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Baire category theorem

BAIRE'S CATEGORY THEOREM
If XX is a non-empty complete metric space, then XX is not of first category in itself. Equivalently, if
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.
In the late nineteenth century, Baire introduced a topological notion of smallness that turned out to be fundamental in analysis.
To make this idea precise, we first recall two basic operations on a set: interior and closure. A point is in the interior of AA if a small open ball around that point stays completely inside AA.
Interior and boundary of a set $A$
Interior and boundary of a set AA
INTERIOR AND CLOSURE
The interior of AA is
A∘:={x∈S:∃r>0 such that B(x,r)⊂A}.\begin{equation*}A^{\circ}:=\{x\in S: \exists r>0\text{ such that }B(x,r)\subset A\}.\end{equation*}
The closure of AA is
A‾:={x∈S: for every r>0, B(x,r)∩A≠∅}.\begin{equation*}\overline{A}:=\{x\in S:\text{ for every }r>0,\ B(x,r)\cap A\neq\varnothing\}.\end{equation*}
We say that a set EE is nowhere dense in XX if every nonempty open set in XX contains a nonempty open ball disjoint from EE.
NOWHERE DENSE SET
A set E⊂XE \subset X is called nowhere dense in XX if
(E‾)∘=∅.\begin{equation*}(\overline{E})^{\circ}=\varnothing.\end{equation*}
A typical example of a nowhere dense set in R\mathbb{R} is the set of integers Z\mathbb{Z}. A countable union of such sets is called of first category.
The dots on the first-category side represent a countable set, such as Z2\mathbb{Z}^2.
FIRST AND SECOND CATEGORY
A set A⊂XA \subset X is called of first category if it can be written as
A=⋃n=1∞En,\begin{equation*}A=\bigcup_{n=1}^\infty E_n,\end{equation*}
where each EnE_n is nowhere dense in XX.
A set is called of second category if it is not of first category.
Baire's theorem says that a complete metric space is never of first category.
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. Assume for contradiction that no set CnC_n contains a nonempty open ball.
So, there exists a closed ball B‾(p1,ε1/2)\overline{B}\left(p_1, \varepsilon_1/2\right) with some radius ε1>0\varepsilon_1>0 and center p1p_1, which is disjoint with C1C_1, i.e.
B‾(p1,ε1/2)∩C1=∅.\begin{equation*}\overline{B}\left(p_1, \varepsilon_1/2 \right) \cap C_1=\emptyset.\end{equation*}
Next, we can choose p2∈B(p1,ε1/4)p_2 \in B\left(p_1, \varepsilon_1/4 \right) and ε2>0\varepsilon_2>0 such that
B‾(p2,ε2)⊂B(p1,ε1/4)andB‾(p2,ε2)∩C2=∅.\begin{equation*}\overline{B}\left(p_2, \varepsilon_2\right) \subset B\left(p_1, \varepsilon_1/4 \right)\qquad \text{and} \qquad\overline{B}\left(p_2, \varepsilon_2\right) \cap C_2=\emptyset.\end{equation*}
Proceeding inductively, we obtain points pnp_n and radii εn>0\varepsilon_n>0 such that the closed balls B‾(pn,εn)\overline{B}\left(p_n,\varepsilon_n\right) are nested, their diameters tend to 00, and
B‾(pn,εn)∩Cn=∅for all n≥1.\begin{equation*}\overline{B}\left(p_n,\varepsilon_n\right) \cap C_n=\emptyset \qquad \text{for all } n \geq 1.\end{equation*}
Since the closed balls are nested and their diameters tend to 00, completeness implies that
⋂n=1∞B‾(pn,εn)≠∅.\begin{equation*}\bigcap_{n=1}^\infty \overline{B}\left(p_n,\varepsilon_n\right) \neq \varnothing.\end{equation*}
Let qq belong to this intersection. For each nn, we have B‾(pn,εn)⊂X∖Cn\overline{B}\left(p_n,\varepsilon_n\right) \subset X\setminus C_n, so q∉Cnq\notin C_n for all nn. Thus
q∉⋃n=1∞Cn=X,\begin{equation*}q \notin \bigcup_{n=1}^\infty C_n = X,\end{equation*}
which is impossible.
This is the desired contradiction to the statement of the theorem. Thus, at least one of the CnC_n must contain an open ball of positive radius. □\Box
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 an open ball BB of positive radius such that sup⁡nsup⁡x∈B∣fn(x)∣<∞\sup_n \sup_{x \in B} |f_n(x)| < \infty.
\quad Proof. For each N∈NN \in \mathbb{N}, define
XN:={x∈X:sup⁡n∣fn(x)∣≤N}.\begin{equation*}X_N := \{x \in X : \sup_n |f_n(x)| \leq N\}.\end{equation*}
Then
XN=⋂n=1∞{x∈X:∣fn(x)∣≤N},\begin{equation*}X_N = \bigcap_{n=1}^{\infty} \{x \in X : |f_n(x)| \leq N\},\end{equation*}
so each XNX_N is closed, because every set {x∈X:∣fn(x)∣≤N}\{x \in X : |f_n(x)| \leq N\} is closed by continuity of fnf_n.
By the pointwise boundedness assumption, every x∈Xx \in X belongs to some XNX_N. Hence
X=⋃N=1∞XN.\begin{equation*}X = \bigcup_{N=1}^{\infty} X_N .\end{equation*}
Since XX is complete and the sets XNX_N are closed, Baire's category theorem implies that some XNX_N contains a nonempty open ball BB. For this ball,
sup⁡nsup⁡x∈B∣fn(x)∣≤N<∞.\begin{equation*}\sup_n \sup_{x \in B} |f_n(x)| \leq N < \infty.\end{equation*}
This proves the claim. □\Box

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