Timeline for What is your favorite proof of Tychonoff's Theorem?
Current License: CC BY-SA 3.0
8 events
| when toggle format | what | by | license | comment | |
|---|---|---|---|---|---|
| May 17 at 17:59 | comment | added | Christopher King | It may be worth pointing out why this doesn't work in the box topology. | |
| Jan 23, 2022 at 6:17 | comment | added | Martin Väth | @GeraldEdgar: The name you mention is spelled W.A.J. Luxemburg | |
| Feb 27, 2013 at 3:38 | comment | added | Benjamin Steinberg | Tychonoff for Hausdorff equals Boolean principal ideal Thm. The ultrafilter proof only needs Boolean principal ideal Thm in the Hausdorff case because you dont need to choose a limit point in each factor. | |
| May 20, 2011 at 16:05 | comment | added | David White | What I'm asking I guess is: Is Tychonoff for Hausdorff spaces equivalent to AC? It seems the answer must be no or else the Boolean Theorem would be equivalent to AC | |
| May 20, 2011 at 16:03 | comment | added | David White | Wait a minute, now I'm confused. I have never heard of the Boolean Algebra Maximal Ideal Theorem, but if it implies Tychonoff in this proof without AC and if Tychonoff implies AC then how can the Boolean Theorem be strictly weaker than AC? | |
| May 20, 2011 at 12:17 | comment | added | Gerald Edgar | W.A.J. Luxembourg noted an interesting consequence of doing the proof this way. If the spaces are Hausdorff, then no additional application of choice is needed. Only the use of AC that produces the nonstandard model ... and for that the Boolean Algebra Maximal Ideal Theorem (strictly weaker than AC) suffices. | |
| May 20, 2011 at 8:42 | comment | added | mathahada | How do you define the associated "non standard space" to a standard space? | |
| May 20, 2011 at 8:39 | history | answered | jasomill | CC BY-SA 3.0 |