Nonprincipal ultrafilters over $\mathbb{N}$

Question

So, I'm given $\mathcal{A}\subseteq \mathcal{P}(\mathbb{N})$ that has the property that for any $\mathcal{A}_0\subseteq \mathcal{A}$ finite, $\cap\mathcal{A}_0$ is infinite. I have to show that there exists a nonprincipal ultrafilter $\mathcal{U}$ over $\mathbb{N}$ such that $\mathcal{A}\subseteq\mathcal{U}$. Here it's what I've done: Let $$D=\{\mathcal{F}\subseteq \mathcal{P}(\mathbb{N})|\mathcal{F} \text{ is a filter and } \mathcal{A}\subseteq\mathcal{F}\}$$

I claim that $D$ is not empty, in fact, if I let $S=\{\cap\mathcal{A}_0|\mathcal{A}_0\subseteq\mathcal{A}\text{ is finite}\}$ then $T=\mathcal{A}\cup S\cup \{S'|S''\subseteq S' \text{ for some } S''\in S\}\in D$

Now, let $\mathcal{F}_1\subseteq\mathcal{F}_2\subseteq ...$ be any chain of elements in $D$. Since each $F_i$ is a filter, then $\cup\mathcal{F}_i\in D$ is an upper bound for the chain. In consequence, I can use Zorn's Lemma to guarantee that exists a maximal element for the set $D$ which we will call $\mathcal{V}$. Now, it's clear that, $\mathcal{V}$ is not only a maximal element of $D$ but it's also a maximal element of the set of filters over $\mathbb{N}$. Therefore, $\mathcal{V}$ is an ultrafilter that contains $\mathcal{A}$, So it only remains to show that $\mathcal{V}$ is nonprincipal. This is where I'm having trouble, I was trying assuming that $\mathcal{V}$ was principal in order to achieve a contradiction to the fact that for any $\mathcal{A}_0\subseteq \mathcal{A}$ finite, $\cap\mathcal{A}_0$ is infinite, but I didn't succed. Can you help me with this? Also, Is this first part of the proof correct? Thanks in advance

Answer 1

Your proof works to find an ultrafilter $\mathcal{U}$ with $\mathcal{A}\subseteq \mathcal{U}$. But as written, you might accidentally get a principal ultrafilter when you appeal to Zorn's Lemma.

An ultrafilter is non-principal if and only if it contains all cofinite sets. So in order to make sure you don't get a principal ultrafilter, you're going to have to throw in the complement of every finite set.

That is, letting $\mathcal{C} = \{X\subseteq \mathbb{N}\mid \mathbb{N}\setminus X\text{ is finite}\}$ (this is the cofinite filter), look at the poset $$D' = \{\mathcal{F}\mid \mathcal{F}\text{ is a proper filter, and } \mathcal{A}\cup \mathcal{C}\subseteq \mathcal{F}\}.$$ You can show that $D'$ is nonempty in exactly the way you showed that $D$ is nonempty, by arguing that the filter generated by $\mathcal{A}\cup \mathcal{C}$ is proper.