What is this topology?

Question

I'm working through a text that is using some topology. It defines the following topology, I'm confused on what it would look like.

In topology, $2^\kappa$ denotes $^\kappa\{0,1\}$, where $2= \{0,1\}$ is given the discrete topology and $2^\kappa$ has the Tychonov product topology.

What would the product topology look like? would it just be $2\times 2\times 2\times \cdots$, $\kappa$ times? But if $2$ has the discrete topology in which every subset of $\{0,1\}$ is open, then every set of function in $2^\kappa$ would be open. Also $\kappa$ may be infinite and really big. What would the product topology look like in that case?

Answer 1

The product topology would correspond to all indexed finite collections $\mathcal F $ of subsets of $\{0,1\} $ in a natural way. That's it's defined as $\prod_{k\in K }U_k $, where $U_k $ is open in $\{0,1\} $ and equals $\{0,1\}$ for all but finitely many $k$.

It's the coarsest topology which makes all the projections continuous.

If $K $ is infinite, it is coarser than the box topology, where every product of open sets is open. In the product topology $2^K $ is compact. Not so in the box topology, which happens to be the discrete topology in this case.

When $K $ is finite, we get the discrete topology for the product topology here.

Answer 2

The set $2^\kappa$ is the set of all functions on $\kappa$ with values in $\{0,1\}$, which can be identified with the collection of all subsets of $\kappa$, identifying a set with its characteristic function.

But taking the function view, the topology is determined by all projections, which in this case are all evaluations $\pi_\alpha: 2^\kappa \to \{0,1\}$, where $\alpha$ ranges over all elements of $\kappa$ and defined by $\pi_\alpha(f)=f(\alpha)$. The topology on $2^\kappa$ is to be the smallest (minimal) topology such that all $\pi_\alpha$ are continuous.

This implies that all sets of the form $\pi_\alpha^{-1}[\{i\}] = \{f \in 2^\kappa\mid f(\alpha)=i\}$, where $\alpha \in \kappa, i \in \{0,1\}$ must be in this topology (as inverse images of open sets in $\{0,1\}$ under the functions that have to be continuous), and one can check easily that all finite intersections of sets of this form form a base for the product topology on $2^\kappa$.

So a set $O \subseteq 2^\kappa$ is open iff for all $f \in O$ we can find a finite subset $\{\alpha_1,\ldots,\alpha_n\}$ of $\kappa$ such that

$$\{g \in 2^\kappa\mid \forall 1 \le i \le n: g(\alpha_i)=f(\alpha_i)\} \subseteq O$$

This gives a quite concrete understanding of all open subsets of $2^\kappa$.