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Aug 27, 2022 at 1:31 comment added Jacob Manaker The Cantor-Bernstein operator is in fact K1, because $f$ and $g$ are injections.
Oct 10, 2019 at 7:08 history edited Dave L Renfro CC BY-SA 4.0
Many small tweaks and edits, and some additional comments/references.
Oct 9, 2019 at 22:51 vote accept Leonid Dworzanski
Oct 8, 2019 at 9:44 history edited Dave L Renfro CC BY-SA 4.0
added 108 characters in body
Oct 8, 2019 at 9:36 history answered Dave L Renfro CC BY-SA 4.0