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Assume ZFC.

Theorem: Given some cardinal K, the is no set of all sets equinumerous to K.

I have been thinking about this for a few days and can't come up with a proof. Intuitively, it seems to me such a set could be used to construct the set of all sets which is not allowed. This would be obvious if there is a set of all cardinals. But I don't think this needs to be relied upon. (Because my book certainly has not given me the tools to prove that statement either way.)

Either complete proofs or hints would be appreciated.

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  • $\begingroup$ Except for $K=0$. $\endgroup$ – bof Jan 5 '16 at 3:29
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Fix a set $X$ of cardinality $\kappa,$ fix some $x\in X.$ Consider the class of sets $$\bigl(X\setminus\{x\}\bigr)\cup\{y\}$$ with $y\notin X.$

Note that this works as long as you can fix $x,$ so works for almost every cardinality $\kappa.$

Added: You should be able to prove this rather straightforwardly from a few axioms. In particular, Pairing, Union, and Comprehension should let you show that all sets described above exist. Then Comprehension will let you show that, if there is a set of all sets of cardinality $\kappa,$ then the class described above is a set, as well. Finally, Pairing, Union, and Replacement will let you show that the class of all sets is again a set, which is impossible.

Added: Alternately, as Noah points out, the class described above helps us demonstrate that the union of "the set of all sets of cardinality $\kappa$" is "the set of all sets," thereby avoiding Replacement.

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  • $\begingroup$ Slightly simpler: just show that every set $x$ is an element of some set of size $\kappa$. (To do this, fix a single set $Y$ of size $\kappa$ and a $y\in Y$; then if $x\not\in Y$, the set $(Y\setminus\{y\})\cup\{x\}$ has size $\kappa$ and contains $x$.) Now, since every set is an element of some set of size $\kappa$, if $X$ is the set of all sets of size $\kappa$ then $\bigcup X$ is the set of all sets. In particular, no need for replacement. $\endgroup$ – Noah Schweber Jan 5 '16 at 5:09
  • $\begingroup$ @Noah: I like it! $\endgroup$ – Cameron Buie Jan 5 '16 at 5:23

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