The Baire Category Theorem

I doubt many of you are familiar with the Baire Category Theorem (BCT) -on second thought, you might; I mean, you have taken the time to read this article (I am guessing that only a subset of a certain group of people read random math/philosophy blogs in their spare time). I have been involved in some work on the BCT lately. I have, honestly, had trouble cracking it. What I mean by ‘cracking’ is that I have not had a clear moment of lucidity – you know? I mean, I have not had one of those moments where it becomes clear and intuitive. Let’s start out by stating some definitions of the BCT and try to pin down what “it” is an why ‘it’ is important to us. Often when the BCT is introduced to the student by way of the Banach-Mazur game. A certain version of this game is as follows (taken from Elementary Real Analysis by Bruckner and Thomson):

Player A is given a subset A $\subset \mathbb{R}$ and player B is given the complimentary set B= $\mathbb{R} \backslash A$ Player A first selects a closed interval $I_{1} \subset \mathbb{R}$ ; then player B chooses closed interval $I_{2} \subset I_{1}$ The players, then, alternate moves. They select a closed interval inside of the previously chosen interval. The way the game proceeds determines a descending sequence of closed intervals

$I_1 \subset I_2 \subset \cdots \subset I_n \subset \cdots$

Where player A chooses those with an odd index and B chooses those with an even index (this is just because of how they picked ):

$A\cap \bigcap\limits_{n=1}^{\infty}I_n \neq\emptyset$

Then, we say, that player A wins and B looses; otherwise, if the intersection would have been empty, B would have won.

What does this say about the BCT? Well, first, we can assert that $\mathbb{R}$ cannot be expressed as a countable union of nowhere dense sets -which can be rephrased as: a union of nowhere dense sets cannot “fill up” and interval. We can take a moment and think of the Cantor argument for the reals being uncountable; Cantor’s theorem that no interval (a,b) can be the range of some sequence. Basically, we can just rephrase the later as: the union of nowhere dense sets cannot complete some interval (a,b). Second, the compliment of a countable union of nowhere dense sets is dense.

So, lets formalize this to ensure greater lucidity.

Defintion #1: Let A be the set of real numbers.

A is said to be of the first category if it can be expressed as a countable union of nowhere dense sets.
A is said to be of the second category if it is NOT of the first category.
A is said to be residual in $\mathbb{R}$ if the compliment $\mathbb{R}\backslash A$ is of the first category.

What does it mean to be in the first category? Mathematicians denote any set that is in the first category as a meager set -that is, the set is said to be small or negligible. Some facts of meager sets include the notion that every meager set has an empty interior, but, you may recall that a nowhere dense set is a set whose closure has an empty interior. The result follows: if something is nowhere dense, then it is necessary that it is meager. However, that does not mean that if something is meager it is nowhere dense. Think about the rationals… they are indeed dense in $\mathbb{R}$ , yet they can be expressed as the countable union of nowhere dense sets.

Let me back-track for a moment. We are including much information on the idea of dense sets, so I feel that we need to make sense of what it means for a set to be dense.

Definition -Dense: The set of real numbers A is said to be dense in $\mathbb{R}$ if for each open interval ${(a,b}$ the set $A\cap{(a,b)}\neq\emptyset$

Another way to think about this is: suppose that A is said to be dense in some metric space ${X}$ . Then $\forall{x}\in {X}\rightarrow x\in{A}$ or ${x}$ is an accumulation point of A. So, either all of our points in ${X}$ are in A or they are all arbitrarily close to some point in A. So, the real numbers are either a rational or arbitrarily close to one.

BCT: Every residual subset of $\mathbb{R}$ is dense in $\mathbb{R}$ . Proof: Consider X a nonempty complete metric space. If

$X= \bigcup\limits_{n=1}^{\infty}A_n$

where $A_n$ is closed. Then at least one $A_n$ contains a non-empty open subset.

