most recent 30 from http://mathoverflow.net 2013-06-19T06:11:35Z http://mathoverflow.net/feeds/user/11771 http://www.creativecommons.org/licenses/by-nc/2.5/rdf http://mathoverflow.net/questions/20882/most-unintuitive-application-of-the-axiom-of-choice/70413#70413 Chris Eagle 2011-07-15T08:32:24Z 2011-07-15T08:32:24Z

<p>Lebesgue measure exists for every Borel set, and is countably additive.</p> <p>I've always found it more surprising that our fuzzy intuitive ideas of area and volume can be pushed as far as they can than that they break when you push even further.</p>

http://mathoverflow.net/questions/59359/choice-function-on-the-countable-subsets-of-the-reals/59362#59362 Chris Eagle 2011-03-23T22:42:29Z 2011-03-23T23:17:24Z

<p>The standard construction of a <a href="http://en.wikipedia.org/wiki/Vitali_set" rel="nofollow">Vitali set</a> only involves making choices from countable subsets of $\mathbb{R}$ (specifically, from sets of the form $(r+\mathbb{Q}) \cap [0,1]$). It is well known that ZF (assuming consistent) does not prove the existence of a non-measurable subset of $\mathbb{R}$, hence it doesn't prove the existence of a Vitali set, and thus doesn't prove your restricted choice principle.</p> <hr> <p><a href="http://consequences.emich.edu/conseq.htm" rel="nofollow">Consequences of the Axiom of Choice</a> is useful for this sort of thing. According to this, form 85 (every collection of countable sets has a choice function) is true while form 79 (every collection of sets of reals has a choice function) is false in model $\mathcal{M} 1$, which I believe is Cohen's original model of ~AC. Thus your countable choice principle for the reals does not imply full choice for the reals.</p>

http://mathoverflow.net/questions/55539/a-random-walk-on-natural-numbers/55541#55541 Chris Eagle 2011-02-15T17:46:59Z 2011-02-15T17:46:59Z

<p>Yes. The possibility of staying at $M$ is irrelevant, so let's ignore it, so that the probability of an increase is 4/9 and the probability of a decrease is 5/9. For the moment, let's also ignore the actual sizes of these increases and decreases and let $X_n$ equal the total number of increases in the first n steps minus the total number of decreases in those steps. Then $X_n$ is a simple random walk on $\mathbb{Z}$ starting at 0 with probaility 4/9 of moving right and 5/9 of moving left. By a standard result on random walks on $\mathbb{Z}$, $X_n$ almost surely tends to $-\infty$. So for any k, we almost surely come to a point when we've had k more decreases than increases.</p> <p>Going back to our original walk, every increase is by a factor of at most 2 and every decrease is by a factor of at least 2. So if we started at M and have had l increases and l+k decreases, then our current position is at most $2^l (1/2)^{l+k} M=(1/2)^k M$. So for k large enough, we must be at 0.</p>

http://mathoverflow.net/questions/54299/about-deformation-retract/54303#54303 Chris Eagle 2011-02-04T11:09:38Z 2011-02-04T11:09:38Z

<p>If $X$ is a subcomplex of $Y$ and the inclusion map is a homotopy equivalence, then $X$ is a deformation retract of $Y$. See for example proposition 0.16 and corollary 0.20 in <a href="http://www.math.cornell.edu/~hatcher/AT/ATpage.html" rel="nofollow">Hatcher</a>.</p>

http://mathoverflow.net/questions/52831/total-order-on-the-powerset/52836#52836 Chris Eagle 2011-01-22T15:28:54Z 2011-01-22T15:48:13Z

<p>No. If this were true, then ZF would prove that "every set can be totally ordered" implies "every set can be well-ordered", which (assuming ZF is consistent) it doesn't. I can't find the original citation for this nonimplication, but it's in Howard and Rubin's "Consequences of the Axiom of Choice" for example.</p>

http://mathoverflow.net/questions/51790/families-of-finite-groups-arising-from-direct-products/51791#51791 Chris Eagle 2011-01-11T22:13:34Z 2011-01-11T22:13:34Z

<p>A finite group is nilpotent iff it is a direct product of $p$-groups.</p>

http://mathoverflow.net/questions/51415/is-it-possible-to-show-that-an-infinite-set-has-a-countable-infinite-subset-wi/51417#51417 Chris Eagle 2011-01-07T17:31:31Z 2011-01-07T17:31:31Z

<p>No. A set which has a countably infinite subset is called <a href="http://en.wikipedia.org/wiki/Dedekind-infinite_set" rel="nofollow">Dedekind-infinite</a>. Clearly every Dedekind-infinite set is infinite; the statement that every infinite set is Dedekind-infinite is not provable in ZF (assuming ZF is consistent, of course). You don't need full AC, though. In fact, the equivalence isn't even as strong as countable choice.</p>

http://mathoverflow.net/questions/50550/asymptotic-density-of-provable-statements-in-zfc/50556#50556 Chris Eagle 2010-12-28T11:54:12Z 2010-12-28T11:54:12Z

