We consider the measurable space $(M,\mathcal{B}(M))$ where $M=\{0,1\}^{\mathbb{N}}=\{\omega_1,\omega_2,\dots\},\omega_i=0$ or $ \omega_i=1$ i.e the space of all binary sequences indexed by the naturals and $\mathcal{B}(M)$ is the product topology (the coarsest topology which which makes all the projections continuous).
Let $\pi_n: \Omega \to \{0,1\}^n$ given by $\omega \to (\omega_1,\omega_2, \dots)$ be a projection and $\mathcal{F_n}=\pi_n^{-1}(\mathcal{B}(\{0,1\}^n))$.
Why is $\mathcal{F}_n$ finite? the way I reason, I am quite convinced that it should be uncountable. Clearly $\mathcal{B}(\{0,1\}^n)$ is the power set on all n sized binary sequences. I mean even $\pi_5^{-1}((0,1,0,1,0))$ is the set of all $ \omega \in \{0,1\}^{\mathbb{N}}$ whose first 5 entries are $0,1,0,1,0$.
I am very confused as to why is $\mathcal{F}_n$ is finite.
If $f:X\to Y$ where $Y$ is a finite set then every subcollection $\mathcal V\subseteq\wp(Y)$ is a finite set so that also $f^{-1}(\mathcal V):=\{f^{-1}(V)\mid V\in\mathcal V\}$ is a finite set.
This can be applied on $\pi_n: \Omega \to \{0,1\}^n$ where $\{0,1\}^n$ is a finite set, and $\mathcal{B}(\{0,1\}^n)\subseteq\wp(\{0,1\}^n)$.