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Cardinal numbers are used to express sizes of sets.

It $c$ is a cardinal number then $2^c$ is also a cardinal number which represents size of the set of all subsets of a set whose cardinality is $c$.

Question: What $3^c,4^c,\cdots$ represent? Are that all equal to $2^c$?

p Groups
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2 Answers2

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First of all, if $a$ and $b$ are two cardinal numbers, then we define $a^b$ to be the cardinality of the set of all functions from $b$ to $a$. In particular, you can show that $2^A$ is the cardinality of the power set of $A$. You can think of $4^A$ as the cardinality of the power set of the disjoint union $A \sqcup A$.


In fact, if $c$ is any infinite cardinal, then we have $3^c=4^c=...=2^c$.

Indeed, we can show that $2^c=5^c$, for instance. Clearly, $2^c ≤ 5^c$ holds. On the other hand, you have $5 ≤ 2^c$, and since $c$ is infinite : $c^2=c$. Therefore, $$5^c ≤ (2^c)^c = 2^{c^2}=2^c.$$ By Cantor-Schröder-Bernstein theorem, we conclude that $2^c=5^c$.

Of course, if $c$ is a finite cardinal with $c \neq \varnothing$, then $2^c=k^c \implies k=2$.

Watson
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Detail: "$c$" is a bad letter to use for a general cardinal, because $c$ is standard notation for a specific cardinal, namely the cardinality of the reals.

If $\kappa$ and $\mu$ are cardinals then $\mu^\kappa$ represents the cardinality of the set of all mappings from a set of size $\kappa$ to a set of size $\mu$. If $\kappa$ is infinite and $\mu\ge2$ is finite then it turns out that $\mu^\kappa=2^\kappa$.

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    I'd add: the cardinality of the set of subsets is not by definition $2^c$ (to use the questioner's notation), and to force it to have this meaning causes confusion. It being $2^c$ is more like a theorem, stemming from the fact that 2 is the cardinality of the set {Yes,No}. Each subset or the original set corresponds to $c$ Yes/No answers to the question "Is this particular item in this particular subset?". – Martin Kochanski Jun 06 '16 at 13:12