Timeline for A simpler proof that compact sets have cardinality continuum?
Current License: CC BY-SA 3.0
20 events
| when toggle format | what | by | license | comment | |
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| Sep 29, 2016 at 8:14 | comment | added | Włodzimierz Holsztyński | There are trade-offs. It's good to have different proofs. | |
| Sep 29, 2016 at 8:12 | comment | added | Boaz Tsaban | That's a very cute argument! I am not sure whether it is simpler than building Cantor's set inside, but it is definitely refreshing. Thanks. | |
| Sep 29, 2016 at 6:01 | history | edited | Włodzimierz Holsztyński | CC BY-SA 3.0 |
a smoother language
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| Sep 29, 2016 at 3:42 | comment | added | Włodzimierz Holsztyński | @Goldstern, thank you--it's so nice to know it! | |
| Sep 29, 2016 at 3:24 | comment | added | Goldstern | @WłodzimierzHolsztyński Cantor's first proof was not the well-known "diagonal" proof (using decimal or ternary or binary expansions), but a sequence of nested intervals, avoiding more and more elements of a given countable sequence. I would call that "straight from the Dedekind axiom". | |
| Sep 29, 2016 at 2:54 | comment | added | Włodzimierz Holsztyński | BTW, in the paaaaast I got a direct proof of $\ |\mathbf R|>\aleph_0,\ $ straight from Dedekind Axiom, without appealing to the Cantor's diagonal method. Perhaps such proofs are known. | |
| Sep 29, 2016 at 2:47 | history | edited | Włodzimierz Holsztyński | CC BY-SA 3.0 |
LaTeX typo
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| Sep 29, 2016 at 2:41 | comment | added | Włodzimierz Holsztyński | @fedja, thank you again for being alert and keeping me honest. | |
| Sep 29, 2016 at 2:33 | comment | added | Włodzimierz Holsztyński | (I apologize for my lack of concentration. Sometimes it happens when things are too obvious, I guess. Excuses, excuses...). | |
| Sep 29, 2016 at 2:30 | history | edited | Włodzimierz Holsztyński | CC BY-SA 3.0 |
"The", not "a"
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| Sep 29, 2016 at 2:25 | comment | added | Włodzimierz Holsztyński | I am simply impatient. I'll edit it right now. The proof is sound. | |
| Sep 29, 2016 at 2:04 | comment | added | fedja | It is still a bit off: removing all isolated points creates new isolated points in general and we start sliding towards the standard argument about the existence of a compact perfect subset if we try to fix that... | |
| Sep 29, 2016 at 0:16 | comment | added | Włodzimierz Holsztyński | @fedja, done. Thank you for catching my silly (unnecessary :) ) mistake. | |
| Sep 29, 2016 at 0:15 | history | edited | Włodzimierz Holsztyński | CC BY-SA 3.0 |
Trying to bee unnecessarily "elegant". I've fixed it/
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| Sep 29, 2016 at 0:13 | comment | added | fedja | OK, I'll wait for the edit :-) | |
| Sep 29, 2016 at 0:12 | comment | added | Włodzimierz Holsztyński | I tried to be too sleek. Let me fix it. | |
| Sep 29, 2016 at 0:10 | history | edited | Włodzimierz Holsztyński | CC BY-SA 3.0 |
LaTeX formatting detail
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| Sep 29, 2016 at 0:10 | comment | added | fedja | Take $C=[-1,0]\cup\{1/n:n\ge 1\}$. Then $K$ is $[-1,0]$ plus one equivalence class to which the whole sequence collapses. The topology gets rather terrible. You cannot separate that class and $0$. Am I missing something? | |
| Sep 29, 2016 at 0:08 | history | edited | Włodzimierz Holsztyński | CC BY-SA 3.0 |
LaTeX typo
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| Sep 28, 2016 at 23:46 | history | answered | Włodzimierz Holsztyński | CC BY-SA 3.0 |