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An Open and Closed Interval in Q

  1. Jan 27, 2013 #1
    Why is the interval ##(-√2,√2)## closed in ##\mathbb{Q}##

    I know why it is open, but do we consider it closed because it has no limit points in ##\mathbb{Q}##, thus vacuously it is closed.
     
  2. jcsd
  3. Jan 27, 2013 #2
    In other words since the boundary of ##(-\sqrt 2,\sqrt 2)## does not exist in ##\mathbb{Q}##, then it is closed.
     
  4. Jan 27, 2013 #3

    jbunniii

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    It certainly has limit points in ##\mathbb{Q}##. For example, 0 is a limit point because the sequence (1/n) has 0 as a limit. Similarly, any rational number contained in ##(-\sqrt{2}, \sqrt{2})## is a limit point of that interval. The important thing is that there are no limit points of ##(-\sqrt{2}, \sqrt{2})## in ##\mathbb{Q}\setminus (-\sqrt{2}, \sqrt{2})##. (Proof?)

    Another way to see that ##(-\sqrt{2}, \sqrt{2})## is closed in ##\mathbb{Q}## is that its complement ##(-\infty, -\sqrt{2}) \cup (\sqrt{2}, \infty)## is open in ##\mathbb{Q}##, for essentially the same reason that ##(-\sqrt{2},\sqrt{2})## is open.
     
    Last edited: Jan 27, 2013
  5. Jan 27, 2013 #4

    jbunniii

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    A third way to see it is that the closure of ##(-\sqrt{2}, \sqrt{2})##, which is nominally ##[-\sqrt{2}, \sqrt{2}]##, is actually the same as ##(-\sqrt{2}, \sqrt{2})## because, as you said, [itex]-\sqrt{2}[/itex] and [itex]\sqrt{2}[/itex] are not elements of [itex]\mathbb{Q}[/itex]. Since the interval equals its closure, it must be a closed set. You can make this argument more rigorous if you know a bit about subspace topologies.
     
    Last edited: Jan 27, 2013
  6. Jan 27, 2013 #5
    Since ##\mathbb{Q}## is dense in ##\mathbb{R}## with infinitely many empty interior points then the boundary of ##(-\sqrt 2, \sqrt2) =[-\sqrt{2}, \sqrt{2}]## which happens to be its closure as well.
     
  7. Jan 27, 2013 #6

    jbunniii

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    I'm not sure what you mean by "empty interior points." As a subset of ##\mathbb{R}##, in fact ##\mathbb{Q}## has no interior points.

    Here is what I was getting at when I referred to subspace topologies. If we view ##\mathbb{Q}## as a subspace of ##\mathbb{R}##, then a subset ##A \subset \mathbb{Q}## is open in ##\mathbb{Q}## if and only if it is possible to write it as ##A = U \cap \mathbb{Q}##, where ##U## is an open subset of ##\mathbb{R}##. Also, a subset ##A \subset \mathbb{Q}## is closed in ##\mathbb{Q}## if and only if it is possible to write it as ##A = F \cap \mathbb{Q}##, where ##F## is a closed subset of ##\mathbb{R}##.

    If I may introduce the notation ##(-\sqrt{2}, \sqrt{2})_\mathbb{R}## to mean ##\{x \in \mathbb{R} : -\sqrt{2} < x < \sqrt{2}\}## and ##(-\sqrt{2}, \sqrt{2})_\mathbb{Q}## to mean ##\{x \in \mathbb{Q} : -\sqrt{2} < x < \sqrt{2}\}##, then we have
    $$(-\sqrt{2}, \sqrt{2})_\mathbb{Q} = (-\sqrt{2}, \sqrt{2})_\mathbb{R} \cap \mathbb{Q}$$
    which shows that ##(-\sqrt{2}, \sqrt{2})_\mathbb{Q}## is an open subset of ##\mathbb{Q}##, and
    $$(-\sqrt{2}, \sqrt{2})_\mathbb{Q} = [-\sqrt{2}, \sqrt{2}]_\mathbb{R} \cap \mathbb{Q}$$
    which shows that ##(-\sqrt{2}, \sqrt{2})_\mathbb{Q}## is a closed subset of ##\mathbb{Q}##.
     
