1

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 $|\bigcup_{\alpha} X_{\alpha}| \leq |\bigcup_{\alpha} Y_{\alpha}|$?

Jxt921
  • 4,458
  • Does "totally ordered by inclusion" mean that $X_0$ is a proper subset of $X_1$ or does it just mean that $X_0\subseteq X_1$? – bof Sep 22 '19 at 14:25
  • @bof Well, I assumed that $X_{\alpha} \neq X_{\beta}$ for $\alpha \neq \beta$ (and similarly for $Y$). – Jxt921 Sep 22 '19 at 14:34
  • In that case it looks like an easy "yes". To start with, we can assume $A$ is well-ordered, can't we? – bof Sep 22 '19 at 14:39
  • 1
    By the way, you should put that assumption in the question, because it's not clear from what you wrote, and it drastically changes the meaning. – bof Sep 22 '19 at 14:43
  • @bof Sorry, edited again. Initial comment was wrong. – Jxt921 Sep 22 '19 at 14:56
  • @bof Dear bof, I'm sorry to bother you, but you seemed to have some ideas regarding a case which might be similar to my later question. I would be grateful if you could visit this question: https://math.stackexchange.com/questions/3366187/a-proof-in-module-theory-a-set-theoretic-consideration-bigcup-i-in-i-b-i – Jxt921 Sep 24 '19 at 10:41

1 Answers1

2

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 $\bigcup Y_\alpha=\omega$.

Eric Wofsey
  • 330,363
  • From the comments it seems clear that, in spite of how it is written in the question, the assumption is that if $\alpha\ne\beta$ are in $A$, then $X_\alpha\ne X_\beta$ and $Y_\alpha\ne Y_\beta$. – Andrés E. Caicedo Sep 22 '19 at 17:16
  • 1
    They later said their initial comment was wrong, though, after editing to the current version. – Eric Wofsey Sep 22 '19 at 17:19