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Intersection of nested subsets

  1. Jun 17, 2005 #1
    I'm just an interested laymn, and I'm trying to improve my knowledge in some areas where I'm weak. To this end, I found that Shilov's Elementary Real and Complex Analysis was highly recommended, and the Dover edition was available for only ten bucks, so how could I go wrong? But it didn't take me long to get stuck.

    On page 18, corollary 1.74, he says that in considering an infinite series of half open intervals of real numbers, each a half sized subset of the previous, the overall intersection can be empty.
    [tex](0,y] \supset (0,\frac{y}{2}] \supset ... \supset (0,\frac{y}{n}] \supset ... [/tex]
    This makes no sense to me. A set can only be a subset of another if it contains elements in common, so the intersection of two such sets cannot be empty. Since the reals are infinitely dense, even the smallest interval, no matter how many times you cut it in half, contains an infinite number of points, right? Isn't the intersection the entire contents of the latest subset, wich is never enpty? Plus, he gets this from Theorem 1.73: given arbitrary real numbers x>0 and y>0, there exists an integer n>0 such that y/n < x. I understand that, but don't see at all how it asserts the corollary. Help! I'm hesitant to go on without resolving this, lest I learn a misunderstandings or reach false conclusions. Does anyone here have this book? Can anyone clear up my confusion?

    Note that this isn't homework, and its not for any class. I'm just curious. Thanks!

    Fixed LaTex: direction of subset symbols
    Last edited: Jun 17, 2005
  2. jcsd
  3. Jun 17, 2005 #2


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    I don't know how to solve your problem- just two comments.
    Are you sure that's what it says? Shouldn't it be
    [tex](0,y] \supset (0,\frac{y}{2}] \supset ... \supset (0,\frac{y}{n}] \supset ... [/tex]
    ?? Maybe I'm crazy.
    The empty set is a subset of every set, and the intersection of the empty set and any other set is empty. :smile: This is because the definition of subset is: set S is a subset of set T iff every member of S is also a member of T. The empty set has no members, so it is a subset of every set.
    Anyway, this may give you an idea of how the statement could be true.
  4. Jun 17, 2005 #3
    Yes, thanks, my bad. I had the subset symbols backwards in LaTex. I fixed it in the original post so it wouldn't confuse others.

    I understand that the empty set is in the intersection, that's a given (actually, I do have a different question related to that, but one thing at a time). What I don't get is how the empty set is the whole of the intersection. Reducing the parent set by half never produces an empty set, no matter how often it's done, right?
  5. Jun 17, 2005 #4
    True, reducing an interval to half of it's size will not reduce it to the empty set. But that doesn't mean that the intersection of all the sets cannot be empty, because the intersection isn't formed by intersecting the sets one at a time.

    Basically, what you're pointing out is that if we take the first n sets in that sequence, then their intersection will be non-empty, and this is true for any n, no matter how large it is. But this still doesn't tell us that the intersection of all of the sets is non-empty; because the intersection of the first n sets just isn't the same thing, no matter how large n is.
  6. Jun 18, 2005 #5


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    I didn't mean that the empty set was in the intersection. I was just commenting on that one statement, pointing out that the intersection of a set and one of its subsets can be empty (when at least one of the sets is empty).
    I don't understand how the intersection is empty either, but I imagine it would have to do with the series being infinite. Have you tried a proof by contradiction? You have chosen some arbitrary y > 0 for your series. Assume there is some real number x that is in the intersection (so x > 0). Wouldn't this imply that x < y/n for all n?
  7. Jun 18, 2005 #6

    matt grime

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    yes, that proof is right. something is in the intersection iff it is in each interval, ie iff it is strictly greater than 0 and less than y/n for all n (y some fixed number) there are no such numbers, hence the intersection is empty.
  8. Jun 18, 2005 #7
    Yes, but I couldn't conclude that there's a contradiction in there. Isn't it also a contradiction to say that for arbitrary y in the positive numbers, there exists a positive n such that y/n is not a positive number? Isn't that also a contradiction? Alternately, given arbitrary positive x and y, is there a positive n so big y/n < x, but no m such that (y/n) m > x? I don't see why I can't define an x that's so small on one hand, if I can define an n so big that y/n can't have anything smaller on the other. If y/n exists it must be positive, so it can be multiplied. Maybe I'n not so sure that such an n exists, rather than saying that such an x cannot exist. But I'm not comfortable with any of those statements either.

    Shilov talks about the greatest lower bound, inf P, and it's clear that this bound does not have to be an element of the set P. Could we define as similar a bound that did have to contained in the set it was bounding, maybe call it cinf P? There's no problem if the interval of P is closed, but does the construct become meaningless if the interval is open? This doesn't feel quite right either.

    Another way of seeing my difficulty with the original idea is that reducing the range of the interval with every subset formed does nothing to reduce the number of elements contained within that interval, and the notion of the empty set is all about the number of elements contained, right? As long as there are an infinite number of points between any two arbitrary points in the interval, no matter how close those points are, then the intersection can never be empty. And given that there are an infinite number of points, removing one and only one point to make the interval open can't make suddenly make the number of points contained go to zero. To me, that seems like writing [itex]\infty - 1 = 0[/itex].

