# Is this use of Zorn's lemma in the proof that every infinite set has a non-principal ultrafilter correct?

I just want to quickly confirm that I am using Zorn's correctly in this short proof.

There is a non-principal ultrafilter on any infinite set $X$.

Consider the filter $\mathcal{F}$ of of cofinite subsets of $X$. And consider the poset $P$ of filters containing $\mathcal{F}$, ordered by inclusion.

Since every chain of filters has their union as an upper bound, By Zorn's lemma, there is a maximal filter $\mathcal{G}$ in $P$, i.e. a maximal filter that contains $\mathcal{F}$. Since a maximal filter is an ultrafilter, it remains to show that $\mathcal{G}$ is non-principal. But $\{x\}^c$ is cofinite and hence an element of $\mathcal{F} < \mathcal{G}$. QED

(edit - I will assume every chain of filters has their union as an upper bound, because this is not what I want to check.)

• That seems almost fine to me. Just one small observation: Zorn applied this way gives you a filter maximal not among all filters, but only among filters containing $\mathcal{F}$. But in fact Zorn guarantees that in an inductive set (that is every chain has an upper bound) any element is bounded by some maximal element, so just apply this to the family of all filters and to the Fréchet filter. – Sonner Sep 18 '15 at 12:13
• Oh, is it true that Zorn gives me a maximal element that is an upper bound, rather than just a maximal element? – Calvin Khor Sep 18 '15 at 12:18
• Given $F$ inductive family and $x\in F$, there exists $a\in F$ such that $a$ is maximal and $x\leq a$. Proof: apply classical Zorn to the set $G=\{y\in F: x\leq y\}\subseteq F$ which is non-empty because it contains $x$. Call $x$ the obtained element (maximal in $G$). If $x\leq z$ for some $z\in F$, then $z\geq x$ and so $z\in G$, but now by maximality in $G$ $y=z$. – Sonner Sep 18 '15 at 12:24
• You can do that in the filter context as well: if the filter $\mathcal{G}$ you obtained is contained in another filter $\mathcal{H}$ then $\mathcal{H}$ contains $\mathcal{F}$ and so by maximality of $\mathcal{G}$ in $P$ you have equality, that is $\mathcal{G}$ is maximal in the family of all filters. – Sonner Sep 18 '15 at 12:26
• Ok, I'm happy with this. Thanks! If you want to compile this into an answer I will gladly accept it. – Calvin Khor Sep 18 '15 at 12:35

That seems almost fine to me. Just one small observation: Zorn applied this way gives you a filter maximal not among all filters, but only among filters containing F. But in fact Zorn guarantees that in an inductive set (that is every chain has an upper bound) any element is bounded by some maximal element, so just apply this to the family of all filters and to the Fréchet filter.

Let me explain that corollary of Zorn. Given F inductive family and $x∈F$, there exists $a∈F$ such that a is maximal and $x≤a$. Proof: apply classical Zorn to the set $G={y∈F:x≤y}⊆F$ which is non-empty because it contains $x$. Call $x$ the obtained element (maximal in $G$). If $x≤z$ for some $z∈F$, then $z≥x$ and so $z∈G$, but now by maximality in $G$ we get $y=z$.

You can do that in the filter context as well: if the filter $\mathcal{G}$ you obtained is contained in another filter $\mathcal{H}$ then $\mathcal{H}$ contains $\mathcal{F}$ and so by maximality of $\mathcal{G}$ in P you have equality, that is $\mathcal{G}$ is maximal in the family of all filters.

The proof looks okay, granted you have shown before that an increasing union of filters is a filter (which is something usually proved within this sort of statement).

Since the nontrivial part in most Zorn-related proofs (or at least the simple ones) is to verify that the partial order satisfies Zorn's condition, it might be worth showing that.

• I intentionally left out parts that I knew were correct in the interest of creating a 'minimal working example', but I see now how that is confusing when asking for a verification! Thanks anyway :) – Calvin Khor Sep 18 '15 at 12:51
• Even if you did write it, are you certain that you wrote it correctly? Why not verify that as well, while you're at it? If you asked whether or not your use of Zorn's lemma is correct, then this is exactly what you're asking about! :-) – Asaf Karagila Sep 18 '15 at 12:52
• You're completely right; I rushed the question out so that I could leave to have dinner (GMT+8) and so I didn't word it properly. Miraculously, Sonner still managed to help me with the part I wasn't sure about. Sorry for the confusion, I'll think about how to reword the question, or perhaps just add the remainder of the proof. – Calvin Khor Sep 18 '15 at 12:58
• Well, math is not a sprint. It's not even a marathon. It's an ultramarathon, several times over. – Asaf Karagila Sep 18 '15 at 13:20