Proposition 1   Let $A$ be a non-empty set, and $E(A)$ the class of all sets equinumerous to $A$ . Then $E(A)$ is a proper class.

Proof. $E(A)\ne \varnothing$ since $A$ is in $E(A)$ . Since $A\ne \varnothing$ , pick an element $a\in A$ , and let $B=A-\lbrace a\rbrace$ . Then $C:=\lbrace y\mid y\mbox{ is a set, and }y\notin B\rbrace$ is a proper class, for otherwise $C\cup B$ would be the ``set'' of all sets, which is impossible. For each $y$ in $C$ , the set $F(y):=B\cup \lbrace y\rbrace$ is in one-to-one correspondence with $A$ , with the bijection $f:F(y)\to A$ given by $f(x)=x$ if $x\in B$ , and $f(y)=a$ . Therefore $E(A)$ contains $F(y)$ for every $y$ in the proper class $C$ . Furthermore, since $F(y_1)\ne F(y_2)$ whenever $y_1\ne y_2$ , we have that $E(A)$ is a proper class as a result.