Can we obtain the Borel $\sigma$-algebra on $[0;1]$ as a limit of finite algebra?

Question

The Borel sets of $[0;1]$ are defined to be the smallest $\sigma$-algebra which contains all the subintervals $\subseteq [0;1]$.

Is there a construction where we can obtain it as a limit of a countable sequence of finite algebra? Or as a limit of limits of countable sequences.

For instance something similar (in form) to : $$\lim_{n \to \infty} \bigcup_{k = 1}^{n} A_k \quad \text{where $A_k$ is a finite algebra}$$

Answer 1

Yes there is a natural and useful construction of $\sigma$-algebras as a "limit". But not as the limit of a sequence, rather the limit of an uncountable collection, indexed by the countable ordinals (something logicians might call an $\omega_1$-sequence, but which the rest of us don't usually call a "sequence", since it's not countable).

First, given $A\subset\mathcal P(X)$, let $S(A)$ consist of all countable unions of elements of $A$ together with the complements of elements of $A$.

Now suppose $A\subset\mathcal P(X)$. By "transfinite recursion" there exists a family $(A_\alpha)_{\alpha<\omega_1}$ of subsets of $\mathcal P(X)$ with $$A_0=A$$and $$A_\alpha=S\left(\bigcup_{\beta<\alpha}A_\beta\right)$$for $\alpha>0$. Then it's not hard to show that$$\sigma(A)=\bigcup_{\alpha<\omega_1}A_\alpha$$is precisely the $\sigma$-algebra generated by $A$. (Of course we haven't defined what we mean by "limit" here, but it's natural to regard this as the limit of $A_\alpha$ as $\alpha\to\omega_1$.)

Proof. Say $\sigma(A)$ is defined as above and let $\sum(A)$ be the $\sigma$-algebra generated by $A$. Note first that if $F$ is a $\sigma$-algebra and $B\subset F$ then $S(B)\subset F$ by definition. So since $A\subset\sum(A)$ it follows by transfinite induction that $A_\alpha\subset\sum(A)$ for every $\alpha<\omega_1$, hence $\sigma(A)\subset\sum(A)$.

So we need only show that $\sigma(A)$ is a $\sigma$-algebra. This is clear. If $E\in\sigma(A)$ then there exists $\alpha < \omega_1$ with $E\in A_\alpha$; hence $X\setminus E\in A_{\alpha+1}$, so $X \setminus E \in \sigma(A)$.

Suppose that $E_1,E_2 \dots \in \sigma(A)$. For every $n$ there exists $\alpha_n<\omega_1$ with $E_n\in A_{\alpha_n}$. Since $\omega_1$ is uncountable there exists $\alpha<\omega_1$ with $\alpha>\alpha_N$ for every $n$; hence $\bigcup E_n\in A_\alpha\subset\sigma(A)$ $\square.$

This is useful, for example as far as I know it's the only way to show that the Borel algebra on $\Bbb R$ has cardinality $\mathfrak c$.

Answer 2

The answer is negative. Let $\mathcal{A_n}$, $n\in\mathbb{N}$ be a collection of finite algebras such that: $$\mathcal{B}=\bigcup_{n=1}^\infty\mathcal{A}_n$$ where $\mathcal{B}$ is the $\sigma-$algebra of all the Borel subsetes of $\mathbb{[0,1]}$. Now, on the one hand we have that $\#\mathcal{B}$ is uncountable, on the other hand $\#\bigcup\limits_{n=1}^\infty\mathcal{A}_n$ is obviously countable - countable union of finite sets.

As for your edit, it is also false that: $$\bigcup_{n=1}^\infty\mathcal{P}(\{1,\dots,n\})=\lim_{k\to\infty}\bigcup_{n=1}^k\mathcal{P}(\{1,\dots,n\})$$ is uncountable, since it is a countable union of finite sets. It is important to see that: $$\bigcup_{n=1}^\infty\mathcal{P}(\{1,\dots,n\})\underset{\neq}{\subset}\mathcal{P}(\mathbb{N})$$ since $\mathbb{N}\in\mathcal{P}(\mathbb{N})$ but $\mathbb{N}\not\in\bigcup\limits_{n=1}^\infty\mathcal{P}(\{1,\dots,n\})$ - obviously, since there is not $n\in\mathbb{N}$ such that $\mathbb{N}\in\mathcal{P}(\{1,2,\dots,n\})$.

Edit/Further Info: What is true is that: $$\bigcup_{n=1}^\infty\mathcal{P}(\{1,\dots,n\})=\{A\mid A\subset\mathbb{N},\ A\text{ is finite}\}.$$