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Infinite sets?

  1. Apr 6, 2007 #1
    1) Is an infinite intersection of open sets not open?
    2) Is an infinite union of closed sets not closed?

    3)But an infinite union of open sets is open.
    4) And an infinite intesection of closed sets is closed.

    This information is all correct?

    If so could you provide reasoning to them especially the first two?

    For 1) Is it because an infinite intersection of open sets could be a single point. This set which contains only a single point and so there is no 'ball' that can be drawn in this set with a non zero radius with the point as its centre. Since that point is the whole set.

    But I have been too specific there may be cases where the infinite intesection may contain more than 1 point.
    Last edited: Apr 6, 2007
  2. jcsd
  3. Apr 6, 2007 #2


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    Technically, they're just axioms. That is, a topology on a set X is a collection T of subsets of X such that:

    1. The whole set X and the empty set are in T.
    2. Any union of subsets in T is in T.
    3. Any finite intersection of subsets in T is in T.

    The sets in T are called the open sets, and their complements are called the closed sets. Equivalently, you can define things in terms of closed sets, in which case "union" and "intersection" would switch places (since the complement of a union is the intersection of the complements, and vice versa), and open sets would then be the complements of closed sets.

    But then you can just ask why we picked this definition. An open set is one which, roughly speaking, has some wiggle room about each point. Now, any point in a union of open sets is in one of the open sets, and so has wiggle room in that open set, and so also in the union. Similarly, a point in a finite intersection of open sets is in every one of the open sets. Thus it has wiggle room in every one of the open sets, and so since there are only finitely many of them, there is one in which it has the least amount of wiggle room, and it will have exactly that much in the intersection. Let me know if that's understandable.

    For example, take the sets (-1/n,1/n) in R. The intersection of all of these as n runs from 1 to N is just (-1/N,1/N). The point 0 has at least "2N of wiggle room" in each of these sets, so has this much in their intersection. However, if we take the infinite union over all natural numbers n, we get {0}, a non-open set. There is no minimum amount of wiggle room about 0 in these open sets, since they get arbitrarily small (this is the key), so even though it is in the intersection, it can't be moved around without moving out of the intersecion, so the intersection isn't open. I don't think I've said wiggle more times in my life.

    You could also think of it in terms of closed sets, and I'll let you try to make this argument. But again, this just explains what motivated the above definition, and at the end of the day, that definition is what you have to answer to.
    Last edited: Apr 6, 2007
  4. Apr 6, 2007 #3


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    Similar to StatusX's example of (-1/n, 1/n) for all positive integers n as an infinite collection of open sets whose intersection is not open:

    [1/n, 1-1/n] for all positive integers n is a collection of close sets whose union, (0, 1), an open set.
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