    Obviously, I'm having a lot of problems finding a way to see this through, and perhaps my difficulty is about the nature of open sets, since that seems to be the distinguishing feature of this problem. But I don't just want to take the conclusion of this theorem on faith. I won't do that. (Sorry if I sound emotional, but this is really bugging me)
  9. Jun 18, 2005 #8


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    We call that the minimum -- written min P. And yes, the minimum does not exist for an open interval. (That's the reason we use sups and infs; they always exist... at least if you use the extended reals)

    Anyways, you need to think in an orthogonal direction. The right question to ask yourself is:

    "What numbers occur in all of then intervals (0, y/n]?"

    What can you say about numbers that don't occur in all of them?
  10. Jun 19, 2005 #9

    matt grime

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    it would be but we aren't saying that, are we? you need to think about this, becuase it really is quite simple

    what's this got to do with anything?

    the simple answer is that you have not understood what an intersection is. once yuo do that it is quite straigh forward. go back and read the answers here and think about it and stop trying to over complicate it.

    to recap.

    let y be greater than 0, indeed, let us fix y=1

    consider the intersectionn

    [tex] \cap (0,1/n)[/tex]

    the intersection indexed over n in N.

    the intersection is empty; what element do you think might be in it?
  11. Jun 19, 2005 #10
    I do appreciate the effort all of you are puting into helping me with this. And I' msure it's reallysimple, but that's part of the problem. It's simplicity makes it fundamental, so I want to be sure and comfortable with the result. And it's not that I don't see the correctness to your arguments. It's just that I can't see the error to counterarguments, so I can't get rid of my reservations. If I could figure out why these aren't really contradictions, I'd feel much better.

    For example, what's the error in saying that for any arbitrary y>0 and n>0, there exist some x>0 such that x < y/n. Isn't that true? If it is true, it contradicts the earlier train of thought. Explicitely, is it because the arbitrary n must be some real actual number, and infinity doesn't meet that criteria, right?

    The x mentioned above, where y = 1. Isn't it true that there exists some real 0 < x < 1/n for arbitrary n? Granted, it's not a number I can easilly name. My problem may not be in the notion of intersection (though I might not really understand that as you suggested) but in the notion of infinity and infinite subsets. Infinity and infinite processes aren't simple for me to understand.

    What if I tried it the other way around? Is there a notation for removing subsets? For now, I'll use an overline like a compliment, to mean all of the original set except the subset. Then what is
    [tex](0,1] \overline{\supset} (\frac{1}{2},1] \overline{\supset} (\frac{1}{4},\frac{1}{2}] \overline{\supset} ... (\frac{1}{n+1},\frac{1}{n}] \overline{\supset} ...[/tex]
    Is that the same thing as the previous problem? Actually, I'm not sure that helps me at all, although the notation n+1 as n approaches infinity is a bit awkward. No, it doesn't help. As long as there exists some x < 1/n for arbitrary n, I've got the same problem.

    Is that my problem? Is the statement "there exists some 0 < x < 1/n for arbitrary n" incorrect?
  12. Jun 19, 2005 #11
    Yes, this statement is incorrect. Given any x > 0, you can always find a value n such that 1/n < x. There is no real number x such that x < 1/n for arbitrary n.
  13. Jun 20, 2005 #12


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    You need to be more careful about how you say things. Pay very close attention to the assignment and order of the quantifiers ('for all' & 'there exists'). There are four options:

    1) For all x > 0, there exists some n > 0 such that x < 1/n.

    2) For all n > 0, there exists some x > 0 such that x < 1/n.

    3) There exists some x > 0 such that, for all n > 0, x < 1/n.

    4) There exists some n > 0 such that, for all x > 0, x < 1/n.

    Here's some simpler examples to hit the idea home:
    For all x, there exists some y such that x = y.
    This means that for every x you choose, you can find some y that is equal to x. This is obviously true. The y that is equal to x is x itself. Let x = 1. There does exist some y such that x = y : 1. Let x = 45. There does exist some y such that x = y : 45. But try it the other way.
    There exists some x such that, for all y, x = y.
    This means that you can find some x that is equal to every y! Obviously not true. 1 doesn't equal every other number. 45 doesn't equal every other number...

    So which of the above do you think is true? Which false? See the difference?

    You might try replacing "for all x" with "for every x that you choose" and replacing "there exists some x" with "you can find some x" to help you get used to the difference between them.
    Another example:
    For all x, there exists some y such that y is x's mother.
    For every x that you choose, you can find some y such that y is x's mother. So it's just saying that everyone has a mother- but it isn't saying that everyone has the same mother. It just says I have a mother, you have a mother, Tom, Dick, and Harry have mothers, etc.
    There exists some x such that, for all y, y is x's mother.
    You can find some x such that, for every y that you choose, y is x's mother. So this says that someone is a son or daughter of every mother, i.e., every mother is this person's mother. So this person would be brother or sister to me, you, Tom, Dick, and Harry.

    Oh, I should make clear that 'there exists some' means there exists at least one. And this doesn't exclude the possibility that the 'some' could be all.
    Last edited: Jun 20, 2005
  14. Jun 20, 2005 #13

    matt grime

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    it isn't true; it is the defining cahracteristic of the real numbers: the only element of the positive reals less than 1/n for all n is 0.

    perhaps if you didn't use arbitrary arbitrarly and stuck to for alls and there exists you'd do bette; read honestrosewater's last pst.

    not in the reals as i have said; it is the archimedean property. this is one of those things that i meant when i should check you understand your terminology.

    yes, that is false.thoguh i'd prefer it "for all n"
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