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Let $I$ be a linearly ordered set and $\{A_i\}, \{B_i\}$ be two collection of sets indexed by $I$, in an order-preserving way (i.e. $A_i\subsetneqq A_j\iff i<j$ and $B_i\subsetneqq B_j\iff i<j$). If for all $i\in I$, ${\rm card} A_i\le {\rm card}B_i$, do we have \begin{equation*} {\rm card}\ \bigcup_{i\in I} A_i \le {\rm card}\ \bigcup_{i\in I} B_i ? \end{equation*}

I guess the answer is yes. At first I wanted to construct an injection $f:\bigcup_{i\in I} A_i\to \bigcup_{i\in I} B_i$ by "gluing". But since $I$ may not be finite or countable, this seems hard to be written rigorously.

Asigan
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I think the inequality is true (assuming Choice, at least). I will make fairly heavy use of cardinal arithmetic/facts.

First, we can assume without loss of generality that $I$ is a limit ordinal, by passing to an appropriate cofinal subset (and because if $I$ has a maximum element, it's obvious). I will argue by proving a nice formula for the cardinality of each union.

Lemma. If $I$ is a limit ordinal and we have sets $A_i$ for each $i \in I$ such that $A_i \subsetneq A_j$ iff $i < j$, then $|\bigcup_i A_i| = \max\{|I|, \sup_i |A_i|\}$.

Proof. Let's write $A = \bigcup_i A_i$. There is an injection $I \to A$ given by sending $i$ to some element of $A_{i + 1} \setminus A_i$, so $|I| \le |A|$. It's also clear that for any $i$, we have $|A_i| \le |A|$. So $\sup_i |A_i| \le |A|$.

Conversely, we have $|A| \le |I| \cdot \sup_i |A_i|$, for example because if $S$ is a set of cardinality $\sup_i |A_i|$, then clearly $A$ injects into $I \times S$: for each $i$, choose an injection $\iota_i: A_i \to S$, and then for each $a \in A$, pick any $i$ such that $a \in A_i$ and send $a$ to $(i, \iota_i(a))$. But by cardinal arithmetic, $|I| \cdot \sup_i |A_i| = \max\{|I|, \sup_i |A_i|\}$, so this finishes the proof.

So now we know that $|\bigcup_i A_i| = \max\{|I|, \sup_i |A_i|\}$ and $|\bigcup_i B_i| = \max\{|I|, \sup_i |B_i|\}$. Since it's clear that $\sup_i |A_i| \le \sup_i |B_i|$, we are done.

I spent a bit of time looking for any reference to the above lemma on MSE, but I have not managed to find any. I am fairly sure it's true though..

Perhaps one reason it's hard to write down a proof is that there are counterexamples if you weaken things a bit - if you just ask for "$i \le j$ implies $B_i \subseteq B_j$", then there is a counterexample given by taking $A_i$ to be proper initial segments of $\omega_1$ and $B_i$ to all be $\omega$, for example. Also the example of "proper initial segments of $\Bbb Q$" gives some indication of why taking $I$ to be a limit ordinal was an important step - in that case the union $A$ is actually much smaller than $I$.

I have no idea what will happen in the absence of choice - that's above my pay grade. Here is some related discussion you may find interesting though.

Izaak van Dongen
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  • Thanks for your answer. However there are two points that I do not understand. The first is why we can assume $I$ is a limit ordinal? The second is, in the proof of the lemma, you wrote "$A_{i+1}\setminus A_i$". Is an element of a limit ordinal always a successor ordinal? – Asigan Mar 06 '24 at 07:35
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    I think Choice, in the form of a maximality principle, is required to show that every total ordering has a cofinal wellordered subset. – BrianO Mar 06 '24 at 08:28
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    @Asigan Re your second question — No: $\omega \in \omega + \omega$. But notice that $\omega + 1 \in \omega + \omega$ too. – BrianO Mar 06 '24 at 08:32
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    @Asigan, BrianO is right. We can assume $I$ is a limit ordinal because every total order has a well-ordered cofinal subset, which in this case can have no largest element, and passing to a cofinal subset won't change the union. I don't ever use "an element of a limit ordinal is always a successor ordinal" (I'd need that to talk about $i - 1$) - just "every element of a limit ordinal has a successor", which is equivalent to being a limit ordinal. Let me know if anything is still unclear :) – Izaak van Dongen Mar 06 '24 at 09:19
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    OK. I think that now I catch the proof. This is a very nice answer, thanks a lot! – Asigan Mar 06 '24 at 15:06