Wednesday, February 08, 2006

What's the most implausible but true mathematical proposition?

Some would say the Banach-Tarski paradox but my attitude has always been that all bets are off when dealing with unmeasurable sets. I think the existence of a number satisfying the main property of Khinchin's constant is much less plasuble.

Addendum: According to this paper, even though the Khinchin property is true for almost all real numbers, nobody has explicitly found a number for which it is true! You can artificially construct a number for which it is true but that doesn't give you an explicit result, and anyway, that's cheating. Experimentally, π gives something near the Khinchin constant, but there's no proof of what happens in the limit.

10 comments:

Craig said...

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 Sierpinkski Carpet, which I found quite amazing.

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!).

sigfpe said...

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.

sigfpe said...

One *closed* 1-D manifold, at least.

JuanPablo said...

Goodstein Theorem! Its worse than anything I found, including Banach Tarsky.

The theorem seems rather simple, until you understand what a G sequence is. But a few computations of a G sequence show the incredible growth of it. ¿Zero...?

Theo said...

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.

Kenny said...

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.

David R. MacIver said...

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.

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.

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.

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).

sigfpe said...

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.

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!

Kenny said...

I'm not certain that replacement is needed for Goodstein's theorem - it just sounds plausible to me that it might be.

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.

Peter Smith said...

Replacement isn't needed for Goodstein's Theorem. Check out my colleague Michael Potter's wonderful book Set Theory and its Philosophy. He proves Goodstein's Theorem on p. 215. (And he doesn't even introduce Replacement until p. 225.)

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