Timeline for Compactness of the Hilbert cube without the Axiom of Choice
Current License: CC BY-SA 3.0
7 events
| when toggle format | what | by | license | comment | |
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| Jan 19, 2013 at 22:34 | comment | added | Goldstern | I don't understand what you are arguing about. Asaf, Andrej and I agree that the version of König's lemma that is needed here does not need AC. | |
| Jan 19, 2013 at 22:15 | comment | added | Asaf Karagila♦ | Clearly, you should understand my comment in the form that is relevant to pointing out that the general case of Koenig's lemma does require choice. I have seen more than enough people thinking that the existence of free ultrafilters is equivalent to the axiom of choice, just because you are using Zorn's lemma. | |
| Jan 19, 2013 at 22:07 | comment | added | Asaf Karagila♦ | Andrej, it depends on your formulation of the lemma. If you formulate it over a countable tree - sure. But if you only state that the tree has $\omega$ levels and each is finite then you do need the axiom of choice. See Jech's "The Axiom of Choice" for details, Chapter 7 somewhere around the end. | |
| Jan 19, 2013 at 22:03 | comment | added | Eric Wofsey | Compactness of $2^\omega$ can also easily be seen by identifying it with the Cantor set, a closed subset of $[0,1]$. | |
| Jan 19, 2013 at 22:00 | history | edited | Goldstern | CC BY-SA 3.0 |
WKL
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| Jan 19, 2013 at 21:59 | comment | added | Asaf Karagila♦ | But Koenig's lemma does require a minute amount of choice. It is not needed in the case of $2^\omega$ because that is a linearly ordered set, and every level is finite, so we get a choice function. | |
| Jan 19, 2013 at 21:54 | history | answered | Goldstern | CC BY-SA 3.0 |