Topology of set-theoretic limits

Question

The convergence of a sequence of subsets of a fixed set $\Omega$ is defined in-terms of set-theoretic limit which does not use any metric or topology on $2^{\Omega}.$ Now I wonder:

What is the smallest topology on $2^{\Omega}$ that induces by these convergent sequences?

For a finite set $\Omega$ this seems to be the discrete (or co-finite) topology. But for infinite sets I have no clue.

Answer 1

We naturally have $2^\Omega = \prod_{\omega \in \Omega} \{0,1\}$ with the product topology, where the topology on $\{0,1\}$ (which corresponds to a singleton of $\Omega$) must be the discrete topology.

Coincidentally, using this natural identification and $f \in \prod_{\omega \in \Omega}\{0,1\} \iff f : \Omega \to \{0,1\},$ then the $\liminf$ and $\limsup$ of any such sequence of sets become their standard definitions when identified as functions.

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Here are some details:

• The natural correspondence of sets (in fact, of lattices) is $$A \in \mathcal{P}(\Omega) \iff \mathbf{1}_A \in \prod_{\omega \in \Omega}\{0,1\}, \\ \text{where } \mathbf{1}_A(\omega) = \begin{cases}1, & \omega \in A \\ 0, & \omega \not\in A\end{cases}$$
• Under this correspondence, given any sequence $(A_n)$ of subsets of $\Omega,$ we can set $$L = \liminf_{n\to\infty} A_n = \bigcup_{n\geq 1} \bigcap_{k\geq n} A_k \\ U = \limsup_{n\to\infty} A_n = \bigcap_{n\geq 1} \bigcup_{k \geq n} A_k$$ and show that $$\mathbf{1}_L = \liminf_{n\to\infty} \mathbf{1}_{A_n} \\ \mathbf{1}_U = \limsup_{n\to\infty} \mathbf{1}_{A_n}$$ where the limits infimum and supremum on the right side is defined by the standard pointwise definition for functions $f : \Omega \to \{0,1\} \subseteq \mathbb{R}$ [Show this.]
• From the previous step, we see that $\{0,1\} \subseteq \mathbb{R}$ must be assigned its subspace topology, which is the discrete topology. Also, since the liminf and limsup in the function space are defined pointwise, convergence in $\prod_{\omega\in \Omega} \{0,1\}$ can be recognized as pointwise convergence, which means it has the product topology. [Depending on your definition of product topology, this might need to be shown, but it's straight-forward]