$|A|:=\{B|B \sim A\}$ is a set?

Question

let: http://en.wikipedia.org/wiki/Equinumerosity

let $A$ a set, I define the cardinality of $|A|:=\{B|B \sim A\}$, but $|A|$ is a set?

Thanks in advance!!

Answer 1

If $A$ is non-empty then the answer is no.

To see this, note that you can fix $a\in A$ and pick any $x\notin A$ and consider $A_x=(A\setminus\{a\})\cup\{x\}$. Show that this is an injection from $V\setminus A=\{x\mid x\notin A\}$ into $|A|$.

Is $V\setminus A$ a set?

Answer 2

No, it is not, unless $A$ is empty. If it were, then (for example) taking a singleton $A$, we would have that $|A|$ is the set of all singletons, which is equinumerous to the set-theoretic universe in question, and then we've encountered Russell's paradox.

There are several work-arounds for this. Typically, if $A$ is well-orderable, then $|A|$ is taken to be the least ordinal equinumerous to $A$; otherwise, $|A|$ is taken to be the set of sets of least rank that are equinumerous to $A$.