The Baire space I – getting off the ground

An area in mathematics concerning a special kind of topological spaces called Baire spaces is an example of a notion that intersects three branches of mathematics, namely topology, analysis and set theory – this intersection in particular is called descriptive set theory. We start off by defining the space, then exploring its properties and ending up looking at its applications throughout analysis and topology. Let $^\omega\omega$ be the set of functions $f:\omega\to\omega$ and $^{<\omega}\omega$ the set of functions $f:n\to\omega$ for $n<\omega$.

Definition 1. Let $n<\omega$ and $s\in{^n\omega}$. Define

$N_s=\{x\in^\omega\omega\mid x\upharpoonright n=s\}$

to be the basic open subsets of $^\omega\omega$.

Examples of these basic open sets are for instance if $s=\left<1,4,2,3,5\right>$, then the elements of $N_s$ are all the infinite sequences starting with $s$. For example, $\left<1,4,2,3,5,0,0,0,\hdots\right>$ is in $N_s$. This then leads up to the construction of the open sets on $^\omega\omega$:

Definition 2. Define $U\subseteq{^\omega}\omega$ to be open if there exist a class $\mathcal{F}$ of basic open sets of $^\omega\omega$ satisfying $U=\bigcup\mathcal{F}$.

From this definition we can then define the Baire space $(^\omega\omega,\mathcal{T})$, where $\mathcal{T}$ is the open subsets of $^\omega\omega$. The only thing we’re missing now is actually to prove that our definition of the Baire space is correct, that the defined open sets really satisfies the properties of a topology. This shows in the following theorem:

Theorem 3. The Baire space $(^\omega\omega,\mathcal{T})$ is a topological space.

Proof. We have to check that the three properties of topologies are satisfied. First of all that $\emptyset\in\mathcal{T}$ and $^\omega\omega\in\mathcal{T}$. For $\emptyset$, just choose $\mathcal{F}$ to be empty in Definition 2. For $^\omega\omega$ notice that $N_{\left<\right>}={^\omega}\omega$. for $\left<\right>$ being the empty sequence.

Next, we show that $\mathcal{T}$ is closed under arbitrary unions. Let $A\subseteq\mathcal{T}$ be a class of open sets. Each of these open sets is a union of basic open sets by Definition 2, so $\bigcup A$ would then be a union of unions of basic open sets, making it itself a union of basic open sets, thus making it open.

Lastly, as the hardest step, we show that $\mathcal{T}$ is closed under finite intersections. We show that $\forall A,B\in\mathcal{T}: A\cap B\in\mathcal{T}$. Let $A=\bigcup\mathcal{F}$ and $B=\bigcup\mathcal{G}$, where $\mathcal{F}$ and $\mathcal{G}$ are classes of basic open sets. Define $\mathcal{H}$ to be the class of basic open sets such that there exist $r,s\in{^{<\omega}}\omega$ satisfying

1. $N_r\in\mathcal{F}$ and $N_s\in\mathcal{G}$
2. $r\subseteq s$ or $s\subseteq r$
3. $t=r\cup s$.

The last two clauses 2 and 3 implies that $r$ and $s$ are comparable – that is, either $r=s\upharpoonright dom(r)$ or $s=r\upharpoonright dom(s)$. Thus would $t$ be the longest of the two, due to clause 3. We will have to show that

$A\cap B=\bigcup\mathcal{H}$.

$\subseteq$“: Let $x\in A\cap B$. Then there is $r$, $s\in{^{<\omega}}\omega$ such that $x\in N_r\in\mathcal{F}$ and $x\in N_s\in\mathcal{G}$. $r$ and $s$ would then be comparable, because they’re restrictions of the same infinite sequence $x$. Let $t=r\cup s$. Then $x\in N_t\in\mathcal{H}$, since it satisfies all above three clauses by construction of $r$ and $s$, meaning $x\in\bigcup\mathcal{H}$.

$\supseteq$“: Let $x\in\bigcup\mathcal{H}$, meaning there is a $t\in{^{<\omega}}\omega$ satisfying $x\in N_t$. Then by clause 1 above, there is $r,s\in{^{<\omega}}\omega$ such that $N_r\in\mathcal{F}$ and $N_s\in\mathcal{G}$, satisfying $x\in N_r$ and $x\in N_s$. Thus $x\in A$ and $x\in B$, meaning $x\in A\cap B$.

$\blacksquare$