equivalence class of equinumerous sets is not a set

Recall that two sets are equinumerous iff there is a bijection between them.

Proposition 1.

Let $A$ be a non-empty set, and $E\mathit{}\mathrm{(}A\mathrm{)}$ the class of all sets equinumerous to $A$. Then $E\mathit{}\mathrm{(}A\mathrm{)}$ is a proper class.

Proof.

$E(A)\ne \mathrm{\varnothing }$ since $A$ is in $E(A)$. Since $A\ne \mathrm{\varnothing }$, pick an element $a\in A$, and let $B=A-\{a\}$. Then $C:=\{y\mid y\text{ is a set, and }y\notin B\}$ 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 $F(y):=B\cup \{y\}$ 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.∎

Remark. In the proof above, one can think of $F$ as a class function from $C$ to $E(A)$, taking every $y\in C$ into $F(y)$. This function is one-to-one, so $C$ embeds in $E(A)$, and hence $E(A)$ is a proper class.