most recent 30 from http://mathoverflow.net 2013-05-20T02:33:03Z http://mathoverflow.net/feeds/user/8913 http://www.creativecommons.org/licenses/by-nc/2.5/rdf http://mathoverflow.net/questions/37272/are-all-sets-totally-ordered Steven 2010-08-31T15:59:29Z 2013-05-19T07:05:28Z

<p>The question is the title. </p> <p>Working in ZF, is it true that: for every nonempty set X, there exists a total order on X ? </p> <p>If it is false, do we have an example of a nonempty set that has no total order?</p> <p>Thanks</p>

http://mathoverflow.net/questions/66462/which-functions-of-one-variable-are-derivatives/66553#66553 Steven 2011-05-31T14:40:59Z 2011-05-31T14:40:59Z

<p>A result that is related to your question (the "almost everywhere" is the difference) :</p> <p>Every Henstock-Kurzweil integrable function on [a,b] is almost everywhere the derivative of a differentiable function, and inversely, any derivative is Henstock-Kurzweil integrable.</p> <p>More here : <a href="http://www.math.vanderbilt.edu/~schectex/ccc/gauge/" rel="nofollow">http://www.math.vanderbilt.edu/~schectex/ccc/gauge/</a></p>

http://mathoverflow.net/questions/22927/why-worry-about-the-axiom-of-choice/22937#22937 Steven 2010-10-21T16:30:26Z 2010-10-21T16:30:26Z

Actually, it is provable in ZF (without any form of the axiom of choice) that a function from R into R is continuous if and only if it is sequentially continuous. (Theorem 4.52 of &quot;Axiom of Choice&quot; of Herrlich). Most of the real analysis, contrarily to popular belief, is still in force without any form of the axiom of choice.