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The Baire Category Theorem (RUDIN)

  1. Sep 27, 2007 #1
    Theorem: If R^k=countable union of closed sets F_n_ then there is one closed F_n_ with nonempty interior.

    Equivalent statement: Let G_n_ be a open dense subset of X then take a countable intersection of these G_n_'s, the intersection is nonempty in fact it is dense in R^k.

    So, i'm really not sure where to begin. This is a special case of Baire's theorem and is given as the last problem on chapter 2 of Rudin's PMA. Rudin's hint tells me imitate the proof of theorem 2.43 which proves that every perfect set in R^k is uncountable.

    Can anyone give me a hint on where to begin with the hint!

    It's easy to show that any finite intersection of dense open subsets is also open and dense in R^k. That's as far as I've gotten.

    Please do not give the problem away.
  2. jcsd
  3. Sep 27, 2007 #2


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    I'm assuming you're looking for a proof of the theorem? Also, in your equivalent statement, what is X? Is it R^k?

    Anyway, to prove the theorem, suppose that all the F_n's have empty interior. Then there exists an open subset of R^k that sits outside F_1. In particular, we can assume it's an open ball B_1. Consider the concentric closed ball C_1 of 1/2 the radius of B_1. Now, because F_2 has empty interior, then we can find an open ball B_2 that's sitting inside the interior of C_1 but outside of F_2. Consider the concentric closed ball C_2 of 1/3 the radius of B_2. Repeat the same process for F_3, and so on. In the end you'll get a decreasing sequence {C_n} of 'nice' closed balls.

    Try to fill in the details and take it from here. I'm going to assume you're familiar with Cantor's intersection theorem.

    Getting the equivalent statement is easy once you have the theorem. Think De Morgan's laws. I'll let you think about how to get from it to the theorem.
    Last edited: Sep 27, 2007
  4. Sep 27, 2007 #3


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    the point is to use completeness of the metric space.
  5. Sep 27, 2007 #4
    you gave me more information than i wanted! HA! It was my choice to peek though ;).

    Pf: Assume each F_n has empty interior.

    Construction of closed compact C_n:

    F_1 has empty interior so F_1 compliment is nonempty and open b/c F_1 is closed, so there is a neighborhood, N(x), contained in F_1 compliment. Let C_1 be the closed nhbd centered around the same pt. as N(x) with radius 1/2 of N(x). Having picked C_1,...C_n, consider F_n+1, if C_n's interior contains no neighborhood that is disjoint with F_n+1, i.e a nhbd that is contained in F_n+1 compliment, then pick any x in F_n+1 and every nhbd of x intersects F_n+1 which means x is a lim pt of F_n+1 and therefore contained in F_n+1 b/c it is closed this means C_n is inside F_n+1, which contradicts the assumption that F_n has empty interior for all n. So there must exist a neighborhood interior to C_n that is contained in F_n+1 compliment, call it N, then there is a nhbd centered at the same point as N with radius 1/2 the radius of N, the closure of this nhbd is C_n+1 and because it is contained in N it is disjoint with F_n+1. The construction is complete.

    If x is in C_n, then x is contained in a nhbd inside C_n-1, moreover this neighborhood and F_n are disjoint. So C_n-1 contains C_n for all n and C_n is contained in F_n compliment.

    The intersection of C_n is contained in the intersection of F_n compliment. The latter is empty because it's compliment is R^k. C_n is compact b/c it is closed and bounded for all n. Cantors intersection thm gives a contradiction.

    why make the radius of C_n (1/n+1)th the amount of the neighborhood it's in? 1/2 is more than enough isn't it? Maybe i'm not seeing something.

    Until the last two lines I thought I had proved the general case but I realized I needed to use Heine-Borel.
    Last edited: Sep 27, 2007
  6. Sep 27, 2007 #5


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    It doesn't matter really, as long as you get lim radii = 0.
  7. Sep 28, 2007 #6
    do you even need the radii to goto zero? As long as they are contained in each other you still get that their intersection must be nonempty as long as every finite subcollection of compacts has nonempty intersection. The contradiction is thus still obtained.
  8. Sep 28, 2007 #7


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    That indeed seems to work. On the other hand, the easiest way to make sure they sit inside each other is to let the radii go to zero.
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