Express the set of all paths with infinite $1$'s using set theoretical representation and elements of another set

Question

Define the collection $\mathscr{A}$ of subsets of $\Omega=\{0,1\}^{\infty}$ as $$\mathscr{A}=\{S\times\Omega:S\subseteq\{0,1\}^{l},l\ge1\}$$ Let all infinite length paths be denoted by $$\omega=(\omega_n)_{n\ge1}$$ Then, define $A$, the set of all paths with infinitely many up($\{1\}$) moves by $$A=\left\{\omega:\sum\mathbb{I}_{\{\omega_n=1\}}=\infty\right\}$$

Express $A$ using sets of $\mathscr{A}$ and 'countably many', $\cup$, 'complementation', $\cap$... (set theoretic representation)

My approach:
Initially, I proceeded by defining $A_1,A_2...\in\mathscr{A}$ as $A_k=S_k\times\Omega$, where $S_k=\{1,1,...$ $_k$ $_{times}$$\}$ and using $\underset{k=1}{\overset{\infty}{\bigcap}}A_k=\{1,1,1...\}$.
But, this excludes the cases the paths with finite $0$'s.
However, if I take $\underset{k=1}{\overset{\infty}{\bigcup}}A_k=\{1,1,1...\}$, this also includes paths that do not satisfy the infinte $1$'s condition.

How must I do this?

Answer 1

Let $A_k:=\left\{\left(\omega_n\right)_{n\geqslant 1}\in \Omega\mid \omega_k=1\right\}$. Then the equality $$ A=\bigcap_{j\geqslant 1}\bigcup_{k\geqslant j}A_k $$ holds (see this thread for a better understanding of the limit superior and inferior of a sequence of sets). In order to see that $A_k$ belongs to $\mathcal A$, write this as $S_k\times\Omega$ where $S_k:=\left\{\left(\omega_i\right)_{i=1}^k\in \{0,1\}^k\mid \omega_k=1\right\}$.