<p>You can make $T_n(n)/n$ converge to any value you like by choosing a suitably silly Gödel numbering. Partition $\mathbb{N}$ (computably) into a set $A$ of density $p$, and set $B$ of density $1-p$, and an infinite set $C$ of density 0. Arrange your numbering scheme so that $A$ corresponds exactly to the ZFC axioms, $B$ corresponds to the negations of the axioms, and $C$ corresponds to all other statements. For this numbering, $T_n(n)/n$ tends to $p$.</p>

http://mathoverflow.net/questions/50165/compactness-theorem-with-preserved-substructure/50169#50169 Chris Eagle 2010-12-22T16:06:25Z 2010-12-22T16:06:25Z

<p>No. Just let the signature contain lots (more than cardinality continuum) of constants and let the axioms of T be that all the constants are different and R holds for all of them.</p>

http://mathoverflow.net/questions/75048/independence-of-being-an-integer Chris Eagle 2011-09-09T23:21:07Z 2011-09-09T23:21:07Z

If it's not an integer, then ZFC can certainly prove that, just by approximating it closely enough.

http://mathoverflow.net/questions/72455/a-question-about-the-axiom-of-choice Chris Eagle 2011-08-09T08:12:05Z 2011-08-09T08:12:05Z

This is not a research-level question.

http://mathoverflow.net/questions/67786/bijection-of-proper-classes/67789#67789 Chris Eagle 2011-06-18T10:18:14Z 2011-06-18T10:18:14Z

How does this prove $\kappa^2=\kappa$ for every $\kappa \ge \aleph_0$? It seems to only work for alephs.

http://mathoverflow.net/questions/29006/counterexamples-in-algebra/29761#29761 Chris Eagle 2011-06-18T08:50:29Z 2011-06-18T08:50:29Z

Every integral domain is a subring of a UFD (for example, its field of fractions). So all you need here is an integral domain which is not a UFD.

http://mathoverflow.net/questions/68100/using-bayes-theorem-gives-a-probability-1 Chris Eagle 2011-06-17T22:00:08Z 2011-06-17T22:00:08Z

This is not a research-level maths question.

http://mathoverflow.net/questions/66543/ordinal-set-omega-von-neumann-definition-and-modern-definition Chris Eagle 2011-05-31T13:19:26Z 2011-05-31T13:19:26Z

With what you call the modern definition, the ordinals themselves are not sets (except $0$), so it doesn't even make sense to ask if the class of ordinals is a set.

http://mathoverflow.net/questions/65513/why-the-triangle-inequality/65517#65517 Chris Eagle 2011-05-20T10:46:04Z 2011-05-20T10:46:04Z

Perhaps Michael means that you need the triangle inequality to show that open balls are open, and so the metric topology is generated by the open balls.

http://mathoverflow.net/questions/65112/incompleteness-in-real-analysis/65136#65136 Chris Eagle 2011-05-17T14:04:59Z 2011-05-17T14:04:59Z

Once you've got quantification over sets, it's easy to define the integers. &quot;$x$ is an integer&quot; means &quot;$x$ is in every subset of $\mathbb{R}$ containing $0$, $1$ and $-1$ and closed under addition&quot;.

http://mathoverflow.net/questions/65112/incompleteness-in-real-analysis/65113#65113 Chris Eagle 2011-05-16T14:02:49Z 2011-05-16T14:02:49Z

The first-order theory of the real ordered field is not categorical in any uncountable cardinal. Such theories are $\omega$-stable, and hence can never define an infinite total order.

http://mathoverflow.net/questions/64617/mathematical-ideas-named-after-places/64631#64631 Chris Eagle 2011-05-11T16:35:39Z 2011-05-11T16:35:39Z

and the &quot;British Rail metric&quot;

http://mathoverflow.net/questions/64272/characterization-of-tychonoff-spaces-in-terms-of-open-sets Chris Eagle 2011-05-08T10:31:29Z 2011-05-08T10:31:29Z

Here's one $\mathbb{R}$-free characterization: a space is Tychonoff iff it has a Hausdorff compactification. Is that the sort of thing you want?

http://mathoverflow.net/questions/63596/how-to-decompose-an-infinite-set-into-two-isomorphic-ones-without-choice Chris Eagle 2011-05-01T11:37:05Z 2011-05-01T11:37:05Z

@Andr&#233;: Why are you presuming that? The usual definition of &quot;infinite&quot; is &quot;not equinumerous with any natural number&quot;. Your property is called &quot;Dedekind infinite&quot;.

http://mathoverflow.net/questions/62929/could-make-a-bijection-between-polynomials-with-integer-coefficients-and-rational Chris Eagle 2011-04-25T14:33:34Z 2011-04-25T14:33:34Z

This is not suitable to this site. Try <a href="http://math.stackexchange.com/" rel="nofollow">math.stackexchange.com</a>.

http://mathoverflow.net/questions/62833/zariskis-main-theorem Chris Eagle 2011-04-24T11:58:35Z 2011-04-24T11:58:35Z

Crossposted at <a href="http://math.stackexchange.com/questions/34836/zariskis-main-theorem" rel="nofollow" title="zariskis main theorem">math.stackexchange.com/questions/34836/&hellip;</a>

http://mathoverflow.net/questions/60666/solvability-of-an-elementary-functional-equation Chris Eagle 2011-04-05T10:34:29Z 2011-04-05T10:34:29Z

$f(x,0)=g(0)$, not $0$.