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> Proof or counterexample: if $\mathcal F\subseteq \mathbb{N}^{(r)}$ is intersecting then $\exists A\subseteq \mathbb{N}: A$ is finite and $\{F\cap A:F\in\mathcal F\}$ is intersecting.
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I know this result is false for $\mathcal F\subseteq \mathcal P( \mathbb{N})$ because you can find a bijection $f:\mathbb{N}^2\rightarrow \mathbb{N}$ and define $(F\_i)\_{i\in\mathbb{N}}$ by $F\_i:=\{ f(i,j), f(j,i) :j\in\mathbb{N} \}$ so that $F\_i\cap F\_j=\{f(i,j),f(j,i)\}$ (or even make $f$ symmetric in its arguments to get a unique value in each singleton).


As I couldn't find a simpler construction, I suspect the result is true in the case of $r$-sets. I want to do something like taking an $A\in\mathcal F$ and considering sets $B,C$ such that $B\cap C$ is disjoint from $A$ (if none exist, we are done). Then include $B$ and find $D\cap E$ disjoint from $A\cup B$, and so on and it feels, based on instinct only, like after $r$ (or possibly $2^r$) iterations, we should be done.


Is the result true, and how can I prove it?



Does an intersecting $r$-family in $\mathbb{N}$ have a finite underlying base set?
It is easy to prove by induction the following stronger statement.



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> If $\cal F$ and $\cal G$ are two families of sets of size at most $r$ and $s$, respectively, that *cross-intersect*, i.e., for all pair of sets $F\in \mathcal F, G\in\cal G$ we have $F\cap G\ne\emptyset$, then $\exists A\subseteq \mathbb{N}: A$ is finite and $\mathcal F\_A=\{F\cap A \mid F\in\mathcal F\}$ and $\mathcal G\_A=\{G\cap A \mid G\in\mathcal G\}$ are also cross-intersecting.
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We prove by induction on $r+s$.  

The claim is obviously true for $r=1$ or $s=1$ with $|A|=1$.  

Pick an arbitrary $G\_0\in \cal G$, and define for each non-empty $I\subset G\_0$ the families $\mathcal F\_I=\{F\setminus I\mid F\in\mathcal F, F\cap G\_0=I\}$ and $\mathcal G\_I=\{G\mid G\in\mathcal G, G\cap I=\emptyset\}$.  

Since the size of sets in $\mathcal F\_I$ is at most $r-1$, we can apply induction for the cross-intersecting $\mathcal F\_I$ and $\mathcal G\_I$ to obtain a set $A\_I$ for them that satisfies the requirements.  

Finally, take $A=G\_0\cup \bigcup\_{I\subset G\_0} A\_I$.


Note that this gives only $|A|<2^{r+s}$, or for your original problem $|A|<4^r$, but with a more careful analysis one can probably also get a polynomial bound.


