Posted in math.LO, tagged ultrafilters on December 14, 2010 |
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We continue our exploration of ultrafilters. Today we’ll discuss the infinite Ramsey theorem, which is the following classical result:
Theorem: Suppose the complete graph on countably many vertices has its edges colored in one of colors. Then there is a monochromatic (i.e. an infinite subgraph all of whose edges are the same color).
The finite Ramsey theorem implies that there is a monochromatic for every positive integer , but this is a strictly stronger result; it implies not only the finite Ramsey theorem but the “strengthened” finite Ramsey theorem, and by the Paris-Harrington theorem this is independent of Peano arithmetic (although Peano arithmetic can prove the finite Ramsey theorem). Indeed, while the standard proof of the finite Ramsey theorem uses the finite pigeonhole principle, the standard proof of the infinite Ramsey theorem uses the infinite pigeonhole principle, which is stronger; this is part of the subject of a post by Terence Tao which is quite enlightening.
Given a non-principal ultrafilter on , any partition of into finitely many disjoint subsets (that is, any coloring) has the property that exactly one of the subsets is in (that is, has “full measure”), and this subset must be infinite. This subset can, in turn, be colored (partitioned), and exactly one of the blocks of the partition is in , and it must again be infinite, and so forth. It follows that a non-principal ultrafilter lets us use the infinite pigeonhole principle repeatedly (in fact this is in some sense what a non-principal ultrafilter is), and since this is exactly what is needed to prove the infinite Ramsey theorem we might expect that we could use a non-principal ultrafilter to prove the infinite Ramsey theorem. Today we’ll describe this proof, and then describe how the infinite Ramsey theorem implies the finite Ramsey theorem, which involves a different use of a non-principal ultrafilter on .
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Posted in math.GN, math.LO, math.PR, tagged ultrafilters on December 9, 2010 |
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Remark: To forestall empty set difficulties, whenever I talk about arbitrary sets in this post they will be non-empty.
We continue our exploration of ultrafilters from the previous post. Recall that a (proper) filter on a poset is a non-empty subset such that
- For every , there is some such that .
- For every , if then .
- is not the whole set .
If has finite infima (meets), then the first condition, given the second, can be replaced with the condition that if then . (This holds in particular if is the poset structure on a Boolean ring, in which case .) A filter is an ultrafilter if in addition it is maximal under inclusion among (proper) filters. For Boolean rings, an equivalent condition is that for every either or , but not both. Recall that this condition tells us that ultrafilters are precisely complements of maximal ideals, and can be identified with morphisms in . If for some set , then we will sometimes call an ultrafilter on an ultrafilter on (for example, this is what people usually mean by “an ultrafilter on “).
Today we will meander towards an ultrafilter point of view on topology. This point of view provides, among other things, a short, elegant proof of Tychonoff’s theorem.
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Recently, I have begun to appreciate the use of ultrafilters to clean up proofs in certain areas of mathematics. I’d like to talk a little about how this works, but first I’d like to give a hefty amount of motivation for the definition of an ultrafilter.
Terence Tao has already written a great introduction to ultrafilters with an eye towards nonstandard analysis. I’d like to introduce them from a different perspective. Some of the topics below are also covered in these posts by Todd Trimble.
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