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Jun 6 at 21:07 comment added Alex Kruckman Nice argument, thanks!
Jun 6 at 20:44 comment added Elliot Glazer We’ll justify $3 \rightarrow 1$ by the contrapositive. Fix $s \in \prod A_i.$ Identify $\bigsqcup A_i \setminus \{s_j: j<\omega\}$ with $\{t \in \prod A_i: \exists ! j (s_j \neq t_j)\}.$ The latter injects into $\mathbb{R}$ and is thus orderable.
Jun 6 at 19:09 comment added Alex Kruckman Can you explain why $|\prod A_i|=|\mathbb{R}|$ implies $\bigsqcup A_i$ countable?
Jun 6 at 13:34 vote accept Noah Schweber
Jun 6 at 10:11 history edited Elliot Glazer CC BY-SA 4.0
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Jun 6 at 10:05 history answered Elliot Glazer CC BY-SA 4.0