Let $A\neq\varnothing$; prove there is no set $S$ containing all sets equipotent to $A$.

My proof outline: Assume such a set $S$ exists. If I can show $\bigcup S$ (which is a set) = "set of all sets", then I have derived a contradiction. But, how do I show that?

Thank you.

  • 1
    $\begingroup$ Hint: Every set is a member of some set equipotent to $A$. $\endgroup$ – Peter Košinár Jul 17 '13 at 18:02
  • $\begingroup$ @PeterKošinár: I have tried to show every set is a member of some set equipotent to A to no avail $\endgroup$ – Alexy Vincenzo Jul 17 '13 at 18:07
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    $\begingroup$ Consider any set $X$. If $X\in A$, we are done (clearly, $A$ is equipotent to itself). If $X\not in A$, we can take any $y\in A$ and replace it by $X$ to obtain a set equipotent to $A$, which contains $X$. $\endgroup$ – Peter Košinár Jul 17 '13 at 18:08
  • $\begingroup$ @Douglas: Why did you change the $\varnothing$ to $\emptyset$? $\endgroup$ – Asaf Karagila Jul 17 '13 at 19:03
  • $\begingroup$ I'll change it back; I shouldn't have changed it (my main purpose was to add a title). [Hmm... although, now that I look, it was \emptyset in the original version.] $\endgroup$ – Douglas S. Stones Jul 17 '13 at 19:05


If you already know that the collection of all sets is not a set, you can show that the collection $W=\{x\mid x\notin A\land x\neq\varnothing\}$ is not a set (how many sets are not in this collection?). Then you can find an injection from $W$ into $S$ by fixing $a\in A$ and considering $(A\setminus\{a\})\cup\{x\}$. What happens if $S$ was a set? Show that the collection of all sets, or even just $W$ is a set as well.


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