 Solution

Define $D_{\mathcal{G}}(A)=\{x:A-x\in \mathcal{G}\}$. The proposed addition of filters on the natural numbers isBecause addition of positive integers stays in $\mathbb{N}$, $D_{\mathcal{G}}(\mathbb{N})=\mathbb{N}$ and $D_{\mathcal{G}}(\varnothing )=\varnothing$. Thus the sum contains the whole set and excludes the empty set. If $A\subseteq B$, upward closure of $\mathcal{G}$ gives $D_{\mathcal{G}}(A)\subseteq D_{\mathcal{G}}(B)$, so upward closure of $\mathcal{F}$ gives upward closure of the sum.

Finally, for every $x$,The conjunction property for $\mathcal{G}$ proved in (i) consequently givesFinite-intersection closure of $\mathcal{F}$ proves the same closure for its sum. All proper-filter axioms hold, so (iv) is always true. No ultrafilter assumption is needed here.