about [itex]n[/itex] being a limit point.

    Actually I think it made sense, but had nothing to do with the question, and therefore made zero sense within the context of this question.
  9. Dec 12, 2011 #8


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    if every Cauchy sequence in a set converges to a point in the set, then the set is certainly closed. I take the definition of closed to be if every convergent sequennce in the set converges to a point inside the set.

    This is not a vacuous statement for the integers.

    But the intuition here is that any Cauchy sequence must after a while just be constant so it must converge to an integer.
  10. Dec 12, 2011 #9
    Again, the textbook has not mentioned sequences yet, so they cannot be used to solve the problem. I am also not looking for any other way to solve it other than directly, because I am fairly certain that it can be done so, and that is an exercise that I have created for myself to make sure that I understand the basic principles of open and closed sets.

    Then please guide me in the correct direction! Why is assuming that [itex]n[/itex] is a limit point of [itex]\mathbb{Z}[/itex] and then noticing that [itex]\mathbb{Z}[/itex] has no limit points not a vacuous statement?
  11. Dec 12, 2011 #10
    I think you're misinterpreting lavinia's comment. Here is the full quote, with context.

    The statement about convergent sequences is not vacuous because there are convergent sequences in the set of integers. The statement about limit points (every limit point of [itex] \mathbb{Z} [/itex] is in [itex] \mathbb{Z} [/itex]) is vacuous because [itex] \mathbb{Z} [/itex] has no limit points.

    Anyway, I think lavinia is actually using the definition for a set to be complete, and then using some proposition that says a complete subset of [itex] \mathbb{R} [/itex] is closed.
  12. Dec 13, 2011 #11


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    - I don't think completeness is needed. A set is closed in a metric space is every convergent Cauchy sequence in the set converges to a point in the set.

    - Correct me if I am wrong but a limit point of a set is a point not in the set but every open neighborhood of the point has non-empty intersection with the set. In a metric space this means that any open ball around the point no matter how small its radius must have non-empty intersection with the set. An infinite sequence of these balls with ever vanishing radii creates a Cauchy sequence in the set that converges to the point. So in a metric space the Cauchy sequence definition and the limit point definition are the same.

    Showing that the integers have no limit points is the same as showing that one can choose a ball small enough around any non-integer so that is does not contain an integer. Why is that a vacuous argument?

    But you are right that the Cauchy sequence argument is far too complicated for this example.
    Last edited: Dec 13, 2011
  13. Dec 13, 2011 #12
    If you want to take that as your definition of closed, okay. But really the property you stated about Cauchy sequences is usual the definition of completeness for a metric space (cf. http://en.wikipedia.org/wiki/Complete_metric_space). Being closed is really a topological property. The definition I've always used is S is closed iff the complement of S is open. One can then show that a set S is closed iff [itex] S = \overline{S} [/itex] iff S contains all its limit points. It can also be shown that our definitions agree in the case of a complete metric space, i.e. given a subset A of a complete metric space X, A is closed iff A is a complete subspace.

    There are varying definitions of limit point. Using the one I'm familiar with (http://en.wikipedia.org/wiki/Limit_point), the point can be in the set. For instance, all points of the interval [0,1] are limit points of the interval.
    Sorry if I wasn't being clear. You're right, that part of the argument (proving that [itex] \mathbb{Z} [/itex] has no limit points) is not vacuous. Once we've done that, we show that [itex] \mathbb{Z} [/itex] is closed by showing it contains all its limit points. Since [itex] \mathbb{Z} [/itex] has no limit points, it vacuously contains all of them.
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