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Open sets

  1. Feb 20, 2009 #1
    we can prove that:
    the intersection of a finite number of open sets is open.

    how about:
    the intersection of any number of open sets?

    it's maybe not open.but how to prove it or the example?

    :smile:
     
  2. jcsd
  3. Feb 20, 2009 #2
    In the subject of topology, the first statement you make is part of the definition of open sets.

    The second statement isn't true. For example in a metric space consider a nested sequence of open balls whose radius goes to zero.
     
  4. Feb 20, 2009 #3
    thanks for reply.
    can i say something like that:
    the intersection of these balls turns into a point.
    and a point is a closed set.
     
  5. Feb 21, 2009 #4

    HallsofIvy

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    The real question is, what definitions do you have to work with? In particular are you assuming a metric space?

    Yes, you have a collection of open sets whose intersection is a closed set. That alone does not prove the intersection is not open!
     
  6. Feb 21, 2009 #5
    wait!
    i thought that it should be metric space, because we need the notion of distance to define it. but
    is there any difference in other space?
     
  7. Feb 21, 2009 #6
    You don't need any notion of distance to define a topology. Halls is right, we need to know whether you're working in a specific space, ie., real analysis, or if this is an introduction to topology. What is the context of your question?
    You seem to be working in a metric space, where "open set" is defined as a set where every member is contained in an open ball that is also contained in the set and "closed" is defined by the complement of a set being open. Just showing that the singleton is closed is not enough to show that it is not open. For example, in all topologies, the entire space is both open and closed and in the discrete topology, every singleton is both open and closed. You have to negate the definition of open and show that the set satisfies the definition of "not open" (An alternative is to assume that the set is open and show that this leads to a contradiction).
     
    Last edited: Feb 21, 2009
  8. Feb 21, 2009 #7

    HallsofIvy

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    "Metric topology" is one specific type of topological space and we define "open sets" in any topological space. As maze said in his first response "the first statement you make is part of the definition of open sets, in general topology." It would have helped if you had given the definition of "open set" you are using. (I know several equivalent definitions even in metric topology.)

    I will guess that the definition of "open set" you are using (the most common) is "a set, A, is open if and only if all members of A are interior points of A". That, in turn, means that for any point p in A, there exist some neighborhood of p that is a subset of A: for some [itex]\delta> 0[/itex], the set [itex]\left{ q| d(p,q)< \delta\right}[/itex] is a subset of A.

    Suppose {[itex]A_{\alpha}[/itex]} is a collection of open sets and let A be their intersection. Let p be a member of A. Then p is a member of [itex]A_\alpha[/itex] for some [itex]\alpha[/itex] and some neighborhood of p is in [itex]A_\alpha[/itex]. Now, is that same neighborhood contained in A?

    You will still need to show that an intersection of open sets may not be open. Yes, you can use the example above, of which {An}= (-1/n, 1/n} so that the intersection is {0}, a closed set. But you still need to show that {0} is not an open set.
     
    Last edited: Feb 21, 2009
  9. Feb 21, 2009 #8
    As a little hint, note that you could construct _any_ nonempty set by taking the union of singleton points, and the infinite union of open sets is open.
     
  10. Feb 21, 2009 #9

    HallsofIvy

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    Still doesn't do it, maze! The "discrete topology", defined with the metric d(x,y)= 0 if x= y, 1 otherwise, has the property that ALL sets are both open and closed. The statement "the intersection of any collection of open sets is open" is TRUE in that space.

    The statement "The intersection of any collection of open sets is open in all topologies" is false but to specify a counterexample you will have to be careful to specify the metric.
     
    Last edited: Feb 21, 2009
  11. Feb 21, 2009 #10
    I didn't want to give away the whole thing, as people here tend to be a bit overzealous about not posting solutions to homework problems. However, if you take a second look at my post, you will find that it can be made completely rigorous if there exists even one set that is not open. The discrete topology is the only topology where this is not true, so it is somewhat irrelevant to the issue at hand.
     
  12. Feb 21, 2009 #11
    so the proof of closure is not enough.i should show that the intersection of these balls is not an open set.

    Thanks to all of you!
     
  13. Feb 22, 2009 #12

    HallsofIvy

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    The most important you will need to specify the metric. Fortunately you only need to give a counterexample. I would recommend showing that the intersection of (-1/n, 1/n) in the real numbers, with the "usual" metric: d(x,y)= |x- y|, is not an open set. I don't think I am giving away too much to state that the intersection is {0} and taking any [itex]\delta> 0[/itex] gives a neighborhood of 0 that is NOT a subset of that set.
     
  14. Feb 22, 2009 #13
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