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Baby rudin theorem 2.36 explanation

  1. Jun 5, 2012 #1
    Theorem 2.36 says that given a collection of compact subsets of a metric space X such that the intersection of every finite subcollection is nonempty, then the intersection of the entire collection is nonempty. The proof is very simple and I easily follow the abstract reasoning. However, I think that there is a deeper intuition behind this example which I cannot quite seem to figure out.

    Could anyone provide a more concrete explanation as to why it is the property of compactness that results in this property?
  2. jcsd
  3. Jun 5, 2012 #2


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    that is not a theorem, that is an equivalent restatement of the definition. rudin has a wonderful way of making trivial things seem hard. i really dislike that book.

    i.e. the definition of X compact says: if U Xi = X, Xi open, then for some finite subset Xj of the Xi, UXj = X.

    But two sets are equal iff their complements are equal, So this says:

    if complem( U Xi) = complem X, then for a finite subset Xj of Xi, complem(UXj) = complem

    But the complement of a union is the intersection of the complements, thus:

    if intersection(complem.Xi) = ø, then for some finite subset Xj of the Xi,
    intersection(complem.Xj) = ø.

    Now the contrapositive of this is:

    if for every finite subset Xj of the Xi we have intersection(complem.)Xj ≠ ø,

    then for the full collection Xi, we have intersection(complemen.Xi) ≠ ø.

    But a set is open iff its complement is closed, so this becomes:

    if Zi is a collection of closed sets, such that: for every finite subset Zj of the Zi we have
    intersectionZj ≠ ø,

    then for the full collection Zi, we have intersectionZi ≠ ø.

    so this is just a really tedious and silly way of rendering a trivial statement as hard to grasp as possible. can't you find a book you like better than that one? or is some analysis professor torturing you by assigning that book to read?
  4. Jun 5, 2012 #3

    The above characterization of compact set is not really that hard/messy after one has already developed some topological abstract

    intuition, and it is very necessary for the proof I know of Tychonoff Theorem, one of the jewels in topology.

  5. Jun 5, 2012 #4

    I agree with this.. although I did like Rudin at one point, I think it was just the thrill of the first chapter being kinda neat.
    I've never really found a decent book for that leve of analysis so I'm lacking a certain amount of groundwork there when I do stuff at a slightly higher level.

    Would you be kind enough to reccomend a better rudin replacement?
    Pretty please :blushing:
  6. Jun 6, 2012 #5


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    we have often recommended analysis books other than Rudin. anything by Sterling K Berberian, Foundations of modern analysis is also hard to read, but much more rewarding than Rudin, and other people younger than I are probably better at suggesting more recent books. Search the thread on books, or the who wants to be a math guy thread.

    In fact it looks to me as if the first 8 chapters of rudin are just at the level of a good single variable calculus book. so i would suggest the two books of spivak, or the two volumes of apostol, or the two volumes, or even just the first volume, of lang's two books, i.e. analysis I. I also like fleming's calculus of several variables.

    But just because I dislike rudin, you may still get some benefit from it, especially if you enjoy the slick way he does some things.

    And DonAntonio is right, that theorem 2.36 is really very easy in a sense. I.e. if you understand complementation well, and contrapositives, those two statements are as I said, exactly equivalent. I just don't care for someone pretending as rudin does that this is a significant result when it is only a trivial restatement, with many double negatives, of the same thing.

    I just think that it should be presented in a way that makes clear that it is trivial, not calling it a theorem. and kelley gives two proofs of tychonoff, one using each version.
    Last edited: Jun 6, 2012
  7. Jun 6, 2012 #6


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    Thanks to DonAntonio's remark, I may have an idea to explain why rudin includes this result, but i cannot explain why he gives no intuition for it, that is his trademark.

    In my whole career I have only used the open set version, but that is because I am using compactness instead of proving it. Once you prove something like Tychonoff's theorem, say for me 45 years ago, you never prove it again, you just apply it.

    So in my experience one essentially always prefers the open set version of compactness when applying the property. But perhaps when proving compactness, one often may want to use proof by contradiction, and that's where the closed set version could come in.

    I.e. if we want to prove compactness by contradiction, as most proofs of Tychonoff seem to do, then we start by assuming we have a family of open sets such that no finite subset covers. I.e. for every finite subcollection there is a point outside those finite sets, or equivalently in the complement of those finite sets, i.e. in all their complements, thus in the intersection of their (closed) complements.

    So the contradiction of the conclusion of compactness is that we have a collection of closed sets, namely the complements of the original family, such that every finite subcollection of these closed sets has a point in common. Then if we can prove all the sets have a point in common, we have proved the contrapositive of compactness, which is equivalent to compactness.

    So Rudin's 2.36 is just the statement of the contrapositive of the definition, but made a little more complicated by also considering the complements of the sets as well as the contrapositive of the logical statement.

    But of course he doesn't say so. Notice that although pages 36-37 of rudin do have some explanation and intuition, pages 38-40 have essentially none, just a list of theorems and proofs.

    His fans consider this "elegant" but I consider it lazy and unhelpful writing. At least it is unhelpful to beginners, although if you already know the material, as professional analysts who prefer this book do, then it may be useful to see everything laid out efficiently and briefly. But I think his book is aimed at learners, of which only a few seem to benefit.
    Last edited: Jun 6, 2012
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