= Solution
Write $\mathfrak c=2^{\aleph_0}$. Its <cofinality> is $\aleph_1$, so choose a strictly increasing sequence of infinite <cardinal numbers> $(\kappa_\xi)_{\xi<\omega_1}$ below $\mathfrak c$ with supremum $\mathfrak c$. Such a sequence is obtained by refining a cofinal sequence and choosing larger cardinals recursively; $\mathfrak c$ is a <singular cardinal>, hence a <limit cardinal>.
By the <axiom of choice>, fix a <bijection> $b:\mathfrak c\to\mathbb R$. Set
$$
A_\xi=b[\kappa_\xi],\qquad
\boxed{\mathcal A=\{A_\xi:\xi<\omega_1\}}.
$$
Every $A_\xi$ is infinite, $|A_\xi|=\kappa_\xi<\mathfrak c$, and distinct indices give distinct <cardinalities>. The <cofinality> of the sequence ensures
$$
\bigcup_{\xi<\omega_1}A_\xi=b\left[\bigcup_{\xi<\omega_1}\kappa_\xi\right]=b[\mathfrak c]=\mathbb R.
$$
Thus $|\mathcal A|=\aleph_1$. Pairwise different sizes means different sizes for distinct members; without that qualification the parenthetical condition would contradict the case $A=B$.
Back to article page
