# Examples of sets whose cardinalities are $\aleph_{n}$, or any large cardinal. (not assuming GCH)

One of the answers to this question indicates that large cardinals are useful for destructive testing of set theory. That aside, and not assuming GCH, are there any sets known that have a cardinality of $\aleph_{n}$, $n>0$, or that of any of the large cardinals? There are a few examples of sets that have these cardinalities on the wikipedia page, but they are meager and few compared to the examples on the page for Beth cardinalities.

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You cannot prove the existence of large cardinals within ZFC. Not even if you assume GCH. As for sets whose cardinalities are $\aleph_n$ well, how about all the ordinals whose cardinality is below $\aleph_n$? That is usually enough. –  Asaf Karagila Nov 13 '10 at 10:54

There are very few examples where you directly prove in ZFC that a certain set must have size $\aleph_n$. This is because most of the sets we construct are defined in terms of power sets. This means that we can compute the size of these sets in terms of ℶ numbers pretty easily, but we can't compute them in terms of ℵ numbers. The difficulty is related to the unprovability of the continuum hypothesis. It turns out that ZFC can say very, very little about the cardinalities of ℶ numbers.
One way to get sets of a fixed cardinality is to talk directly about well orderings. For example, $\aleph_1$ is exactly the set of order types of well orderings of $\omega$ (regardless of what $\beth_1$ is).
For large cardinals, there is no way to explicitly compute their cardinality. For example, any inaccessible cardinal number $\kappa$ will have the property that $|\kappa| = \aleph_\kappa$, so you will not be able to make progress by trying to compute its ℵ number.
I can't seem to make $\beth$ appear. If anyone can, please feel free to edit my post. Thanks, –  Carl Mummert Nov 13 '10 at 14:24
One workaround is to write ℶ<sub>1</sub> (or &#x2136;<sub>1</sub>). –  Tsuyoshi Ito Nov 14 '10 at 0:19