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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!!

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    $\begingroup$ Think of the simple case where $A$ is a singleton. What are the singletons in set theory? Do they form a set? $\endgroup$ – Marc van Leeuwen Aug 19 '13 at 8:07
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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?

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  • $\begingroup$ ok.. therefore I must define the cardinality as the en.wikipedia.org/wiki/Von_Neumann_cardinal_assignment so $|A|$ is a set.. is correct? $\endgroup$ – mle Aug 19 '13 at 8:09
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    $\begingroup$ @Soviet: You don't have to do that. But if the axiom of choice is assumed, that it's quite a sensible and reasonable thing to do. Historically, it was about 30 years from von Neumann's cardinal assignment, until they figured out how to define cardinals internally in $\sf ZF$ using Scott's trick. $\endgroup$ – Asaf Karagila Aug 19 '13 at 8:15
  • $\begingroup$ ah wow I don't know the Scott's trick.. thanks soo much.. :) $\endgroup$ – mle Aug 19 '13 at 8:21
  • $\begingroup$ Are there some (e)books or online papers about Scott's trick ?? thanks in advance!! $\endgroup$ – mle Aug 19 '13 at 8:31
  • $\begingroup$ @Soviet: I posted two links below Cameron's answer (which references Scott's trick, albeit not by name) on this page, they talk about cardinal assignment. $\endgroup$ – Asaf Karagila Aug 19 '13 at 8:47
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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 $A$.

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    $\begingroup$ The second paragraph has appeared on the site before: here and here. $\endgroup$ – Asaf Karagila Aug 19 '13 at 8:08

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