*Note: This post is copied over from an old WordPress blog, with two posts. The posts were moved here on June 23, 2018.*

Let \(X\) be a set, and let \(\mathcal{U}\) be a family of subsets of \(X\). Prove that \(\mathcal{U}\) is an ultrafilter on \(X\) if and only if for every partition of \(X\) into three sets \(S_1\), \(S_2\), and \(S_3\), exactly one \(S_i\) lies in \(\mathcal{U}\).

The point, I guess, is that the most fundamental property of ultrafilters is their ability to select one set out of any partition to be “significant” and declare the rest unimportant. This matches the intuition of an ultrafilter being a “perfect locating scheme” (Wikipedia). Except for the very surprising fact that it really only needs to be a perfect scheme; the “locating” part is implied!

Of course, a disadvantage of defining ultrafilters in this way is that I can’t come up with any analogous definition of a filter. Filters, being imperfect locating schemes, need to be explicitly specified to be closed under intersection and so on.

*Thanks to Marion Scheepers for the math problem (a variant of what I presented above).*

*Edited on .*