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Aug 2, 2018 at 10:56 comment added Robert Furber @AryehKontorovich That was loose wording by me (I have edited the answer). What I mean is the existence of such a cardinal being consistent.
Aug 2, 2018 at 10:54 history edited Robert Furber CC BY-SA 4.0
Rephrased some lax wording about consistency of existence of measurable cardinals
Aug 2, 2018 at 10:46 comment added Aryeh Kontorovich What does it mean for a cardinal to be consistent?
Aug 2, 2018 at 10:29 comment added Robert Furber @AryehKontorovich On my side it looks like my counter-edit won, so there is no problem with the answer as it is now.
Aug 2, 2018 at 10:28 comment added Aryeh Kontorovich I see. Are you able to roll back my edit?
Aug 2, 2018 at 10:10 history edited Robert Furber CC BY-SA 4.0
Added a new reference for Ulam's theorems.
Aug 2, 2018 at 10:06 comment added Robert Furber @AryehKontorovich The $\aleph_1$ is intentional. It's a strict inequality. Otherwise the definition of $\kappa$-additivity doesn't work in the definition of a measurable cardinal (the measure is only additive for disjoint families strictly smaller than $\kappa$). Sets of cardinality $< \aleph_1$ are exactly countable sets.
S Aug 2, 2018 at 10:05 history edited Robert Furber CC BY-SA 4.0
Edit to correct a mistaken impression that prompted another edit.
Aug 2, 2018 at 10:02 comment added Aryeh Kontorovich I made an edit, please confirm.
Aug 2, 2018 at 10:02 review Suggested edits
S Aug 2, 2018 at 10:05
Aug 2, 2018 at 9:11 comment added Aryeh Kontorovich Many thanks for the detailed and very helpful answer, @Robert! I've upvoted and would upvote again several times if I could.
Aug 1, 2018 at 22:28 history answered Robert Furber CC BY-SA 4.0