tag:blogger.com,1999:blog-11295132.post113941209358966518..comments2014-10-20T08:43:52.658-07:00Comments on A Neighborhood of Infinity: What's the most implausible but true mathematical proposition?Dan Piponihttps://plus.google.com/107913314994758123748noreply@blogger.comBlogger10125tag:blogger.com,1999:blog-11295132.post-1145137888887510002006-04-15T14:51:00.000-07:002006-04-15T14:51:00.000-07:00Replacement isn't needed for Goodstein's Theorem. ...Replacement isn't needed for Goodstein's Theorem. Check out my colleague Michael Potter's wonderful book <I>Set Theory and its Philosophy</I>. He proves Goodstein's Theorem on p. 215. (And he doesn't even introduce Replacement until p. 225.)Peter Smithhttp://www.blogger.com/profile/01319650511844414592noreply@blogger.comtag:blogger.com,1999:blog-11295132.post-1140035386440166322006-02-15T12:29:00.000-08:002006-02-15T12:29:00.000-08:00I'm not certain that replacement is needed for Goo...I'm not certain that replacement is needed for Goodstein's theorem - it just sounds plausible to me that it might be.<BR/><BR/>Somehow mentioning stationary sets reminded me that one can prove in ZFC that an Aronszajn tree exists. This isn't as bad as a Suslin tree, but it's still pretty bad. It's a tree of uncountable height (thus, with uncountably many elements), but each branch is only countable, and each level is only countable. In a Suslin tree, not only is each level of the tree countable, but so is every antichain, which is just totally ridiculous, but consistent with ZFC.Kennyhttp://www.blogger.com/profile/12226268498253877151noreply@blogger.comtag:blogger.com,1999:blog-11295132.post-1139684885963092142006-02-11T11:08:00.000-08:002006-02-11T11:08:00.000-08:00Goodstein's Theorem is one of my favourite counter...Goodstein's Theorem is one of my favourite counterintuitive results too. I like the version for 'hydra games' which makes it seem even weirder to me. <BR/><BR/>Actually, the independence of the Axiom of Choice bothers me too. From time to time I can't help thinking about how to construct a function, f, from the power set of the reals to the reals so that f(A) is in A, even though I know I can't construct such an f. It just seems like I ought to be able to if I just try hard enough, damn it!sigfpehttp://www.blogger.com/profile/08096190433222340957noreply@blogger.comtag:blogger.com,1999:blog-11295132.post-1139672000988742802006-02-11T07:33:00.000-08:002006-02-11T07:33:00.000-08:00Are you really sure you need replacement? Because,...Are you really sure you need replacement? Because, while the argument uses ordinals, you can do a perfectly good job working with ordinals as order types rather than von Neumann ordinals - it's uglier, but it works fine.<BR/><BR/>I've been trying to ome up with an answer to this question, but I'm totally failing at the moment. I think the problem is that I have a very flexible intuition, so that by the time I've understood something I don't find it implausible any more. <BR/><BR/>Possibly some stuff about stationary sets - e.g. that you can find an uncountable set of pairwise disjoint stationary subsets of aleph_1. Stationary sets are weird. One probably wouldn't expect to have much intuition about this sort of thing anyway, but it's just very different from what you'd expect from having looked at 'sensible' order topologies.<BR/><BR/>In a similar style, maybe the well ordering theorem. But then, I find it quite intuitive (after all, there are lots of order types, so you just need to use the axiom of choice to pick new elements to fill up your ordering until you've exhausted the set).David R. MacIverhttp://www.blogger.com/profile/12893796777558636623noreply@blogger.comtag:blogger.com,1999:blog-11295132.post-1139618837633087962006-02-10T16:47:00.000-08:002006-02-10T16:47:00.000-08:00Goodstein's Theorem is true alright - at least, as...Goodstein's Theorem is true alright - at least, as true as anything in second-order arithmetic. But it's definitely not provable in PA. I think it's not even provable in ZC - one needs replacement, possibly to guarantee that there is a set of all countable well-orderings.Kennyhttp://www.blogger.com/profile/12226268498253877151noreply@blogger.comtag:blogger.com,1999:blog-11295132.post-1139618345155111272006-02-10T16:39:00.000-08:002006-02-10T16:39:00.000-08:00And, at least according to one logician I've spoke...And, at least according to one logician I've spoken with, the result about G sequences _isn't even true_ (or at least not provable) in Peano Arithmetic. The result intimately requires the well-ordering of the countable ordinals.Theohttp://www.blogger.com/profile/10449695704601441213noreply@blogger.comtag:blogger.com,1999:blog-11295132.post-1139616187628486022006-02-10T16:03:00.000-08:002006-02-10T16:03:00.000-08:00Goodstein Theorem! Its worse than anything I found...Goodstein Theorem! Its worse than anything I found, including Banach Tarsky.<BR/><BR/>The <A HREF="http://mathworld.wolfram.com/GoodsteinsTheorem.html" REL="nofollow">theorem</A> seems rather simple, until you understand what a <A HREF="http://mathworld.wolfram.com/GoodsteinSequence.html" REL="nofollow">G sequence</A> is. But a few <A HREF="http://en.wikipedia.org/wiki/Goodstein's_theorem" REL="nofollow">computations</A> of a G sequence show the incredible growth of it. ¿Zero...?JuanPablohttp://www.blogger.com/profile/08262232661385198744noreply@blogger.comtag:blogger.com,1999:blog-11295132.post-1139515053477533132006-02-09T11:57:00.001-08:002006-02-09T11:57:00.001-08:00One *closed* 1-D manifold, at least.One *closed* 1-D manifold, at least.sigfpehttp://www.blogger.com/profile/08096190433222340957noreply@blogger.comtag:blogger.com,1999:blog-11295132.post-1139515032079089842006-02-09T11:57:00.000-08:002006-02-09T11:57:00.000-08:00I'm confused. There's only one 1-D manifold and su...I'm confused. There's only one 1-D manifold and surely it's not hard to embed in a Sierpinski Carpet. I must be missing something.sigfpehttp://www.blogger.com/profile/08096190433222340957noreply@blogger.comtag:blogger.com,1999:blog-11295132.post-1139472214995323382006-02-09T00:03:00.000-08:002006-02-09T00:03:00.000-08:00Last week our department was visited by Bob Devane...Last week our department was visited by Bob Devaney who gave a seminar in which he said that you can embed any closed curve in the <A HREF="http://mathworld.wolfram.com/SierpinskiCarpet.html" REL="nofollow">Sierpinkski Carpet</A>, which I found quite amazing. <BR/><BR/>The theorem was actually a bit more involved, but the gist of it was the bit about embedding closed 1-dimensional manifolds (I should really pay more attention!).Craighttp://www.blogger.com/profile/15677380778049757767noreply@blogger.com