  8. Jan 27, 2013 #7
    let me understand something:
    You said: "The important thing is that there are no limit points of ##(-\sqrt{2}, \sqrt{2})## in ##\mathbb{Q}\setminus (-\sqrt{2}, \sqrt{2})##."

    what you saying is that the rationals are the limit points of the reals. I always thought that the reals tend to solve the problem of convergence of the rationals:

    think ##\pi## and the sequence. 3, 31/10, 314/100, 3141/1000....

    ?
     
  9. Jan 27, 2013 #8

    jbunniii

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    Yes, but if we are talking about ##(-\sqrt{2}, \sqrt{2})## as a subset of ##\mathbb{Q}##, then there are no irrational numbers available, either in ##(-\sqrt{2}, \sqrt{2})## or in its complement, ##\mathbb{Q}\setminus(-\sqrt{2}, \sqrt{2})##. The sequence 3, 31/10, 314/100, 3141/1000, ... does not have a limit in this space even though the elements are getting closer and closer together. To use the appropriate technical jargon, this is a Cauchy sequence but not a convergent sequence: the limit to which it would have to converge, namely [itex]\pi[/itex], is missing from the space because it is irrational. In a complete space such as [itex]\mathbb{R}[/itex], all Cauchy sequences converge, but [itex]\mathbb{Q}[/itex] is not complete, meaning it has "holes" (all the irrationals are missing), so not every Cauchy sequence converges.
     
  10. Jan 27, 2013 #9

    jbunniii

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    In fact, the exact opposite is true: the reals are the limit points of the rationals.
     
  11. Jan 27, 2013 #10
    Thanks. I fully understand why the interval is both open and closed.
    But new questions arose from your answers. :)

    My previous statement about the density of ##\mathbb{Q}## came from a wikipedia page:
    http://en.wikipedia.org/wiki/Boundary_(topology). Please read the first line after the examples.

    I guess I should specify the Metric Space that we are working with. respect to the statement:

    ∂(-√2,√2) = [-√2,√2] is true in ##\mathbb{R}## but not in ##\mathbb{Q}## as ∂(-√2,√2) = ∅. Am I correct?

    btw Thank you for explaining the subspace topology. There is a good wiki article about it.
     
  12. Jan 27, 2013 #11

    jbunniii

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    That's a good sign!
    OK, that would be this line: "These last two examples illustrate the fact that the boundary of a dense set with empty interior is its closure." However, the context here is that they are working with subsets of [itex]\mathbb{R}[/itex].

    This is an important point: you can't talk about whether a set is open or closed, or what its boundary or interior are, without specifying which space you are working with. Here are some concrete examples.

    Let us define, as I did in a previous post, [itex](-\sqrt{2}, \sqrt{2})_\mathbb{Q} = \{x \in \mathbb{Q} : -\sqrt{2} < x < \sqrt{2}\}[/itex] and [itex](-\sqrt{2}, \sqrt{2})_\mathbb{R} = \{x \in \mathbb{R} : -\sqrt{2} < x < \sqrt{2}\}[/itex]. Also define [itex][-\sqrt{2}, \sqrt{2}]_\mathbb{Q} = \{x \in \mathbb{Q} : -\sqrt{2} \leq x \leq \sqrt{2}[/itex] and [itex][-\sqrt{2}, \sqrt{2}]_\mathbb{R} = \{x \in \mathbb{R} : -\sqrt{2} \leq x \leq \sqrt{2}\}[/itex].

    Of course, [itex](-\sqrt{2}, \sqrt{2})_\mathbb{Q} = [-\sqrt{2}, \sqrt{2}]_\mathbb{Q}[/itex], but [itex](-\sqrt{2}, \sqrt{2})_\mathbb{R} \neq [-\sqrt{2}, \sqrt{2}]_\mathbb{R}[/itex].

