Let $\kappa$ be an infinite cardinal and let ${\cal S}$ be a collection of non-empty subsets of $\kappa$ such that $|{\cal S}| = \kappa$. A map $f:{\cal S} \to \kappa$ is said to be a *choice function* if $f(S)\in S$ for all $S\in{\cal S}$.

This post is about *injective choice functions*. For the existence of an injective choice function, it is easy to see that Hall's condition is necessary:

For any subset ${\cal T}\subseteq {\cal S}$ we have $|{\cal T}| \leq |\bigcup {\cal T}|$.

However, this condition is not sufficient: Pick $\kappa = \omega$ and set $${\cal S} = \big\{\{n\}:n\in\omega\big\} \cup\big\{\omega\big\}.$$ It is easily verified that Hall's condition holds for ${\cal S}$, but ${\cal S}$ has no injective choice function. Informally speaking, the problem with ${\cal S}$ is that it has lots of small sets.

For any infinite cardinal $\kappa$ and any collection ${\cal S}$ of non-empty subsets of $\kappa$ such that $|{\cal S}| = \kappa$ we set ${\cal S}_{<\kappa} = \{S\in{\cal S}: |S|<\kappa\}$.

**Question**. If Hall's condition holds for ${\cal S}$ and $|{\cal S}_{<\kappa}|<\kappa$, does ${\cal S}$ have an injective choice function?