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How to simplify the following:

$$2^{\aleph_0}(\aleph_0+\aleph_0)^{2^{\aleph_0}}$$

Thank you for every help.

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2 Answers 2

up vote 3 down vote accepted

Use that

(i) For cardinals $\kappa$ and $\lambda$ with $\kappa \le \lambda$ and $\lambda$ infinite you have that $\kappa + \lambda = \lambda$ and hence $\aleph_0 + \aleph_0 = \aleph_0$

(ii) For cardinals $\kappa \le \lambda$ where $\lambda$ is infinite you have $\kappa \cdot \lambda = \lambda$

(iii) For an infinite cardinal $\lambda$ and $2 \le \kappa \le \lambda$ you have $\kappa^\lambda = 2^\lambda$

Then

$$2^{\aleph_0}(\aleph_0+\aleph_0)^{2^{\aleph_0}} \stackrel{(i)}{=} 2^{\aleph_0}\aleph_0^{2^{\aleph_0}} \stackrel{(ii)}{=} \aleph_0^{2^{\aleph_0}} \stackrel{(iii)}{=} 2^{2^{\aleph_0}}$$

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Note that both $(ii)$ and $(iii)$ require the axiom of choice to hold in their stated form. It can be shown that the axiom of choice is not needed for the cardinals mentioned in this particular question. –  Asaf Karagila Dec 13 '12 at 15:13
    
And now all three require the axiom of choice. –  Asaf Karagila Dec 13 '12 at 19:46
    
@AsafKaragila Actually: no. Dependent choice is enough to prove (i). –  Matt N. Dec 13 '12 at 19:52
    
No. It's not. I have a counterexample, although it's not trivial enough for me to type from an iPhone. Soon... –  Asaf Karagila Dec 13 '12 at 19:58
    
@AsafKaragila Meanwhile, here's an outline of the proof: Prove that for any $\alpha \in \mathbf{ON}$ there is a unique limit ordinal $\beta$ and $n \in \omega$ such that $\alpha = \beta + n$. This proof can be done in (DC). Since $\lambda \le \lambda + \kappa \le \lambda + \lambda$ we want to show $\lambda + \lambda \le \lambda$. This we do by defining an injection $f: \lambda \times \{0\} \cup \lambda \times \{1\} \to \lambda$ as $(\alpha, i) \mapsto \beta + 2n + i$ where $\alpha = \beta + n$. –  Matt N. Dec 13 '12 at 20:09

Hint: Prove that $(\aleph_0)^{2^{\aleph_0}}=2^{2^{\aleph_0}}$.

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