rank(⋃a)∈rank(a)
Let
υ and
α respectively be the ranks of
⋃a and
a, and assume for contradiction that
a∈υ.
Take
B∈A∈a. Then
B∈⋃a. Also,
A∈a so by the definition of rank have
A∈Vα. Since
α∈υ then
Vα⊆Vυ, so also
A∈Vυ. Since
B∈A∈Vυ then by transitivity
B∈Vυ.
Thus the rank of
B is at least
υ+1, but by definition the rank of
B is
υ; contradiction.