This is an Exercise from an older exam I am currently practicing on. I am struggling a bit with the concept of Polish spaces. So I am glad for any help or guidance regarding if the proof is correct, well structured or maybe not going far enough. Also if someone has a different solution I would be very interested in that. Thanks in advance! The question is following:
We define $$\Omega =\{(\omega_n)_{n \in \mathbb{N}} | \omega_n \in \{0,1\}, n \in \mathbb{N} \}$$ the Space of all sequences with members in $\{0,1\}$ . We have $d: \Omega \times \Omega \rightarrow [0,1]$ defined by $$d(\omega,\sigma) = \sum_{n=1}^{\infty}2^{-n}\mathbb{1}_{\{\omega_{n} \neq \sigma_{n}\}}$$ (a) Show that $(\Omega,d)$ is a complete metric set
(b) Show that $Q = \{\omega = (\omega_{n})_{n \in \mathbb{N}} \in \Omega | \exists N \in \mathbb{N} \forall n\geq N : \omega_{n} = 0 \}$ is dense in $\Omega$ with $d$
Proof:
(a)
Every Cauchy Sequence of points in $\Omega$ needs to have a limit that is also in $\Omega$
Let $(\omega^{l})_{l \in \mathbb{N}} \subset \Omega$ be a Cauchy Sequence s.t. $\forall \epsilon > 0 \exists N \in \mathbb{N}:\forall k,l \geq N$ and$ \forall n \in \mathbb{N}: d(\omega_{n}^{k}, \omega_{n}^{l}) < \epsilon$
this means $$\sum_{n=1}^{\infty}2^{-n}\mathbb{1}_{\{\omega_{n}^{l} \neq \omega_{n}^{k}\}} \leq \sum_{n=1}^{\infty}\mathbb{1}_{\{\omega_{n}^{l} \neq \omega_{n}^{k}\}} < \epsilon \implies \omega_{n}^{k} = \omega_{n}^{l} = \omega_{n}^{*} \land (\omega_{n}^{*} := \omega_{n}^{N} \in \Omega) $$
Using from the definition of Cauchy Sequences that $\forall \epsilon > 0$ and $\forall k \geq N$ s.t. $d(\omega_{n}^{k},\omega_{n}^{*}) < \epsilon$ holds, I can follow from that, that there exists a limit $$\omega_{n}^{*} = \lim_{l \to \infty} \omega_{n}^{l}$$ which is as proven before in $\Omega$. Following from that I have shown that $(\Omega, d)$ is a complete metric set
(b)
I start with $\omega = (\omega_n)_{n \in \mathbb{N}} \in Q$ and $\sigma = (\sigma_n)_{n \in \mathbb{N}} \in \Omega$ I know that $Q$ includes all sequences whose members after a certain time only take the value zero. For $\sigma$ there are two possibilities. After a certain time all members are either zero (1) or one (2). If I am not mistaken this should follow from (a) and the convergence of the sequence? Proceeding:
$$d(\omega, \sigma) \leq \sum_{n=1}^{N-1}2^{-n}\mathbb{1}_{\{\omega_{n} \neq \sigma_{n}\}} + \sum_{n=N}^{\infty}2^{-n}\mathbb{1}_{\{0 \neq \sigma_{n}\}}$$
Fokussing on the first sum I define an $\epsilon > 0$ and choose $\sigma_{1},...,\sigma_{N} \in Q$ s.t. for an $N$ big enough $\sum_{n=1}^{N-1}2^{-n}\mathbb{1}_{\{\omega_{n} \neq \sigma_{n}\}} \leq \epsilon/2$. We can do this since the sum of $2^{-n}$ itself is smaller or equal to one.
For the second sum we consider (1) and (2). It is clear that with (1) this sum is equal to zero and: $$d(\omega, \sigma) \leq \sum_{n=1}^{N-1}2^{-n}\mathbb{1}_{\{\omega_{n} \neq \sigma_{n}\}} + \sum_{n=N}^{\infty}2^{-n}\mathbb{1}_{\{0 \neq \sigma_{n}\}} \leq \epsilon/2 + 0 < \epsilon \implies Q \text{ dense in } \Omega $$
With (2) we choose the $N \in \mathbb{N}$ large enough such that the sum which is equal to $2^{1-N}$ is smaller than $\epsilon/2$. Following:
$$d(\omega, \sigma) \leq \sum_{n=1}^{N-1}2^{-n}\mathbb{1}_{\{\omega_{n} \neq \sigma_{n}\}} + \sum_{n=N}^{\infty}2^{-n}\mathbb{1}_{\{0 \neq \sigma_{n}\}} < \epsilon/2 + \epsilon/2 = \epsilon \implies Q \text{ dense in } \Omega $$
Just to complete it: since there exists $Q$ a countable and dense set in $\Omega$ and $(\Omega,d)$ is a complete metric set, it is also a polish space.