This question is not about the logical relationships between Choice, Zorn's and the Ultrafilter Lemma, but pedagogical. I am teaching a class and want to go as directly as possible from Choice to a non-principal ultrafilter on $\mathbf{N}$.

The Axiom of Choice is much easier to understand than Zorn's Lemma and the definition of a non-principal ultrafilter on the natural numbers is also pretty easy to understand. However, the proof of Zorn's Lemma using the axiom of choice is relatively technical and introducing partially ordered sets and chains just for this purpose feels a little time-consuming.

Question. Is there a sneaky construction of a non-principal ultrafilter on $\mathbf{N}$ that directly uses the axiom of choice but avoids introducing the statement of Zorn's Lemma?

  • $\begingroup$ I always find it amusing to read "construction that uses AoC" $\endgroup$ – Hagen von Eitzen Feb 23 '17 at 17:26
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    $\begingroup$ @Hagen: Well, that's a standard usage, though. Modulo the choice function you are constructing something (via transfinite recursion usually). $\endgroup$ – Asaf Karagila Feb 23 '17 at 17:29

Of course.

  • Fix a choice function on $\mathcal{P(P(\Bbb N))}$.
  • Start with a non-principal filter, say the co-finite filter.
  • By transfinite induction, each time choose a set not in the filter constructed thus far which can be added to the filter; at limit steps take unions. If no such set exists, you have an ultrafilter.

    Namely, if $\cal F_\alpha$ was defined, either $\{A\subseteq\Bbb N\mid A\cup\mathcal F_\alpha\text{ can be extended to a filter}\}$ is empty, in which case we have an ultrafilter on our hands; or we can use our fixed choice function to choose a set from there, and define $\cal F_{\alpha+1}$ to be the filter extending $\cal F_\alpha$ and the chosen set.

  • Argue that the construction must stabilize at some point.
  • Celebrate by having a beer, or some tea if you prefer.

One key lemma here is that if $\cal F$ is a filter on $X$, and $A\subseteq X$, the $\mathcal F\cup\{A\}$ generates a filter if and only if for all $B\in\cal F$, $A\cap B$ is non-empty.

One important remark here is that the axiom of choice "by itself" is nearly useless. It is mostly with transfinite recursion that we can actually use it to construct something.

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    $\begingroup$ I'm not sure how exactly you are using the choice function. Are you well-ordering your set in order to induct? If so, doesn't this basically involve the proof of Zorn's Lemma? $\endgroup$ – Thompson Feb 23 '17 at 17:35
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    $\begingroup$ You use it to choose a set from those eligible to be added to the filter at each step. And yes, this is "somewhat like the Zorn's lemma proof" because all these proofs are essentially the same. $\endgroup$ – Asaf Karagila Feb 23 '17 at 17:36
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    $\begingroup$ I don't know what you already covered with your students. Have you covered compactness of first-order logic? Have you covered Tychonoff's theorem? There are other proofs that you can use. You asked for one specifically using AC. This is how you do it. There's a reason AC and Zorn's lemma are equivalent, and you are correct to claim that this proof is essentially a proof of Zorn's lemma inside the proof that every chain of filters has an upper bound. Because, well, again, these things are more-or-less the same. $\endgroup$ – Asaf Karagila Feb 23 '17 at 17:38
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    $\begingroup$ Well, do they know about ordinals? They should, because that's how you'd prove that there are only $2^{\aleph_0}$ Borel sets, so "most sets" are not Borel, even the Lebesgue measurable ones. If they know a bit about transfinite recursion, this should be a breeze even by handwaving standards. If ordinals are unavailable either, then you're in for a pickle. You can still handwave your way through, but it might not catch properly (in which case, you can give them this as an axiom: every filter can be extended to an ultrafilter, and as an exercise prove that an ultrafilter is non-principal ... $\endgroup$ – Asaf Karagila Feb 23 '17 at 17:46
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    $\begingroup$ ... if and only if it extends the cofinite filter). If they already took some topology class, there are topological proofs that you can give them. If they took a logic class, there are logic-based proofs that you can give them as well. It might be reasonable to expect measure theory students to have some rudimentary knowledge in topology, but I guess that very much depends on the university. $\endgroup$ – Asaf Karagila Feb 23 '17 at 17:48

Such a sneaky construction will presumably smuggle in the essence of a proof of either (1) Zermelo's well-ordering theorem or (2) Zorn's Lemma. Let's go with option (1), and let's assume as little prerequisite knowledge as possible.

Given a set of sets $\mathcal{S}$ and $A\in\mathcal{S}$, let $A\downarrow\mathcal{S}$ denote the union of all $B\in\mathcal{S}$ that are proper subsets of $A$.

Given a choice function c on $X=\mathcal{P}(\mathcal{P}(\mathbb{N}))$, call a set $\mathcal{W}$ of filters on $\mathbb{N}$ "well" if every $F\in\mathcal{W}$ has the property that $F$ is the smallest (proper) filter containing $$(F\downarrow\mathcal{W})\cup \{c(X\setminus(F\downarrow\mathcal{W}))\},$$ if such a filter exists. If it does not exist, we instead require $F$ to be the smallest filter containing $$(F\downarrow\mathcal{W})\cup \{\mathbb{N}\setminus c(X\setminus(F\downarrow\mathcal{W}))\}.$$

An example well set of size 3 should clarify to your students the meaning of this definition. Follow up with example well sets of "length" $\omega$ and $\omega+1$.

Now let $\mathbb{W}$ denote the set of all well sets of filters. Assert that $U=\bigcup\bigcup\mathbb{W}$ is an ultrafilter on $\mathbb{N}$. Give the easy proof that $U$ is an ultrafilter if it is a filter. It's not so easy to prove that $U$ is a filter; hand wave and leave the details an as optional or extra-credit project. This will not be completely satisfying, but at least you've given an explicit definition of $U$.


A free ultrafilter gives a non-measurable subset of the Cantor set in a standard way. There are models of ZF with DC where all subsets of the Cantor set are measurable. In such models no free ultrafilter can exist. So some form of (uncountable) choice is needed to "construct" (via recursion) a free ultrafilter.

Asaf's construction is also how would go about it: enumerate all (infinite) subsets of $\mathbb{N}$ in a well-ordered sequence and check them sequentially.

I don't think it's really more intuitive than Zorn though. If you're teaching a set theory course, Zorn will probably be useful more generally (e.g. to extend bases in linear spaces, construct algebraic completions etc.)

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    $\begingroup$ I don't think this really answers the question - the OP isn't asking whether choice is needed to construct a free ultrafilter, but rather whether there is a more intuitive argument than via Zorn's lemma. $\endgroup$ – Noah Schweber Feb 23 '17 at 19:18

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