Can there be a fragile family on $\mathcal{P}(\omega)$?

Question

By a fragile family on $\mathcal{P}(\omega)$ (this is a made-up term), I mean a countable family $F\subseteq\mathcal{P}(\omega)$ of pairwise distinct infinite sets such that for any $n\in\omega$, the family $F':=\{X_i\smallsetminus\{n\}\mid X_i\in F\}$ is no longer pairwise distinct. That is, for some elements $X_i\neq X_j$ in $F$, $X_i\smallsetminus\{n\}=X_j\smallsetminus\{n\}$. The family is fragile, because once you take away anything from a member, two members will clash. Can there be a fragile family on $\mathcal{P}(\omega)$?

I wonder if this is a special instance of some more general theorem, so I am also curious about various generalizations, for example for families on $\mathcal{P}(\kappa)$ for uncountable $\kappa$, or varying the size of $F$, or instead of removing a singleton, we remove an infinite subset from each element in the family.

Answer 1

Yes, there are lots of these. In my opinion the simplest is the set of all finite subsets of $\omega$: given $n$, consider $\{n, n+1\}$ versus $\{n+1\}$.

Demanding that the sets in question be infinite doesn't change the picture: consider e.g. the set of $X\subseteq\omega$ such that $X$ differs only finitely much from the set of even numbers. For each $n$, let $E=\{2k: k\in\omega\}$ and $E_n=E\Delta \{n\}$. Then $E\not=E_n$ but $E\setminus\{n\}=E_n\setminus\{n\}$.

I do not at the moment see a way to modify the notion of fragility to avoid this sort of example.

Answer 2

Take any infinite $A\subseteq\omega$ and let $F=\{A\}\cup\{A\mathbin{\triangle}\{n\}:n\in\omega\}$. More generally, for each $n$, you could pick a pair of sets to be in $F$ which differ only on $n$, and the resulting $F$ will be countable.