If we assume that $A_n$ do not contain a non-empty open subset (we are proving by contradiction), then all we do is proceed by constructing a sequence that converges to a point that is not in $A_n$ so that it contradicts our hypothesis of what X is (which is a complete metric space, that is, that every Cauchy sequence in X converges to a point that is also in X).

So, suppose we have $A_1$ , by our assumption, $A_1$ does not contain an open set J, but its compliment $A_1^{c}$ must. Thus, there must exist an $x_1\in X$ and some c so that for 0<c,<1, we have,

$B_{c_{1}}(x_{1})=(x_1-c_{1},x_1+c_{1})\subset A_1^{c}$

The same goes for some $x_2$ not in $A_2$ so it must be that $x_2 \in A_2^{c}$ . More importantly, the point $x_2$ must be in $({A_2^{c}} \cap (B_{c_{1}}(x_{1}))$ , and we can find some smaller interval in $B_{c_{1}}(x_{1})$ such that $c_2<1/2$ so that

$B_{c_{2}}(x_{2}) \subset ({A_2^{c}} \cap B_{c_{2}}(x_{2}))$

We can continue this process to create a whole bunch of nested balls,

$B_{c_{1}}(c_2)\supset B_{c_{2}}(x_2)\supset B_{c_{3}}(x_3)\supset\cdots\supset B_{c_{n}}(x_{n})$

The sequence that he have created is Cauchy –where $|x_n-x_m|<\frac{1}{x_N}$ for $n,m>\mathbb{N}$ So, there must be some point, by way of completeness, so that $x_{\alpha}\in X$ (specifically, $x_{\alpha}\in B_{c_{n}}(x_{n})$ ) where $x_{n} \rightarrow x_{\alpha}$ for $n\rightarrow\infty$ but this commits us to a contradiction; we previously assumed, by hypothesis, that $x_{\alpha}\not\in A_{n}$

This contradicts the fact that,

$X=\bigcap\limits_{n=1}^{\infty}A_n$

We can now rephrase the BCT as:

Theorem #1 (BCT): Let $A_1, A_2, A_3, \ldots$ be at most a countable sequence of subsets of a complete metric space (or topological space) X. if $\bigcup\limits{n=1}^{\infty}A_n$ contains some open set J (i.e. a ball), then at least on of the $A_n$ is dense in a sub interval J’ or (a sub ball B’).

Think of this from the contrapositive: the countable union of nowhere dense sets cannot contain a ball. The reason for this is fairly obvious if one looks at what it means to be nowhere dense while considering our description of the Banach-Mazur game. Basically, our intersection is empty for all our subintervals chose.

From Theorem #1, we have the following corollary:

Corollary #1: Suppose that A is a subset of some X where X is of the second category. We have,

$X= \bigcup\limits_{n=1}^{\infty}A_n$

Where $A_n$ are closed. Then, it follows, that there is some $A_n$ that must contain an open interval.

Proof: by the BCT, at least the is some $A_n$ that is not nowhere dense, so the closure of $A_n$ cannot have an empty interior (see proof of BCT) -so there must be some (c-x,c+x) in $A_n$ . Since $A_n$ is closed, the open interval is in $A_n$ .

What is nice about the BCT is that it provides us with the conception about how big sets are -so big that if we have a residual set where we have removed all nowhere dense sets, then you would still be able to find some interval that contains any point of any interval in the set in question.

Nonetheless, while I am able to prove these theorems, actually seeing the BCT and its implications is fairly difficult for me. My best occurs when I consider the fact that,

$X= \mathbb{R}\backslash\bigcup\limits_{n=1}^{\infty}Q_n$

where $Q_n$ is a sequence of nowhere dense sets, so X, then, is a complete metric space and is residual. So, A, from our Banach-Mazur game, selects $I_1\subset [c-x,c+x]$ while avoiding any part of $Q_n$ and because we can continue this, X must contain a point of ANY interval. Thus, the intersection of residual sets must always be dense.

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John Piccolo

"…..no firm demonstration can be made from the success of hypotheses." -G.W. Leibniz

The Baire Category Theorem

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