Dense. A set $A \subset X$ is dense in $X$ if $\overline{A}=X$. Equivalently, $A$ is dense in $X$ if $A \cap N \neq \emptyset$ for any non-empty open set $N$ in $X$.
Nowhere dense. A set $A \subset X$ is nowhere dense in $X$ if $\text{int}(\overline{A})=\emptyset$.
Meagre. A meagre set is a countable union of nowhere dense sets.
1st category. Meagre.
2nd category. Not meagre.
Residual set. The complement of a meagre set.
Nowhere dense. A set $A \subset X$ is nowhere dense in $X$ if $\text{int}(\overline{A})=\emptyset$.
Meagre. A meagre set is a countable union of nowhere dense sets.
1st category. Meagre.
2nd category. Not meagre.
Residual set. The complement of a meagre set.
Theorem 1.1 (Baire Category Theorem I)
In a complete metric space, the intersection of countably many dense, open subsets is still dense.
In a complete metric space, the intersection of countably many dense, open subsets is still dense.
Proof: Take a sequence of nested open and closed balls, alternatingly. The centres of the closed balls give a Cauchy subsequence, if you shrink the balls fast enough, so there is a limit in the intersection by completeness. Equivalently, the intersection of a sequence of nested compact sets is non-empty in a complete metric space. $\square$
Theorem 1.2 (Baire Category Theorem II)
In a non-empty complete metric space $X$, the union of countably many closed, nowhere dense sets does not cover the whole space. Equivalently, if $X$ is the union of countably many closed sets, then one of those closed sets contains an open ball.
In a non-empty complete metric space $X$, the union of countably many closed, nowhere dense sets does not cover the whole space. Equivalently, if $X$ is the union of countably many closed sets, then one of those closed sets contains an open ball.
Proof: (using 1.1) Suppose $X=\bigcup^\infty F_n$, where the $F_n$ are closed. Set $U_n=X\backslash F_n$, which is open. Then $\bigcap^\infty U_n = \emptyset$, so one of the $U_n$ is not dense, from Theorem 1.1, which implies that one of the $F_n$ contains an open ball. $\square$
Proof of 1.1 assuming 1.2: This is fairly easy, using the observation that the complement of a nowhere dense set is a dense set, and vice-versa. In 1.1 we do not assume that $X$ is empty, so observe that if $X$ does happen to be empty, the intersection of any sequence of sets is dense, and so we are done. Suppose $U_n$ is a sequence of dense, open sets. Then $F_n=X\backslash U_n$ is a sequence of nowhere dense, closed sets. $\bigcap U_n = \bigcap (X\backslash F_n)= X\backslash \bigcup F_n \neq \emptyset$, where inequality follows from 1.2. We have currently only found one point in the intersection of the $U_n$, but this is another one of those cases where we are actually already done with no extra effort. Simply consider the subspace of $X$ given by the restriction of $X$ to any open ball to force the point we found to be where we want it. Thus, the intersection of the $U_n$ is in fact still dense. $\square$
Note that if you have any sequence of sets $F_n$ such that $X=\bigcup^\infty F_n$, you can apply BCT to their closures to conclude that they can't all be nowhere dense, strengthening the above statement slightly.
Remark 1.3.
The Baire Category Theorem is equivalent to the Axiom of Dependent Choice.
Applications of the Baire Category Theorem
| Result | Space | How? |
|---|---|---|
| Every complete metric space with no isolated points is uncountable. | Any complete space | Singleton sets are nowhere dense |
| There exist continuous nowhere differentiable functions. | $C[0,1]$, $\|.\|_\infty$ | The set of functions differentiable at some point is meagre. $Y_n = \{f \mid \exists x \in [0,1], \left|\frac{f(y)-f(x)}{y-x}\right| \leq n$, $\forall y \in [0,1]$, $y \neq x\}$ |
| Principle of uniform boundedness | Any Banach space. | $F_n=\{ x \in X \mid \|Tx\| \leq n, \forall T \in \tau\}$ |
| An infinite-dimensional Banach space has uncountable dimension | Any i.d. Banach space | $F_n=\text{span}\{e_1,\ldots,e_n\}$ |
| Open Mapping Theorem | $X$, $Y$ Banach spaces. $T$ surjective, continuous, linear. | $Y=\bigcup^\infty T(nB_x)$ |
| $f(nx) \to 0$ for each $x>0$, then $f(x) \to 0$ | $\mathbb{R}$ | $F_N=\{x \mid f(nx)\leq \varepsilon$, $\forall n \geq N\}$ |
Here is some interesting mathematics that we didn't cover in lectures, but which is easily accessible using the notions we have developped. Baire introduced a classification of functions on the real line as a way of measuring how discontinuous they are.
Define the set of Baire class zero functions to be the set of continuous functions. Define the set of Baire class $n$ functions to be the set of functions that are the pointwise limit of a sequence of Baire class $n-1$ functions.
The Baire class $n$ is a vector space[citation needed]. Several of the results that we have encountered (and a couple of extra-curricular, but relevant ones) can be rephrased rather pleasingly in terms of Baire classes, and $G_\delta$ language.
Theorem 1.4 The set of points of discontinuity of a Baire class one function is meagre. (See Ex2Q4+12.)
Theorem 1.5 The set of points of continuity of any function is a $G_\delta$ set. (See Ex2Q11.)
Theorem 1.6 The derivative of a differentiable function is Baire class one.
Culminating in a result which I promised someone on stackexchange I would get around to proving:
Theorem 1.7 The graph of a derivative is connected. (More generally, so is the graph of any function that is simultaneously Baire class one, and Darboux.)
Question: what's an example of a Baire function that is not in any Baire class? Note that some sources extend to the definition of Baire classes to include a countable ordinal in the index. Perhaps the indicator function of a non-measurable set might do the trick?
[I am planning to come back to this post to include proofs of these results, and perhaps add some extra results. Example sheet 2 was a rough time for most of us, but it's incredibly rich - I'm gonna mine it for theorems harder than any Minecrafter ever mined for diamonds.]
