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Apr 15 at 11:51 history edited Andrej Bauer CC BY-SA 4.0
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Apr 15 at 11:50 comment added Andrej Bauer @WillSawin: Yes, that would make sense. I will fix it (and the timestamp year, thanks to Anurag).
Apr 15 at 10:29 comment added Will Sawin If the goal is for sequence-avoiding to be stronger than uncountable, maybe the definition should just say that $A$ is inhabited and for every sequence $\mathbb N \to A$ there exists an element of $A$ that is not a term of the sequence? Otherwise the empty set is sequence-avoiding but not uncountable.
Apr 15 at 7:41 comment added Anurag Sahay It seems like your edit timestamp has the wrong year.
Apr 15 at 6:35 history edited Andrej Bauer CC BY-SA 4.0
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Apr 15 at 6:33 comment added Andrej Bauer @WillSawin: An interesting thought, but note that the usual proofs relying on countable choice that $\mathbb{R}$ is sequence-avoiding break down if we use $\mathbb{N} \to 1 + \mathbb{R}$. The problem is, given $a : \mathbb{N} \to \{\star\} + \mathbb{R}$, how do we make any progress in constructing a real $x$ that avoids $a$ when $a$ is just producing $\star$'s? In fact, I think in the effective topos the reals will not be sequence-avoiding in this sense.
Apr 15 at 1:03 history edited LSpice CC BY-SA 4.0
Typos, while this is on the front page
Apr 15 at 1:01 comment added Will Sawin Should the maps in the definition of sequence-avoiding be from $\mathbb N$ to $1+A$, for the same reason as in the definition of countable?
Apr 14 at 23:57 history edited Aaron Meyerowitz CC BY-SA 4.0
SPELLING
Sep 21, 2023 at 5:25 history edited Andrej Bauer CC BY-SA 4.0
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S Sep 21, 2023 at 5:25 history rollback Andrej Bauer
Rollback to Revision 4 - Edit approval overridden by post owner or moderator
Sep 21, 2023 at 4:37 history suggested Jim Kingdon CC BY-SA 4.0
The reals are a set in CZF per comments elsewhere on this question or https://mathstodon.xyz/@MartinEscardo/110944257074282172
Sep 21, 2023 at 4:28 review Suggested edits
S Sep 21, 2023 at 5:25
Aug 29, 2023 at 15:40 comment added Anon Thank you very much for clearing things up! I don't know why I was thinking subcountable and countable are equivalent. This answer really helped.
Aug 29, 2023 at 15:18 vote accept Anon
Aug 24, 2023 at 6:13 comment added Andrej Bauer @DavidRoberts: Thanks, fixed.
Aug 24, 2023 at 6:12 comment added Andrej Bauer @LSpice: I improved the reference, although it is generally known that countable choice suffices for uncountability of reals, so that's by no means the only reference available.
Aug 24, 2023 at 6:11 history edited Andrej Bauer CC BY-SA 4.0
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Aug 24, 2023 at 4:02 comment added David Roberts The theorem giving the subcountability of the Dedekind reals and citing Hamkins' work, is the "no surjection" the other way around? Also, in the second of the two caveats, with the nested intervals method, it mentions with excluded middle and countable choice: is this meant to be without excluded middle?
Aug 23, 2023 at 23:53 history edited Sam Hopkins CC BY-SA 4.0
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Aug 23, 2023 at 23:53 comment added LSpice You mention "Bishop's book on constructive analysis", but he seems to have at least 2. Did you mean "Foundations of constructive analysis" (MSN)?
Aug 23, 2023 at 23:52 history edited LSpice CC BY-SA 4.0
Proofreading
Aug 23, 2023 at 23:41 history answered Andrej Bauer CC BY-SA 4.0