This question stems from Dick Lipton's recent blog post on the Axiom of Choice. I asked there but got no takers. I promise I'm not an inept Googler, but I couldn't find a satisfactory answer. I suspect universal in this context means computable by a universal Turing machine, or something close to that, but I'd like to know for sure.

$\begingroup$ I added the axiomofchoice and settheory tags. $\endgroup$ – Joel David Hamkins Apr 19 '10 at 20:19
In that argument, he just means that g is defined on all the twoelement subsets that may arise in the argument, that is, for a given family F of fourelement sets, g(A) should be defined on any twoelement set A that is a subset of a fourelement set in F.
The reason he needs to assume that is that he cannot allow that we need to choose the choice function g itself to be used with each separate A. There are many choice functions that work for families of 2element sets, and different choices of g will give rise to different functions on the families of fourelement sets. The way the argument works is that you make one choice of the function g that works on all the two element sets that arise, and then you define the choice function on the given family of fourelement sets by the clever construction in the article.
In particular, he is not using some technical meaning of universal.

$\begingroup$ Although an epsilonterm would do the trick as a universal choice operator. $\endgroup$ – Harry Gindi Apr 19 '10 at 20:33

2$\begingroup$ Harry, in that terminology, what the argument the questioner links to concerns is a proof that if you have an epsilon choice function defined on all pairs, then you can define from it (without any choice) a choice function that works on all families of fourelement sets. (But interestingly, not families of threeelement sets!) But actually, there is no need in the argument for the proper class degree of universality provided by the epsilon term idea, so it is a stronger result not to use this. $\endgroup$ – Joel David Hamkins Apr 19 '10 at 20:40

$\begingroup$ I was just saying that is what a universal choice operator "usually" means. =) $\endgroup$ – Harry Gindi Apr 19 '10 at 21:00