Timeline for Can there be two continuous real-valued functions such that at least one has rational values for all x?
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7 events
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| Sep 10, 2010 at 21:55 | comment | added | Joel David Hamkins | I am saying merely that it is consistent with the axioms of set theory that you could extend the result to $\aleph_1$ many functions, but of course, this is only possible when $\aleph_1\lt 2^{\aleph_0}$, which is to say, when the Continuum Hypothesis fails. The result can never hold for continuum $2^{\aleph_0}$ many functions, as your example shows. | |
| Sep 10, 2010 at 21:52 | comment | added | user5810 | There are more than omega_1 reals. (when it doesn't work with omega_1 functions) | |
| Sep 10, 2010 at 21:44 | comment | added | mathahada | Hey Joel. Why it doesn't work with uncountably many functions? It is enough to have f(x) = ax for any real a. | |
| Sep 10, 2010 at 20:58 | comment | added | Joel David Hamkins | It is part of Cichon's diagram (see en.wikipedia.org/wiki/Cichon_diagram). This seems to be one of the standard results in the field of cardinal characteristics. It is probably in the survey article by Andreas Blass: math.lsa.umich.edu/~ablass/set.html | |
| Sep 10, 2010 at 20:51 | comment | added | user5810 | +1. @Joel, do you know where I could find a proof of that consistency result? (I suppose what I'm really interested in is a respectable article stating it, since I probably wouldn't understand a proof.) | |
| Sep 10, 2010 at 20:44 | comment | added | Joel David Hamkins | +1. This way of thinking about it shows that it is consistent with $ZFC+\neg CH$ that you can't even do it with $\aleph_1$ many functions (or more, even), since it is known to be consistent with $\neg CH$ that the ideal of meager sets has more than countable additivity. | |
| Sep 10, 2010 at 20:17 | history | answered | Mikhail Bondarko | CC BY-SA 2.5 |