The set-theoretic result used in the proof I want to know about is the following:
Let $(B_i)_{i \in I}$ be a family of sets such that, for all $i,j \in I$, either $B_i \subseteq B_j$ or $B_j \subseteq B_i$ holds and, for all $i \in I$, $|B_i| \leq |Y|$. Then $|\bigcup_{i \in I} B_i| \leq |Y|$.
Which conditions must be true (if any) for this to hold, and why? I tried to put a well-order on both $I$ and $Y$, but didn't succeed even with that. This seem to be implicitly assumed in the following proof (the book is Abstract Algebra by Grillet).
If this inequality is not true in the general case, I wonder if there is a condition in the proof that make it true.
The claimed set-theoretic result is wrong. For instance, the first uncountable ordinal $\omega_1$ is a union of a well-ordered family of countable ordinals. So I guess that the proof contains a gap in the assurance that the family $\mathcal S$ is closed with respect to the unions of chains of its elements. Maybe it can be fixed by relaxing condition $|B|\le |Y|$ for the family $\mathcal S$ and remarking in the last sentence of the proof that $Y=X$ implies $|B|\le |X|$.