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(A) the class of all sets equinumerous to A. Then E(A) is a proper classMathworldPlanetmath.


E(A) since A is in E(A). Since A, pick an element aA, and let B=A-{a}. Then C:={yy is a set, and yB} is a proper class, for otherwise CB would be the “set” of all sets, which is impossible. For each y in C, the set F(y):=B{y} is in one-to-one correspondence with A, with the bijection f:F(y)A given by f(x)=x if xB, and f(y)=a. Therefore E(A) contains F(y) for every y in the proper class C. Furthermore, since F(y1)F(y2) whenever y1y2, 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 yC 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 classMathworldPlanetmathPlanetmath of equinumerous sets is not a set
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