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Two sets are said to be 'equipotent' if there is a bijection between them. For a given set $A,$ consider the class $\Bbb{A}$ of all sets those are equipotent with $A.$ Is $\Bbb{A}$ form a set?

My answer is "No" unless $A=\emptyset.$ In order to prove this, my idea is to use the fact that class of all singleton sets is not a set.

1) Is my conclusion correct?
2) Is there any better (direct) way to prove this?

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Your conclusion is right.

There are several ways to prove it, and I'm not entirely sure which one you have in mind based on your idea, but here's a hint that follows that idea: what is the cardinality of $\{x\}\times A$, for any set $x$?

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  • $\begingroup$ My idea is, if $A$ is non-empty, then $(A\setminus\{a\})\cup\{x\}$ is equipotent with $A$ for any singleton $\{x\}$ and some $a\in A.$ But your idea is way clear than it. $\endgroup$ – Bumblebee Jun 21 '17 at 5:27

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