# Finite pair-splitting family of $\mathbb{N}$

This is a kind of "dual" of an older question.

Is there a finite family $${\frak F}\subseteq {\cal P}(\mathbb{N})$$ such that for all $$a\neq b\in\mathbb{N}$$ there is $$S\in{\frak F}$$ with $$|S\cap \{a,b\}| = 1$$?

$$\newcommand\F{\mathfrak F}\newcommand\P{\mathfrak P}\newcommand\N{\mathbb N}\newcommand\om{\omega}$$No. Let $$\P$$ be the partition of $$\N$$ generated by $$\F$$.
Detail: If $$\F=\{S_1,\dots,S_n\}$$, then the members of the partition $$\P$$ are all the sets of the form $$\bigcap_{j=1}^n S_j^{\om_j}$$, where $$(\om_1,\dots,\om_n)\in\{0,1\}^n$$, $$S_j^0:=S_j$$, and $$S_j^1:=\N\setminus S_j$$.
Then $$\P$$ is finite and hence at least one member $$P$$ of $$\P$$ is infinite. Taking any distinct $$a$$ and $$b$$ in $$P$$, we see that $$|S\cap\{a,b\}|\in\{0,2\}$$ for each $$S\in\F$$.
Detail: If $$\F=\{S_1,\dots,S_n\}$$ and $$P=\bigcap_{j=1}^n S_j^{\om_j}$$ for some $$(\om_1,\dots,\om_n)\in\{0,1\}^n$$, then for each $$j\in\{1,\dots,n\}$$ we have $$|S_j\cap\{a,b\}|=2$$ if $$\om_j=0$$ and $$|S_j\cap\{a,b\}|=0$$ if $$\om_j=1$$.
The same argument works if $$\N$$ is replaced by any set $$X$$ such that $$|X|>2^{|\F|}$$.