I want to prove that, given a non-empty set $X$ then its equipotence class $[X]_{\asymp}$ (the class of all sets which are in bijection with $X$) is a proper class, using Morse-Kelley framework.
I tried the following:
- Given an element $a \in X$, I defined a functional relation: $$\begin{align} F:V &\longrightarrow V \\ x &\longmapsto (X\setminus \{a\} ) \cup \{x\}\end{align}$$
- Now if we restrict $F$ onto $V\setminus X$ we have that $F:V\setminus X\longrightarrow [X]_{\asymp}\ $ is an injective functional.
- Since $V\setminus X$ is still a proper class, then $[X]_{\asymp}$ needs to be a proper class, otherwise by the axiom of replacement we'd have that $V\setminus X$ is a set.
Is is correct? Is there a more straightforward way? Thanks
Assume for a contradiction that the equipotence class $[X]_{\asymp}$ is a set. Then by the Axiom of Union, we have that $U = \bigcup [X]_{\asymp}$ is also a set.
But for any set $s \in V$, we can find a set $S$ equipotent with $X$ such that $s \in S$, and therefore $s \in U$ (we can prove this using your idea, simply by setting $S = X \setminus \{a\} \cup \{s \}$). By the Axiom of Extensionality, we get that $U = V$. But $V$ is a proper class, a contradiction.
As for your proof: the underlying idea is certainly correct. The correctness of the execution depends on two things.
First, you should establish that $V \setminus X$ is indeed a proper class. You or your textbook may have done this already, of course. Next, the conventional formulation of Replacement states that if the domain of a class function is a set, its range is also a set. But you wish to go from the assumption that the range is a set to the conclusion that so is $V \setminus X$: you do note that $F$ is injective, but there is a logical leap here. Again, you or your textbook may have taken care of this already.