# $|A \cup B| = \mathfrak{c}$, then $A$ or $B$ has cardinality $\mathfrak{c}$.

$A$ 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.

• A (countable) union of sets with the same infinite cardinality has that cardinality.
– user296602
Mar 2, 2016 at 3:06
• Doesn't that require some choice though?
– AJY
Mar 2, 2016 at 3:07
• @AJY yes, but not CH=continuum hypothesis.
– user228113
Mar 2, 2016 at 3:12
• @AJY No - it is consistent with ZF that the reals are the union of two sets of strictly smaller cardinality. Mar 2, 2016 at 3:13
• Do you know the fact that if $A$ and $B$ are infinite, then $|A\cup B|=\max(|A|,|B|)$? Mar 2, 2016 at 4:12

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 $\sum_i a_i\lt\prod_i b_i.$