Skip to content

Baire category theorem

Theorem (Baire's category theorem)
If XX is a non-empty complete metric space such that
X=⋃n=1∞Cn,\begin{equation*}X = \bigcup_{n=1}^{\infty} C_n ,\end{equation*}
where the sets CnC_n are closed. Then at least one of them contains an open ball of positive radius.
Baire category theorem is a fundamental result in Functional Analysis about the characterization of complete metric spaces with various applications.
The dots on the first-category side represent a countable set, such as Z2\mathbb{Z}^2.
In the late nineteenth century, Baire introduced in his doctoral dissertation a notion of size for subsets of the real line which has since provided many fascinating results. In fact, his careful study of functions led him to the definition of the first and second category of sets. Roughly speaking, sets of the first category are 'small,' while sets of the second category are 'large.'
The word “category” in the name of Baire’s theorem is explained by the following terminology: sets that are countable unions of nowhere dense sets are called first category sets and all other sets are called second category sets. Baire’s theorem asserts that a complete metric space is not a first category set.
Before the formulation of Baire's category theorem we need the two definitions.
Definition (Dense set)
A set AA is called dense in a metric space if the closure of AA is the whole space.
Definition (Nowhere dense set)
A set EE is called nowhere dense in a metric space XX if every nonempty open set in XX contains a ball of a positive radius without points of EE.
Let's consider examples of dense sets on R\mathbb{R}.
1. Rational numbers Q\mathbb{Q}
Between any two real numbers, you can always find a rational number. For example, between 1 and 2, you have 32,43,53\frac{3}{2}, \frac{4}{3}, \frac{5}{3}, etc.
1. Any finite set a1,a2,...,an{a_1, a_2, ..., a_n}. For any point in this set, there's an open interval around it containing no other points from the set
2. The set Z\mathbb{Z} of integers. Between any two integers, there's an open interval containing no integers.

The foolowing proof is taken from [Bogachev] pages 15

Two first nested balls in the proof of Baire's category theorem.
\quad Proof. We can assume that the first set C1≠∅C_1 \neq \emptyset since they cannot all be empty and dropping any empty sets does no harm. Let's assume the contrary of the desired conclusion, namely for any open ball UU there is an open ball Bn⊂UB_n \subset U disjoint with CnC_n for any nn.
So, there exist a closed ball B(p1,ε1/2)B(p_1, \varepsilon_1/2) with some radius ε>0\varepsilon > 0 and center p1p_1, which is disjoint with C1C_1, i.e. B(p1,ε1/2)∩C1=∅B\left(p_1, \varepsilon_1/2 \right) \cap C_1=\emptyset.
Next, we can choose p2∈B(p1,ε1/4)p_2 \in B\left(p_1, \varepsilon_1/ 4 \right), such that B(p2,ε2)∩C2=∅B\left(p_2, \varepsilon_2\right) \cap C_2=\emptyset.
So, inductively there is a sequence pi,i=1,…,kp_i, i=1, \ldots, k and ball B(pk,ε1/2k)B\left(p_k, \varepsilon_1 / 2^k\right) and B(pk,ε1 2k)∩Ck=∅B\left(p_k, \varepsilon_1 \ 2^k \right) \cap C_k=\emptyset.
Thus, we have a sequence {pk}\left\{p_k\right\} in XX. Since d(pk+1,pk)<ε1/2kd\left(p_{k+1}, p_k\right)<\varepsilon_1 / 2^k this is a Cauchy sequence, in fact
d(pk,pk+l)<ε12k\begin{equation*}d\left(p_k, p_{k+l}\right)< \frac{\varepsilon_1}{2^k}\end{equation*}
Since XX is complete the Cauchy sequence converges to a some point q∈Xq \in X, but by construction the point qq not belonging to the union of CnC_n. So we get the condrattion
q∈Xbutq∉⋃k=1nCn\begin{equation*}q \in X \quad \text{but} \quad q \notin \bigcup_{k=1}^n C_n\end{equation*}
This is the desired contradiction to the statement of theorem. Thus, at least one of the CnC_n must have an open ball of positive radius. □\Box
Theorem (The Uniform Boundedness Principle)
Suppose that XX is a complete metric space and continuous functions fn:X→Rf_n : X \to \mathbb{R} are such that for every x∈Xx \in X, the sequence {fn(x)}\{f_n(x)\} is bounded. Then there exists a closed ball BB of some positive radius such that sup⁡nsup⁡x∈B∣fn(x)∣<∞\sup_n \sup_{x \in B} |f_n(x)| < \infty.
\quad Proof. Let XN:={x∈X:sup⁡n∣fn(x)∣≤N}X_N := \{x \in X : \sup_n |f_n(x)| \leq N\}. These sets are closed by the continuity of fnf_n. By hypothesis,
X=⋃N=1∞XN.\begin{equation*}X = \bigcup_{N=1}^{\infty} X_N .\end{equation*}
According to Baire's theorem, some XNX_N has inner points and hence contains a closed ball BB of a positive radius. □\Box

References