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How much is ${\aleph_0}^{\aleph _ 0}$?

On the left I can find ${2}^{\aleph_0}\le {\aleph_0}^{\aleph _ 0}$ but on the right I can not found someone that is $\le$.

In general, how do I use Cantor-Bernstein to find equalities of cardinalities.


marked as duplicate by Najib Idrissi, Asaf Karagila, Lost1, user63181, user61527 Feb 18 '14 at 18:50

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    $\begingroup$ Do you mean $|\aleph_0^{\aleph_0}|$? $\endgroup$ – Dustan Levenstein Feb 18 '14 at 17:01
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    $\begingroup$ An idea to prove they are both equipotent to $[0,1]$: for the LHS one, use binary digits decomposition, for the RHS, use decomposition in continued fractions. There are some tricks not to forget, but it should work. $\endgroup$ – Jean-Claude Arbaut Feb 18 '14 at 17:11
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    $\begingroup$ Weekly edition of What is $\aleph_0$ powered to $\aleph_0$? $\endgroup$ – Najib Idrissi Feb 18 '14 at 17:40

Cantor-Bernstein says that $\lambda\le \mu$ and $\mu\le \lambda$ (which is the same as $\lambda\ge \mu$) imply $\lambda=\mu$ for cardinals $\lambda,\mu$. So the strategy for proving equalities of cardinals is always to find an upper and a lower bound for them.

Now let us do this in your example:

$$2^{\aleph_0}\le \aleph_0^{\aleph_0} \le \left(2^{\aleph_0}\right)^{\aleph_0} = 2^{\aleph_0\cdot \aleph_0} = 2^{\aleph_0}$$

Therefore Cantor-Bernstein implies $\aleph_0^{\aleph_0}=2^{\aleph_0}$.

The latter is exactly the cardinality of the real numbers.


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