On $\mathcal{A}$ with $\forall C\in\mathcal{P}_{\infty}(\mathbb{N})\exists A\in\mathcal{A}:A\cap C\in\mathcal{P}_{\infty}(\mathbb{N})$

After doing some more research, I saw that the most natural notion to use here is that of an ideal in the set-theoretical sense. In fact, if we define $\mathcal{A}':=\{A'\subseteq\mathbb{N}\ |\ \exists A\in\mathcal{A},F\subseteq\mathbb{N}\text{ finite}:\ A'\subseteq A\cup F\}$, then $\mathcal{A}'$ is a set-theoretical ideal on $\mathbb{N}$. Furthermore, my question is then equivalent to whether $\mathcal{A}'=\mathcal{P}(\mathbb{N})$ (or simply whether $\mathbb{N}\in\mathcal{A}'$).

And in this context, we can come up with the following counter-example: Take $\mathcal{A}$ to be the set formed by the infinite sets of an asymptotical density of $0$, then $\mathcal{A}'=\mathcal{A}\cup\{F\subseteq\mathbb{N}\text{ finite}\}$ is the ideal formed by all the sets of an asymptotical density of $0$. Then if $C\subseteq\mathbb{N}$ is some infinite set, we clearly can find $A\in\mathcal{A}$ with $A\subseteq C$ and in particular $A\cap C=A\in\mathcal{P}_{\infty}(\mathbb{N})$. But no $A\in\mathcal{A}$ contains $\{N,N+1,...\}$, as that would mean that $A$ would have an asymptotical density of $1$.