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Dec 23, 2016 at 18:26 vote accept Noah Schweber
Sep 10, 2016 at 19:31 comment added Noah Schweber Ah, yes, silly me. This is very neat! Of course this isn't a ZFC proof, but still +1!
Sep 10, 2016 at 3:37 comment added vzoltan If a set $X \subset \omega^\omega$ is non-escaping, then it means that there exists a real $f \in \omega^\omega$ so that every element of $X$ is $\leq^* f$, in other words $X \subset \{r: r \leq^*f\}$ where the latter is a Borel set which is non-escaping.
Sep 10, 2016 at 3:32 comment added Noah Schweber Why must $F(S)$ be covered by a Borel nonescaping set?
Sep 10, 2016 at 1:25 history answered vzoltan CC BY-SA 3.0