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$A$ and $B$ are sets. $|A \cup B| = \mathfrak{c}$, prove that $A$ or $B$ has cardinality $\mathfrak{c}$.

This is an exercise problem from my textbook. It's easy if I assume CH to be true. But how can I prove this without it? I don't know how to eliminate the case of $A$ and $B$ both has cardinality larger than that of natural numbers and strictly smaller than that of reals.

Andrés E. Caicedo
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leafpile
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1 Answers1

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It has been suggested in the comments that you should use the theory of well-ordered sets or Zorn's lemma. Here, instead, is a simple proof using the axiom of choice directly.

I assume that you are familiar with the Cantor-Bernstein theorem. Because of that, it suffices to prove that $|A|\ge\mathfrak c$ or $|B|\ge\mathfrak c.$ I also assume you know that $|\mathbb R\times\mathbb R|=|\mathbb R|=\mathfrak c$; thus we may assume that $A\cup B=\mathbb R\times\mathbb R.$

Case 1. If $A$ is disjoint from some horizontal line in the plane $\mathbb R\times\mathbb R,$ then $B$ contains a horizontal line, and so $|B|\ge\mathfrak c.$

Case 2. If $A\cap L\ne\emptyset$ for every horizontal line $L,$ then by the axiom of choice there is a set $S\subseteq A$ such that $|S\cap L|=1$ for every horizontal line $L.$ Clearly $|S|=\mathfrak c$ and so $|A|\ge\mathfrak c.$

P.S. This argument shows that, if $a\lt c$ and $b\lt c,$ then $a+b\lt c\cdot c.$ This is a special case of Kőnig's theorem: if $a_i\lt b_i$ for each $i,$ then $\sum_i a_i\lt\prod_i b_i.$

bof
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