Timeline for Compactness of the Hilbert cube without the Axiom of Choice
Current License: CC BY-SA 3.0
12 events
| when toggle format | what | by | license | comment | |
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| Jan 21, 2013 at 16:19 | history | edited | Alexey Muranov | CC BY-SA 3.0 |
Remove ambiguities from the proof, use a more convenient metric; Post Made Community Wiki
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| Jan 21, 2013 at 14:47 | history | edited | Alexey Muranov | CC BY-SA 3.0 |
"compliment" -> "complement"
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| Jan 21, 2013 at 14:42 | history | edited | Alexey Muranov | CC BY-SA 3.0 |
fix "intersection" -> "union"
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| Jan 21, 2013 at 13:00 | history | edited | Alexey Muranov | CC BY-SA 3.0 |
fix math braces
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| Jan 21, 2013 at 12:54 | history | edited | Alexey Muranov | CC BY-SA 3.0 |
Format and be more explicit
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| Jan 21, 2013 at 9:55 | history | edited | Alexey Muranov | CC BY-SA 3.0 |
take "the first" x_1,...
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| Jan 21, 2013 at 9:44 | history | edited | Alexey Muranov | CC BY-SA 3.0 |
Give proof details.
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| Jan 21, 2013 at 8:23 | history | edited | Alexey Muranov | CC BY-SA 3.0 |
update: the argument does not seem to work
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| Jan 20, 2013 at 13:05 | comment | added | Asaf Karagila♦ | Alexey far from it. The axiom of choice for families of finite sets do not imply the axiom of choice for families of countable sets, or even countable choice for that matter. | |
| Jan 20, 2013 at 12:57 | comment | added | Alexey Muranov | @Goldstern, probably you are right in general (is the Axiom of Choice for families of finite sets equivalent to the usual one?). In the case of Hilbert cube, the $\epsilon$-nets can be constructed explicitly, so that there will be no problem. | |
| Jan 20, 2013 at 10:05 | comment | added | Goldstern | You have to be careful how you formulate this proof. The implication "Every totally bounded complete metric space is compact" uses AC, as far as I can see. | |
| Jan 20, 2013 at 1:46 | history | answered | Alexey Muranov | CC BY-SA 3.0 |