Is cardinality continuous?

Question

Let the underlying set theory be ZFC. Let $x_1 \subseteq x_2 \subseteq \dots$ and $y_1 \subseteq y_2 \subseteq \dots$ be ascending sequences of sets such that, for every $n \in \{1,2,\dots\}$, $|x_n| = |y_n|$. Is it the case that $\big|\cup_{n =1}^{\infty}x_n\big| = \big|\cup_{n =1}^{\infty}y_n\big|$? If this is not generally true, is it possible to characterize all those—or at least some interesting—cases for which this does hold? Is there a standard terminology for these cases? Can this be generalized to transfinite sequences? Does the answer change if we require that the sequences be strictly increasing, i.e. for every $n \in \{1,2,\dots\}$, $x_n \subsetneq x_{n+1}$ and $y_n \subsetneq y_{n+1}$?

Answer 1

Yes, this is always true. If $x_1\subseteq x_2\subseteq\dots$ and $x=\bigcup x_n$, then $|x|=\sup_n|x_n|$, and in particular $|x|$ is uniquely determined by the sequence of cardinalities $|x_n|$. Clearly $|x|\geq|x_n|$ for all $n$ so $|x|\geq\sup_n |x_n|$. Conversely, $|x_n|\leq \sup|x_n|$ for all $n$ so $|x|\leq \aleph_0\cdot \sup|x_n|=\sup|x_n|$ as long as $\sup |x_n|$ is infinite (and if it is finite then the result is trivial).

Note that if you consider increasing sequences with possibly uncountable index sets then this is no longer true. For instance, with index set $\omega_1$, if you let $x_\alpha=\omega+\alpha$ and $y_\alpha=\omega$ for all $\alpha<\omega_1$, then $|x_\alpha|=|y_\alpha|=\aleph_0$ for each $\alpha$ but $\left|\bigcup x_\alpha\right|=\aleph_1$ while $\left|\bigcup y_\alpha\right|=\aleph_0$. If you require the sequences to be strictly increasing then it is true though: letting $\kappa$ be the cofinality of the index set, the argument above shows that $|x|\geq\sup|x_i|$ and $|x|\leq \kappa\cdot\sup|x_i|=\max(\kappa,\sup|x_i|)$ but also $|x|\geq\kappa$ if the $x_i$ are strictly increasing (since looking at a cofinal well-ordered subsequence gives at least one new element of $x$ for each term in the subsequence), so $|x|=\max(\kappa,\sup|x_i|)$.