User cory knapp - MathOverflow

Non-principal ultrafilters on ω

2010-02-20T05:00:08Z

I thought I had heard or read somewhere that the existence of a non-principal ultrafilter on $\omega$ was equivalent to some common weakening of AC. As I searched around, I read that this is not the case: neither countable choice nor dependent choice are strong enough.

This leads me to two questions:

Where would I find a proof that DC is not strong enough to prove the existence of a non-principal ultrafilter on $\omega$?

Is the assumption that there exists a non-principal ultrafilter on $\omega$ strong enough to show DC or countable choice? I.e. is "there exists a non-principal ultrafilter on $\omega$" stronger than either of countable or dependent choice?

Answer by Cory Knapp for Cocktail party math

2010-02-10T07:29:02Z

When trying to talk about specific results, I really like talking about Cantor's Theorem (or at least, the special case of $2^{\aleph_0}$), and then, if they're willing to accept that, talk about ordinals a bit. If the audience isn't taking it, I'll generally talk about some arbitrary graph problem that comes to my head.

But when asked, I typically try to approach it from the more philosophic perspective of "what mathematicians do"-- study abstract structure. (Or at least, that's the approach I take to math), and try to explain what that means, providing some examples here and there. The definition of group shows up a lot, since there are easy to understand examples of groups. I also view myself as a bit of an artist, so I tend to use analogies that deal with music and painting.

Coherent spaces

2009-11-28T15:55:32Z

In Proofs and Types, Girard discusses coherent (or coherence) spaces, which is defined as a set family which is closed downward ($a\in A,b\subseteq a\Rightarrow b\in A$), and binary complete (If $M\subseteq A$ and $\forall a_1,a_2\in M (a_1\cup a_2\in A)$, then $\cup M\in A$)

It was informally related to topological spaces. Anyway, I have a couple pretty general questions: Are they particularly useful outside of type theory? Perhaps more specifically, do coherent spaces show up in topology?

The last one raises up a philosophical question I've been pondering: Why is it that some structures seem to show up all over the place, while others that seem like they "should" be more or less equivalently useful don't seem to show up much at all? An example would be matroids versus topologies. I feel, morally, that matroids should be more useful than they seem to be.

The last question probably doesn't have any sort of solid answer, but it would be nice to hear some thought from people with a stronger background.

Cheers and thanks, Cory

Edit: After thinking about this some more, it has occurred to me that coherent spaces are a sort of "dual" to ultrafilters. I really don't have the background to be terribly formal, but, let me try to explain:

Let $(X,C)$ be a coherent space, and call the elements of $C$ "open" (I think the analogy is justified, because adding $X$ to $C$ makes it a topology), then the closed sets form an ultrafilter. The one problem is that the closure under intersection is a bit strong (the set of closed sets is closed under arbitrary intersections). On the other hand, if $(X,U)$ is an ultrafilter, the set of complements of open sets almost forms a coherent space-- but the conditions on unions is just a little too weak.

So, my next question is: Has this link been explored at all? Is there even anything there to explore?

Thanks again.

Answer by Cory Knapp for Famous mathematical quotes

2009-11-29T20:37:35Z

My favorite math quote will probably always be Paul Gordan's response to Hilbert's proof of his Basis Theorem: "This is not Mathematics. This is Theology."

Along with his redaction after he came to accept the method: "I have convinced myself that even theology has its merits."

Comment by Cory Knapp

2010-06-09T01:42:53Z

It took me a long time to realize that was false as well... Still being an undergrad, I often catch myself trying to use that "theorem".

Comment by Cory Knapp

2010-04-29T16:15:09Z

This is not at all an answer to your question, but a friend of mine suggested a possible model-theoretic proof of the infinitude of Mersenne primes some time last year. It seemed to reduce the problem to a harder model theory problem (hence the fact that this proof was nver finished.) It seems to be a similar sort of thing: there's a nice model theoretic way of looking at a problem, and a lot of the work has already been done by model theorists somewhere.

Comment by Cory Knapp

2010-03-04T03:15:24Z

I was really glad when my analysis professor first showed this example; I had the same realization that you did.

Comment by Cory Knapp

2010-03-04T03:11:10Z

Thanks, Andrej; These are all really fun... now to pick them apart and internalize them.

Comment by Cory Knapp

2010-02-24T01:00:18Z

I read the "of course" in that sentence to be something to counter what was said previously: "(Some statement about irreducible modules); of course [i.e. however,], every irreducible module is completely reducible.

Comment by Cory Knapp

2010-02-20T06:38:02Z

Thanks, indeed, François. That gives a more solid answer to my first question.

Comment by Cory Knapp

2010-02-20T05:32:25Z

Eep! Thanks... Not sure how I missed that.

Comment by Cory Knapp

2010-02-20T05:26:02Z

Thanks. The article led me in exactly the direction I was looking.

Comment by Cory Knapp

2010-02-06T06:09:45Z

@Steve: There is only one morphism between two objects, but I fail to see how this means the morphisms are irrelevant. A preorder only has only one morphism between any pair of objects, but I would hardly say that there is no relationship between objects and morphisms. @david: There will be more than one morphism with the same "name", but these are distinct morphisms. In your example 1:1->0 and 1:0->1 are actually different morphisms. The category formed this way seems boring: As a picture, it will always be K_n with directed edges. (This gives a hint as to what a functor looks like.)

Comment by Cory Knapp

2009-12-10T06:31:57Z

Thanks! That really informative, and gives me a place to look for them in the future.

Comment by Cory Knapp

2009-12-09T16:28:56Z

I'm assuming omnia (literally all, I think) is used to mean something along the lines of "sum".

Comment by Cory Knapp

2009-12-07T19:15:23Z

Speaking some German, my experience is probably useless here, but the only German texts I can consistently read without much difficulty are math texts. On the other hand, I can almost understand the gist of French texts while knowing "no" French.

Comment by Cory Knapp

2009-12-06T12:15:46Z

I've never actually seen the second definition explicitly, although I've used it implicitly often enough. I don't completely see how that's a clearer exposition, though.

Comment by Cory Knapp

2009-11-30T05:42:57Z

@Georges Elencwajg: "Bewiessen" was a indeed typo, but I was correcting the word have (the original said "er have..." Perhaps it is my inadequate German, but "er habe.." still seems to be an incorrect conjugation.

Comment by Cory Knapp

2009-11-29T21:29:03Z

I thought the justification for the "under 40" requirement for the Fields medal was to encourage future research. This is still a bit of a cop out, but it's worth noting.