All Questions

As discussed in this question, in general a Borel surjection $f:\mathbb{R}\rightarrow\mathbb{R}$ may not have a Borel right inverse, namely a $g$ such that $f\circ g=id$, although there is always a ...

Given $\mathcal{A}\subseteq\omega^\omega$, the Banach-Mazur game with payoff set $\mathcal{A}$ consists of players $1$ and $2$ alternately playing nonempty finite strings of naturals with player $1$ ...