# Nonprincipal ultrafilters over $\mathbb{N}$

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

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.
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.