Timeline for Who wins infinite Hex?
Current License: CC BY-SA 3.0
19 events
| when toggle format | what | by | license | comment | |
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| Jul 6, 2023 at 8:25 | comment | added | Burak | @GeoffreyIrving: Closed (indeed, all Borel) subsets of a Polish space have the perfect set property. So for these subsets being uncountable automatically implies being of size continuum. | |
| Jul 4, 2023 at 23:00 | comment | added | Geoffrey Irving | As a nit: shouldn’t it be an enumeration of cardinality c subsets, not uncountable subsets? Otherwise without the continuum hypothesis some of them might fill up. | |
| Oct 1, 2017 at 23:18 | comment | added | Delio Mugnolo | @PyRulez Good question! Actually, I wanted to ask the same: If an answer can be found (without using the Axion of Choice), then Brouwer's fixed point theorem in separable infinite dimensional spaces could be proved constructively. | |
| Sep 29, 2017 at 1:21 | vote | accept | Christopher King | ||
| Sep 28, 2017 at 15:14 | comment | added | Christopher King | @StevenStadnicki what if you require Bob to have some sort of "constructive" strategy, so he can't make use of the axiom of choice? | |
| Sep 27, 2017 at 23:21 | history | edited | Burak | CC BY-SA 3.0 |
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| Sep 27, 2017 at 23:14 | comment | added | Steven Stadnicki | @EricWofsey Oh, that makes perfect sense - I'd missed that the definition itself(!) needs AC. Thank you! | |
| Sep 27, 2017 at 23:10 | history | edited | Burak | CC BY-SA 3.0 |
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| Sep 27, 2017 at 23:07 | comment | added | Eric Wofsey | In fact, a well-ordering of $\mathfrak{c}$ is exactly what this answer needs to work, so this shows in ZF that no matter what, neither player has a winning strategy (for rather different reasons, depending on whether a well-ordering of $\mathfrak{c}$ exists!). | |
| Sep 27, 2017 at 22:54 | comment | added | Eric Wofsey | @StevenStadnicki: Without AC, then it is possible that neither player can win even if the players are cooperating. Indeed, a victory for either player gives a well-ordering of a set of cardinality $\mathfrak{c}$, and it is consistent with ZF that no such well-ordering exists. | |
| Sep 27, 2017 at 22:52 | history | edited | Burak | CC BY-SA 3.0 |
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| Sep 27, 2017 at 22:45 | history | undeleted | Burak | ||
| Sep 27, 2017 at 22:41 | history | deleted | Burak | via Vote | |
| Sep 27, 2017 at 22:25 | history | edited | Burak | CC BY-SA 3.0 |
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| Sep 27, 2017 at 22:22 | comment | added | Steven Stadnicki | This answer needs AC - I wonder if the ultimate answer depends on choice. | |
| Sep 27, 2017 at 22:18 | history | edited | Burak | CC BY-SA 3.0 |
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| Sep 27, 2017 at 22:13 | history | edited | Burak | CC BY-SA 3.0 |
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| Sep 27, 2017 at 22:05 | history | edited | Burak | CC BY-SA 3.0 |
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| Sep 27, 2017 at 22:02 | history | answered | Burak | CC BY-SA 3.0 |