By using ultrafilter lemma, can we find a function $f:2^{\mathbb N}\to\mathbb Z_2$ such that $f(n_1,n_2,\dots,n_k,0,0,\dots)=(\sum_{i=1}^k n_i)\mod 2$, and $f(a+b\mod 2)=f(a)+f(b)\mod 2$?
It's equivalent to say whether there is a set $S\subset\mathcal P(\mathbb N)$. Where $S$ contain all finite sets with odd cardinality and its complement contain all finite sets with even cardinality; And for $A,B\subset\mathbb N$, their symmetric difference $A\triangle B\in S^c$ if and only if $\{A,B\}\subset S$ or $\{A,B\}\subset S^c$.
This set seems to exist, but I don't know how to use ultrafilter lemma, because the Boolean algebra is not the same as $\mathcal P(\mathbb N)$.
I prove it without ultrafilter lemma, but using Hamel basis of linear space, I wonder if there is any proof to avoid Hamel basis.
Consider $2^\mathbb N$ as a $\mathbb Z_2$-linear space, extend $e_n=(0,\dots,0,1,0,\dots)$ to the hamel basis $\{e_n\}\cup\{e'_\alpha\}$, any linear functional $f:2^\mathbb N\to \mathbb Z_2$ such that $f(e_n)=1$ will do by define the value of $f(e'_\alpha)$.