Cardinality of a totally ordered union

Let $(X_{\alpha})_{\alpha \in A}, (Y_{\alpha})_{\alpha \in A}$ be families of sets such that for $\{X_{\alpha} \mid \alpha \in A \}$ and $\{Y_{\alpha} \mid \alpha \in A \}$ are totally ordered by inclusion. We have $(X_{\alpha},Y_{\alpha}) \neq (X_{\beta},Y_{\beta})$ for $\alpha \neq \beta$. Suppose that $|X_{\alpha}| \leq |Y_{\alpha}|$ for any $\alpha \in A$. It is true that $|\bigcup_{\alpha} X_{\alpha}| \leq |\bigcup_{\alpha} Y_{\alpha}|$?

Solution 1:

No. For instance, let $A=\omega_1$, let $X_\alpha=\alpha$, and let $Y_\alpha=\omega$ for all $\alpha$. Then $|X_\alpha|\leq|Y_\alpha|$ for all $\alpha$ since $X_\alpha=\alpha$ is always countable but $\bigcup X_\alpha=\omega_1$ has larger cardinality than $\bigcup Y_\alpha=\omega$.