$Fn(\kappa \times \omega , \omega)$ has the countable chain condition.

I am struggling with the following question:
Let $\kappa$ be a regular uncountable cardinal. Show that $Fn(\kappa \times \omega , \omega)$ has the countable chain condition.

Where $Fn(I,\omega)$ is the partial order of all finite partial functions $p: I \rightarrow \omega$ with extension relation superset. (For an infinite index set $I$)

• Looks like $Fn(I, \omega)$ up to isomorphism only depends on $|I|$, so can't we replace $\kappa \times \omega$ with just $\kappa$? – Adayah Jun 14 '18 at 7:20
• @Adayah: Yes, we can, but we don't, because it's simpler to use $\kappa\times\omega$ later on in life. – Asaf Karagila Jun 14 '18 at 7:23
• Do you know the $\Delta$-system lemma? – Asaf Karagila Jun 14 '18 at 7:23
• Yes, we discussed the $\Delta$-system lemma a few weeks ago. – GMiiX Jun 14 '18 at 7:24
• Well, then, use the $\Delta$-system to show that if you have uncountably many conditions, then uncountably many of them agree on their common domain, and are disjoint otherwise. – Asaf Karagila Jun 14 '18 at 7:27

Suppose that $\{p_i\mid i\in I\}$ is an uncountable family of conditions, by the $\Delta$-system lemma, there is an uncountable $J\subseteq I$ such that $\{\operatorname{dom} p_j\mid j\in J\}$ form a $\Delta$-system.
Suppose that the root of the system is $A$ which is a finite subset of $\kappa\times\omega$. There are only countably many functions from $A$ to $\omega$, so there is an uncountable $J'\subseteq J$ such that $\{p_j\mid j\in J'\}$ all agree on their common domain. And of those, any two are compatible.