Past exam of the logic and set theory course of the University of Oxford / 2008 / B / 7 / 2 / Solution

Map $S\subseteq \omega$ to its indicator function $1_S:\omega \to \{0,1\}$. Its inverse maps a binary sequence $u$ to $\{n\in \omega :u(n)=1\}$. These maps are mutually inverse, soHere the left superscript denotes the set of functions from $\omega$ into $2$. It is a space of binary sequences, not an ordinal square.