# 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.

Title | equivalence class^{} of equinumerous sets is not a set |
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Canonical name | EquivalenceClassOfEquinumerousSetsIsNotASet |

Date of creation | 2013-03-22 18:50:34 |

Last modified on | 2013-03-22 18:50:34 |

Owner | CWoo (3771) |

Last modified by | CWoo (3771) |

Numerical id | 4 |

Author | CWoo (3771) |

Entry type | Result |

Classification | msc 03E10 |