If X is a set and A is a subset of X containing at least two elements, then certainly for any element $a \in A$, the principal ultrafilter of $a$ contains the principal filter of A (which is NOT an ultrafilter). Are all ultrafilters containing the principal filter of A of this form, or are there non-principal examples? (I'm assuming axiom of choice etc.)
[Math] Ultrafilters containing a principal filter
set-theoryultrafilters
Related Solutions
See the Prime ideal theorem.
The existence of ultrafilters on every Boolean algebra (which implies non-principal ultrafilters on ω, since these come from ultrafilters on the Boolean algebra P(ω)/Fin) is a set-theoretic principle that follows from AC and is not provable in ZF (if ZF is consistent), but which does not imply full AC. Thus, it is an intermediate weaker choice principle.
Your statement about ultrafilters on ω appears to be even weaker, since it is such a special case of the Prime Ideal Theorem.
Nevertheless, I believe that the method of forcing shows that it is consistent with ZF that there are no non-principal ultrafilters on ω. I believe that some of the standard models of ¬AC, built by using symmetric names for adding Cohen reals, have DC, and hence also ACω, but still have no nonprincipal ultrafilters on ω. In this case, neither DC nor ACω would imply the existence of such ultrafilters.
I'm less sure about finding models that have ultrafilters on ω, but not on all Boolean algebras. But I believe that this is likely the case. These models would show that your principle is strictly weaker even than the Prime Ideal Theorem.
The product $a\times b$ of two ultrafilters is an ultrafilter if and only if, for every function $f$ from the underlying set of $a$ into $b$ (that's not a typo for "into the underlying set of $b$"), there is a set $A\in a$ such that $\bigcap_{x\in A}f(x)\in b$. One way for this to happen is for the underlying set of $a$ to be small enough and $b$ to be complete enough, as in Joel's answer. Notice, though, that the condition is, despite its appearance, symmetrical between $a$ and $b$. In particular, if $b$ lives on $\omega$ while $a$ is countably complete, then the condition is satisfied because $f$ will be constant on some set in $a$ (because countably complete ultrafilters are closed under intersection of continuum many sets). [Archaeologists may be interested to know that this characterization of ultrafilters whose product is ultra occurs on page 22 of my 1970 Ph.D. thesis, which is, thanks to patient scanning, available from my web page.]
Best Answer
If a filter $X$ contains any set $A$, then it contains the principal filter of $A$. Thus you are really asking: for which subsets $A$ of a set $X$ can a free ultrafilter contain $A$?
It is a standard exercise to see that such free ultrafilters exist iff $A$ is infinite.
Let me briefly sketch the proof:
If an ultrafilter contains a finite union of sets $A_1,\ldots,A_n$, then it contains $A_i$ for at least one $i$. Thus an ultrafilter which contains any finite set is principal.
If $A$ is infinite, consider the family $F$ of subsets which either contain $A$ or have finite complement. Then $F$ satisfies the finite intersection condition, so is a subbase for a filter $\mathcal{F}$ (see e.g. Exercise 5.2.5 of http://alpha.math.uga.edu/~pete/convergence.pdf). Then it can be extended to an ultrafilter which, since it contains the Frechet filter, is free.