# The Baire space II – the Baire Category Theorem

Having established the Baire space, we will here state and prove a version of the Baire Category Theorem, which is an important tool used within functional analysis, as well as also having applications in other parts of analysis and topology. After proving the theorem, we will state a few results in which the proofs rely on the theorem, to get a feeling of its power.

Firstly I will state two essential properties of the Baire space, which I perhaps will come back to another time and prove – for now I will simply use them to proving the Baire Category Theorem. Here stated in a lemma:

Lemma. The Baire space $({^\omega}\omega,\mathcal{T})$ is a complete metric space, where the metric $d$ is defined as

$d(x,y)=\left\{\begin{array}{ll}1/2^n&,n<\omega\text{ is least such that }x(n)\neq y(n)\\ 0&,x=y\end{array}\right.$.

This lemma then also means that any statement involving open sets can equivalently be stated in terms of the metric alone and vice versa. Although it is merely stated as a lemma for the purpose of this blog post, the completeness in particular is not trivial. Taking this lemma as a fact, we can prove the theorem:

Theorem (Baire Category Theorem). Let $({^\omega}\omega,\mathcal{T})$ be the Baire space and define $\left_{n<\omega}\subseteq\mathcal{T}$ to be a sequence of dense open subsets of $^\omega\omega$. Then

$\mathfrak{D}:=\bigcap_{n<\omega}D_n$

is dense in the Baire space as well.

Proof. We show that $\forall U\in\mathcal{T}: U\cap\mathfrak{D}\neq\emptyset$. Fix $U\in\mathcal{T}$. Since $D_0$ is dense and open, $D_0\cap U$ is nonempty and open, so we can choose $s_0\in{^{<\omega}}\omega$ such that $\bar{N}_{s_0}\subseteq D_0\cap U$ and $\forall x\in N_{s_0}:d(x_0,x)<1$ for some $x_0\in D_0\cap U$ (chosen with the axiom of choice). Then since $D_1$ is open and dense, $D_1\cap N_{s_0}$ is nonempty and open, and again we pick $s_1\in{^{<\omega}}\omega$ such that $\bar{N}_{s_1}\subseteq D_1\cap N_{s_0}$ and $\forall x\in N_{s_1}: d(x_1,x)<2^{-1}$. Continue recursively in this manner, generating a sequence $\left_{i<\omega}$, which is Cauchy due to our choices of the $s_i$‘s. Since the Baire space is complete, the sequence converges to some $x'$. Now, for any $n<\omega$,

$x'\in\bar{N}_{s_{n+1}}\subseteq N_{s_n}$,

since closedness ensures containment of the limit point $x'$; thus meaning $x'\in\mathfrak{D}$. Since furthermore $N_{s_n}\subseteq U$ for every $n<\omega$ by construction, $\mathfrak{D}$ is dense in the Baire space.

$\blacksquare$

Now, to provide some overview as to how this relates to other areas of mathematics, the most important implications lie within functional analysis:

Theorem (Open Mapping Theorem). Let $\mathcal{A}$, $\mathcal{B}$ be Banach spaces and $f:\mathcal{A}\to\mathcal{B}$ be a surjective continuous linear map. Then $f$ is an open map, i.e. $f(U)$ is open for every open set $U$.

Theorem (Closed Graph Theorem). Let $\mathcal{A}$, $\mathcal{B}$ be Banach spaces and $f:\mathcal{A}\to\mathcal{B}$ a linear map. Then $f$ is continuous if and only if its graph is a closed subset of $\mathcal{A}\times\mathcal{B}$ with the product topology.

However, it also leads to interesting results within real analysis –  for instance:

Theorem. Let $f\in C^\infty(\mathbb{R})$ satisfying $\forall x\in\mathbb{R}\exists n<\omega: f^{(n)}(x)=0$. Then $f$ is a polynomial.