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1. • CommentRowNumber1.
   • CommentAuthorTodd_Trimble
   • CommentTimeMar 10th 2016
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   Author: Todd_Trimble
   Format: MarkdownItexI added a brief section on Cantor's theorem for posets to [[Cantor's theorem]], which in one form says that for posets $X$ there can be no surjective poset map $X \to 2^X$ (taking $2 = \{0 \leq 1\}$). You might find it amusing to try to prove this yourself in a pleasant way. I found one proof (you can find it [here](https://ncatlab.org/toddtrimble/published/Cantor's+theorem+for+posets)), but it's possible I was working too hard for it. :-)

   I added a brief section on Cantor’s theorem for posets to Cantor’s theorem, which in one form says that for posets $X$ there can be no surjective poset map $X \to 2^X$ (taking $2 = \{0 \leq 1\}$).

   You might find it amusing to try to prove this yourself in a pleasant way. I found one proof (you can find it here), but it’s possible I was working too hard for it. :-)

2. • CommentRowNumber2.
   • CommentAuthorDavidRoberts
   • CommentTimeMar 10th 2016
   • (edited Mar 10th 2016)
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   Author: DavidRoberts
   Format: MarkdownItexYou have a $2_Y$ in there at one point. More seriously, I presume a similar proof would hold where we take the exponential _as Heyting algebras_? Not sure off the top of my head how different these are. A minor suggestion: why not link to Yanofsky's paper? Would be helpful for the reader.

   You have a $2_Y$ in there at one point.

   More seriously, I presume a similar proof would hold where we take the exponential as Heyting algebras? Not sure off the top of my head how different these are.

   A minor suggestion: why not link to Yanofsky’s paper? Would be helpful for the reader.

3. • CommentRowNumber3.
   • CommentAuthorTodd_Trimble
   • CommentTimeMar 10th 2016
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   Author: Todd_Trimble
   Format: MarkdownItexFixed the typo, and added the reference. Thanks. Sorry, I'm not sure what your second paragraph is suggesting. Could you spell it out more?

   Fixed the typo, and added the reference. Thanks.

   Sorry, I’m not sure what your second paragraph is suggesting. Could you spell it out more?

4. • CommentRowNumber4.
   • CommentAuthorDavidRoberts
   • CommentTimeMar 10th 2016
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   Author: DavidRoberts
   Format: MarkdownItexHmm, was wondering if Cantor's theorem holds in the category of Heyting algebras. This obviously doesn't help for the application in any way (fixed point theorems, leading to CSB), since that's a result that requires classical logic.

   Hmm, was wondering if Cantor’s theorem holds in the category of Heyting algebras. This obviously doesn’t help for the application in any way (fixed point theorems, leading to CSB), since that’s a result that requires classical logic.

5. • CommentRowNumber5.
   • CommentAuthorTodd_Trimble
   • CommentTimeMar 10th 2016
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   Author: Todd_Trimble
   Format: MarkdownItexSo the "Cantor's theorem" I was talking about takes place in a cartesian closed category, e.g. a topos or $Pos$; in each of those cases we are inquiring $X$ such that $X \cong 2^X$ for suitable $2$. The reason I was confused by #2 is that Heyting algebras do not form a cartesian closed category. But maybe you have a different formulation in mind.

   So the “Cantor’s theorem” I was talking about takes place in a cartesian closed category, e.g. a topos or $Pos$; in each of those cases we are inquiring $X$ such that $X \cong 2^X$ for suitable $2$. The reason I was confused by #2 is that Heyting algebras do not form a cartesian closed category. But maybe you have a different formulation in mind.

6. • CommentRowNumber6.
   • CommentAuthorMike Shulman
   • CommentTimeMar 10th 2016
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   Author: Mike Shulman
   Format: MarkdownItexIsn't Knaster-Tarski a pretty big hammer for the conclusion that every monotone map $2\to 2$ has a fixed point? (-:

   Isn’t Knaster-Tarski a pretty big hammer for the conclusion that every monotone map $2\to 2$ has a fixed point? (-:

7. • CommentRowNumber7.
   • CommentAuthorTodd_Trimble
   • CommentTimeMar 10th 2016
   • PermaLink
   Author: Todd_Trimble
   Format: MarkdownItexMike: of course! :-) In some sense that preamble I gave is a little silly, and perhaps a distraction to what I really wanted to record, which was Cantor's theorem for posets, which I don't think is an entirely obvious result. (Or maybe it's more obvious than I think? You be the judge (-: ).

   Mike: of course! :-)

   In some sense that preamble I gave is a little silly, and perhaps a distraction to what I really wanted to record, which was Cantor’s theorem for posets, which I don’t think is an entirely obvious result. (Or maybe it’s more obvious than I think? You be the judge (-: ).

8. • CommentRowNumber8.
   • CommentAuthorDavidRoberts
   • CommentTimeMar 10th 2016
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   Author: DavidRoberts
   Format: MarkdownItexAh, OK, I didn't realise. Thanks!

   Ah, OK, I didn’t realise. Thanks!

9. • CommentRowNumber9.
   • CommentAuthorspitters
   • CommentTimeMar 10th 2016
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   Author: spitters
   Format: MarkdownItexI haven't checked this carefully, but I'd imagine this follows from [Lawvere's theorem](http://math.andrej.com/2007/04/08/on-a-proof-of-cantors-theorem/)?

   I haven’t checked this carefully, but I’d imagine this follows from Lawvere’s theorem?

10. • CommentRowNumber10.
    • CommentAuthorTodd_Trimble
    • CommentTimeMar 10th 2016
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    Author: Todd_Trimble
    Format: MarkdownItexBas, that's the point: it doesn't! At least, not in any way obvious to me. Lawvere's theorem implies that (in a cartesian closed category) if there is a point-surjective map of the form $Y \to X^Y$, then every endomap on $X$ has a fixed point. A typical application is to rule out an isomorphism $Y \cong X^Y$ because $X$ is known to have endomaps with no fixed point (e.g. $\neg: \Omega \to \Omega$ when $X = \Omega$). But for $2 = \{0 \leq 1\}$ in $Pos$, of course every endomap on $2$ does have a fixed point! So that method doesn't work here; we have to look for something else.

    Bas, that’s the point: it doesn’t! At least, not in any way obvious to me.

    Lawvere’s theorem implies that (in a cartesian closed category) if there is a point-surjective map of the form $Y \to X^Y$, then every endomap on $X$ has a fixed point. A typical application is to rule out an isomorphism $Y \cong X^Y$ because $X$ is known to have endomaps with no fixed point (e.g. $\neg: \Omega \to \Omega$ when $X = \Omega$). But for $2 = \{0 \leq 1\}$ in $Pos$, of course every endomap on $2$ does have a fixed point! So that method doesn’t work here; we have to look for something else.

11. • CommentRowNumber11.
    • CommentAuthorspitters
    • CommentTimeMar 10th 2016
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    Author: spitters
    Format: MarkdownItexThanks. Sorry for the noise.

    Thanks. Sorry for the noise.