How do you prove the Baire Category Theorem?

Answer 1

The Baire Category Theorem is, in my opinion, one of the most incredible, influential, and important results from any field of mathematics, let alone topology. It is known as an existence theorem because it provides the necessary conditions to prove that certain things must exist, even if there aren't any examples of them that can be shown. The theorem was proved by René-Louis Baire in 1899 and is a necessary result to prove, amongst other things, the uniform boundedness principle and the open mapping theorem (two of the three most fundamental results from functional analysis), the real numbers being uncountable, and the existence of continuous, yet nowhere differentiable, functions from R to R.

The proof is quite long and involves some pretty advanced math, so to help with the reader's comprehension there is a list of symbols and their meanings at the end of this proof. Also, I've added many related links with definitions and explanations of the terms used in this proof.

The Baire Category Theorem:

If B, D is a nonempty, complete metric space, then the following two statements hold:

1) If B is formulated as the union of countably-many subsets, C1, C2, …, Cp, then at least one of the Cp is somewhere dense.

2) If A1, A2, …, Ap are countably-many, dense, open subsets of B, then ∩pAp is dense in B, i.e. Cl(∩pAp) = B

Proof:

1) If the first statement is false, then there is a countable family {Cp}, p Є P, of subsets of B such that B = ∪pCp, but (Cl Cp)o = Ø for each p Є P. Therefore, for each p, Cl Cp≠ B. Select b1 Є B - Cl C1. There is a positive number m1 < 1, since B - Cl C1 is open, such that N(b1, m1) ⊂ B - Cl C1. Now we set G1 = N(b1, m1/2). Then Cl G1 ⊂ N(b1, m1); hence Cl G1 ∩ Cl C1 = Ø.

Since G1 is a nonempty, open subset of B, that means G1 ⊄ Cl C2. So, choose a b2 Є G1- Cl C2. Since G1 - Cl C2 is open, there is an m2 > 0 such that N(b2, m2) ⊂ G1 - Cl C2. This time we'll require m2 < 1/2 and then set G2 = N(b2, m2/2). Then G2 ⊂ G1 and Cl G2 ∩ Cl C2 = Ø.

If we continue on like this, requiring m3 < 1/3, m4 < 1/4, etc., we'll obtain a decreasing sequence of mp-neighborhoods, G1 ⊃ G2 ⊃ G3 ⊃ … ⊃ Gp ⊃ … such that Cl Gp ∩ Cl Cp = Ø and mp < 1/p. Then Cl G1 ⊃ Cl G2 ⊃ Cl G3 ⊃ … ⊃ Cl Gp ⊃ … and d(Gp) --> 0.

I'm going to use a result from another theorem in topology, not proven here, which says that if G1 ⊃ G2 ⊃ G3 ⊃ … ⊃ Gp ⊃ … , d(Gp) --> 0, and ∩pGp ≠ Ø, then the metric space B, D is complete. Therefore, ∩p Cl Gp ≠ Ø. So, if we pick a g Є ∩pGp, then g Є Cp for some p, since ∪pCp = B. However, that would imply that g Є Cl Cp ∩ Cl Gp which is impossible because Cl Cp and Cl Gp are disjoint. So, for 1), Q.E.D.

2) To start, we're going to suppose that {Ap}, p Є P, is a countable family of dense, open subsets of B. To prove that ∩pAp is dense, all that we need to prove is that every neighborhood of any element of B meets ∩pAp. In other words, for any selected g Є Band any m > 0, we'll show that N(g, m) ∩ (∩pAp) ≠ Ø.

If we set T = Cl N(g, m/2), then T ⊂ N(g, m). Now we'll show that T ∩ (∩pAp) ≠ Ø. We know that T is a subspace of the closed metric space, B, D, and that Titself is closed. So, using an earlier theorem that won't be proved here, T is a complete metric space. If we set Gp = T - Ap which is equal to Tp ∩ (B - Ap), we see that the intersection of two closed subsets of B, Gp is closed in both B and T.

Now suppose Gp is somewhere dense. Then there is an element t Є T and a number q > 0 such that N(t, q) ∩ T ⊂ Cl Gp ∩ T = Gp. Therefore, N(t, q) ∩ (T - Gp) = Ø. We can see that t Є T = Cl N(g, m/2). Therefore N(t, q) meets N(g, m/2) at some point z. We then choose q' > 0 such that N(z, q') ⊂ N(t, q) ∩ N(g, m/2). However, since Ap is dense, N(z, q') intersects Ap at a point we'll call z' . Well, then it must be that z' Є N(t, q) ∩ T ⊂ Gp. But, Gp= T - Ap, hence z' Є T - Ap. That implies then that z' ∉ Ap which is a contradiction. Therefore Gp must be nowhere dense in T.

So, by the first statement of the theorem, 1), which we already proved, T ≠ ∪pGp. Thus, there is n element s Є T - ∪pGp. Therefore, since Gp = T - Ap, then s Є T ∩ (∩pAp) and so T∩ (∩pAp) ≠ Ø meaning N(g, m) ∩ (∩pAp) ≠ Ø.

Q.E.D.

List of symbols:

R - The set of real numbers, including rational, irrational, positive, and negative numbers, as well as 0. Not including complex numbers having an imaginary part other than 0i, where i is the imaginary number √(-1).

B, D - The metric space of set B with metric D.

∩pAp - The intersection of all of the subsets A1, A2, …, Ap.

Cl - The closure of whatever set is written after it.

p Є P - p is an element of the set P.

P - The set of all positive integers, not including 0. This set is often referred to as the set of natural numbers and is labeled N, but since at times the natural numbers are said to include 0, I've labeled this set P to avoid ambiguity. Not to mention, I've used N within my label for neighborhood.

∪pCp - The union of all of the of the subsets C1, C2, …, Cp.

( )o - The interior of whatever set is in the parentheses.

Ø - The empty set; i.e. the set with nothing in it.

N(b1, m1) - The neighborhood of point b1 within distance m1.

⊂ - … is a subset of …

⊃ - … is a superset of …

d( ) - The diameter of whatever set is in the parentheses.

--> 0 - The limit of whatever comes before the symbol "-->" goes to 0.