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Prove the existence of infinite number of infinite cardinals

1) $\alpha$, such that $\alpha<\alpha^\aleph$

2) $\beta$, such that $\beta=\beta^\aleph$

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You might want to look at… for some tips on asking questions here (even if this isn't homework). – Gerry Myerson Apr 19 '12 at 6:36

The first problem is quite a bit harder than the second.

For $1$), use König's theorem. We still then need to show that, for example, there are infinitely many cardinals of cofinality say $\omega$, but that part is not hard. For suppose we start at the infinite cardinal $\kappa_0$. Let $\kappa_1=2^{\kappa_0}$, $\kappa_2=2^{\kappa_1}$, and so on, and let $\kappa_\omega=\bigcup \kappa_n$. Then $\kappa_\omega$ has a right cofinality, and König's theorem applies. For the next one, start at $\kappa_\omega$.

For $2$), we can use for $\beta$ anything of shape say $\kappa^\aleph$, where $\kappa$ is an infinite cardinal.

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Actually just taking $\alpha=\kappa^{+\omega}$ would ensure that $\alpha$ has countable cofinality. – Asaf Karagila Apr 19 '12 at 18:55
Yes, I am taking (possibly) larger hops than necessary. – André Nicolas Apr 19 '12 at 18:56

Using $\aleph$ as a general cardinal might be ambiguous (there are places where it is used particularly for $2^{\aleph_0}$), the answer remains the same regardless to the intended use of $\aleph$.

  1. Recall that for every infinite $\kappa$ we have $\kappa<\kappa^{\operatorname{cf}(\kappa)}$. Simply show that there are infinitely many cardinals whose cofinality is $\aleph$. You can show that there exists a sequence $\alpha_0<\alpha_0^\aleph<\alpha_1<\ldots$ by starting the construction of $\alpha_{n+1}$ from the construction of $\alpha_n^\aleph$.

  2. Cardinal exponentiation has the property $\left(\kappa^\lambda\right)^\mu=\kappa^{\lambda\cdot\mu}$. Take $\alpha$ from the previous part and take $\beta=\alpha^\aleph$, now we have $\beta^\aleph=\alpha^{\aleph\cdot\aleph}=\alpha^\aleph=\beta$.

Note that for the second part you don't really need the first part, however it is easy to show that there are infinitely many $\beta$ if you use the fact that $\alpha_n^\aleph\neq\alpha_k^\aleph$ for $n\neq k$.

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If anyone is interested, there's a short proof of the OP's question in the case of $\aleph_{0}$ at Like Asaf Karagila's argument, it shows slightly more than what was asked by showing the existence of arbitrarily large such cardinals, and not just infinitely many such cardinals. – Dave L. Renfro Apr 19 '12 at 15:30

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