Timeline for A simpler proof that compact sets have cardinality continuum?
Current License: CC BY-SA 3.0
17 events
| when toggle format | what | by | license | comment | |
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| Oct 8, 2016 at 17:27 | history | edited | Boaz Tsaban | CC BY-SA 3.0 |
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| Oct 7, 2016 at 9:55 | vote | accept | Boaz Tsaban | ||
| Oct 5, 2016 at 11:18 | answer | added | Uri Bader | timeline score: 4 | |
| Oct 1, 2016 at 16:52 | comment | added | Boaz Tsaban | @BillJohnson: When I first wrote my question, by "building a Cantor set inside" I meant what you just wrote. But later I realized our compact set may have isolated points, so what you wrote is not immediately "more" than what I asked for. | |
| Sep 30, 2016 at 14:00 | comment | added | Bill Johnson | @BoazTsaban: I want more when I teach real analysis, so I prove that every complete metric space with no isolated points contains a Cantor set (start with two disjoint closed balls of radius at most one and inside the interior of each take two disjoint closed balls of radius at most one half and continue). | |
| Sep 30, 2016 at 8:49 | comment | added | Emil Jeřábek | @R.. Yes, exactly. | |
| Sep 29, 2016 at 15:42 | comment | added | R.. GitHub STOP HELPING ICE | Well I'm confused. Assuming the continuum hypothesis, any uncountable set of real numbers has cardinality continuum. Is the statement that, even omitting that assumption, you can prove a compact set does not have cardinality somewhere in between? | |
| Sep 29, 2016 at 12:39 | comment | added | LSpice | @EmilJeřábek, I think that R..'s question may have had to do with the fact that the title asks for "A simpler proof that compact sets have cardinality continuum", omitting the necessary uncountability hypothesis that is included in the post. | |
| Sep 29, 2016 at 9:54 | history | edited | Boaz Tsaban | CC BY-SA 3.0 |
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| Sep 29, 2016 at 8:23 | comment | added | Boaz Tsaban | @BillJohnson: Do you use Petrov's argument below, or do you have your own preferred variation? In the latter case, it would be useful if you could share it with us. | |
| Sep 29, 2016 at 6:01 | comment | added | Emil Jeřábek | @R.. The statement is correct as is. Your version is wrong, as there are countably infinite compact sets. | |
| Sep 29, 2016 at 1:28 | comment | added | R.. GitHub STOP HELPING ICE | What is the actual statement you're trying to prove? Did you mean infinite and not uncountable? | |
| Sep 28, 2016 at 23:46 | answer | added | Włodzimierz Holsztyński | timeline score: 3 | |
| Sep 28, 2016 at 21:57 | answer | added | fedja | timeline score: 10 | |
| Sep 28, 2016 at 21:47 | answer | added | Fedor Petrov | timeline score: 9 | |
| Sep 28, 2016 at 21:46 | comment | added | Bill Johnson | When I teach real analysis I do it by building a Cantor set inside. That is basically only one step away from building a one to one function from $\{0,1\}^{\aleph_0}$ into the set. | |
| Sep 28, 2016 at 21:25 | history | asked | Boaz Tsaban | CC BY-SA 3.0 |