Timeline for Proof of the axiom of choice for finite sets in ZF
Current License: CC BY-SA 4.0
13 events
when toggle format | what | by | license | comment | |
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May 8 at 7:22 | comment | added | Andrej Bauer | @DominicvanderZypen: all is well, you seem to be past the first stage :-) | |
May 8 at 6:48 | comment | added | Dominic van der Zypen | @AndrejBauer I am only an amateur mathematician - please cut me some slack :) | |
May 8 at 6:46 | comment | added | Dominic van der Zypen | You guys don't seem to like contradiction. In Switzerland we have direct democracy, people can vote on stuff, loads of debates, so a contradictory lifestyle is in my DNA. | |
May 7 at 20:55 | comment | added | Andrej Bauer | Horrible, horrible unecessary use of proof by contradition. Straight induction would have worked. | |
May 7 at 14:10 | comment | added | Sam Hopkins | You also really don’t need to phrase the proof as by contradiction. Anyways, the question itself is off topic here so it doesn’t need a perfect answer. | |
May 7 at 12:46 | history | edited | Dominic van der Zypen | CC BY-SA 4.0 |
simplified the argument after suggestion by user @LSpice
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May 7 at 12:45 | comment | added | Dominic van der Zypen | Thanks @LSpice - will use the shortcut you mention | |
May 7 at 12:12 | comment | added | LSpice | I think it doesn't matter for your last paragraph that $n_0 > 1$, as opposed to $n_0 > 0$. That is, you don't need to handle the $n_0 = 1$ case separately. | |
May 7 at 11:55 | history | edited | Sam Hopkins | CC BY-SA 4.0 |
added 14 characters in body
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May 7 at 11:22 | comment | added | Dominic van der Zypen | It was a pleasure - and welcome to MathOverflow! | |
May 7 at 9:37 | comment | added | Andrew S. | Great answer, thank you so much!) | |
May 7 at 9:37 | vote | accept | Andrew S. | ||
May 7 at 9:30 | history | answered | Dominic van der Zypen | CC BY-SA 4.0 |