    Then, AS SUBSETS OF [itex]\mathbb{Q}[/itex]:
    * [itex](-\sqrt{2}, \sqrt{2})_\mathbb{Q}[/itex] is both open and closed
    * Every point of [itex](-\sqrt{2}, \sqrt{2})_\mathbb{Q}[/itex] is an interior point
    * Every point of [itex](-\sqrt{2}, \sqrt{2})_\mathbb{Q}[/itex] is a limit point, and there are no other limit points of this set
    * The boundary of [itex](-\sqrt{2}, \sqrt{2})_\mathbb{Q}[/itex] is empty.

    Whereas, AS SUBSETS of [itex]\mathbb{R}[/itex]:
    * [itex](-\sqrt{2}, \sqrt{2})_\mathbb{Q}[/itex] is neither open nor closed
    * [itex](-\sqrt{2}, \sqrt{2})_\mathbb{Q}[/itex] has no interior points
    * The set of limit points of [itex](-\sqrt{2}, \sqrt{2})_\mathbb{Q}[/itex] is [itex][-\sqrt{2}, \sqrt{2}]_\mathbb{R}[/itex]
    * The boundary of [itex](-\sqrt{2}, \sqrt{2})_\mathbb{Q}[/itex] is also [itex][-\sqrt{2}, \sqrt{2}]_\mathbb{R}[/itex].

    And, AS SUBSETS OF [itex]\mathbb{R}[/itex]:
    * [itex](-\sqrt{2}, \sqrt{2})_\mathbb{R}[/itex] is open but not closed
    * Every point of [itex](-\sqrt{2}, \sqrt{2})_\mathbb{R}[/itex] is an interior point
    * The set of limit points of [itex](-\sqrt{2}, \sqrt{2})_\mathbb{R}[/itex] is [itex][-\sqrt{2}, \sqrt{2}]_\mathbb{R}[/itex]
    * The boundary of [itex](-\sqrt{2}, \sqrt{2})_\mathbb{R}[/itex] is [itex]\{-\sqrt{2}, \sqrt{2}\}[/itex].

    Finally, AS SUBSETS OF [itex]\mathbb{R}[/itex]:
    * [itex][-\sqrt{2}, \sqrt{2}]_\mathbb{R}[/itex] is closed but not open
    * The set of interior points of [itex][-\sqrt{2}, \sqrt{2}]_\mathbb{R}[/itex] is exactly [itex](-\sqrt{2}, \sqrt{2})_\mathbb{R}[/itex]
    * The set of limit points of [itex][-\sqrt{2}, \sqrt{2}]_\mathbb{R}[/itex] is [itex][-\sqrt{2}, \sqrt{2}]_\mathbb{R}[/itex]
    * The boundary of [itex][-\sqrt{2}, \sqrt{2}]_\mathbb{R}[/itex] is [itex]\{-\sqrt{2}, \sqrt{2}\}[/itex].
     
  13. Jan 27, 2013 #12
    Cool. For instance, the boundary of the singleton S = {0} in ℝ under the discrete metric is the ∅ set.

    But if we specify the topology as being that {0} has no open subsets, then the ∂S is S itself.

    Am I correct?
     
  14. Jan 27, 2013 #13

    jbunniii

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    Under the discrete metric, yes. Under the standard euclidean metric, the boundary would be {0}.

    Well, {0} has only two subsets: itself and ∅. We don't get a choice about ∅: it is open in any topology. So I guess you mean we specify the topology such that {0} is not open. Then you are correct: any open set containing {0} must also contain other points of ##\mathbb{R}##, so 0 is a boundary point. However, there might also be other boundary points. For example, if we use the trivial topology such that the only open sets are [itex]\emptyset[/itex] and [itex]\mathbb{R}[/itex], then every point of [itex]\mathbb{R}[/itex] is a boundary point of {0}.
     
  15. Jan 27, 2013 #14
    Nice

    So in this case int{0} = ∅
     
  16. Jan 27, 2013 #15

    jbunniii

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    Right.
     
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