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# equivalence class of equinumerous sets is not a set

###### 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.

$E(A)\neq\varnothing$ since $A$ is in $E(A)$. Since $A\neq\varnothing$, pick an element $a\in A$, and let $B=A-\{a\}$. Then $C:=\{y\mid y\mbox{ is a set, and }y\notin B\}$ 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 $F(y_{1})\neq F(y_{2})$ whenever $y_{1}\neq y_{2}$, 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.

## Mathematics Subject Classification

03E10*no label found*

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