How to simplify the following:
$$2^{\aleph_0}(\aleph_0+\aleph_0)^{2^{\aleph_0}}$$
Thank you for every help.
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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
$$2^{\aleph_0}(\aleph_0+\aleph_0)^{2^{\aleph_0}} \stackrel{(i)}{=} 2^{\aleph_0}\aleph_0^{2^{\aleph_0}} \stackrel{(ii)}{=} \aleph_0^{2^{\aleph_0}} \stackrel{(iii)}{=} 2^{2^{\aleph_0}}$$
Rudy the Reindeer
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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
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And now all three require the axiom of choice. – Asaf Karagila Dec 13 '12 at 19:46
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@AsafKaragila Actually: no. Dependent choice is enough to prove (i). – Rudy the Reindeer Dec 13 '12 at 19:52
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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
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@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 $\alpha = \beta + n$. – Rudy the Reindeer Dec 13 '12 at 20:09
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But DC won't let you to climb beyond $\omega_1$. – Asaf Karagila Dec 13 '12 at 20:12
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Now, we say that $A$ is $\aleph_1$-amorphous if it is uncountable and cannot be written as a disjoint union of two uncountable sets. It is consistent with DC that an $\aleph_1$-amorphous set exists. Now that if $A$ is such set then $|A|<|A|+|A|$ because $A\times2$ can be split into two uncountable sets. Now we have $|A|<|A|\cdot2<|A|\cdot3$ as well by the same argument. So we have $|A|<|A|\cdot2$ but $|A|+|A|\cdot2=|A|\cdot3\neq|A|\cdot2$. – Asaf Karagila Dec 13 '12 at 20:14
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@AsafKaragila I am enjoying this discussion. Though, instead of a counter example I'd be much more thrilled to be pointed out where in the proof of (i) the axiom of choice is used. – Rudy the Reindeer Dec 13 '12 at 20:17
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Of course I am already trying to figure it out on my own (while at the same time doubtful that it really needs AC) – Rudy the Reindeer Dec 13 '12 at 20:18
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Matt, this is true for ordinals. Not for every two non-well ordered sets. You assume the axiom of choice when you claim this can be used for general cardinals. – Asaf Karagila Dec 13 '12 at 20:18
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@AsafKaragila But a cardinal is an ordinal. – Rudy the Reindeer Dec 13 '12 at 20:20
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...And here is the problem. Cardinals are ordinals in the presence of choice. Without choice $2^{\aleph_0}$ is not a cardinal, therefore cardinal exponentiation is...? – Asaf Karagila Dec 13 '12 at 20:22
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@AsafKaragila I don't understand. Without AC a set might not have a cardinal that is in bijection with it. But cardinals are ordinals with or without AC, by definition: a cardinal is the smallest ordinal in bijection with the set. – Rudy the Reindeer Dec 13 '12 at 20:28
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This is like a discussion with someone convinced that zero is not a natural number... Without the axiom of choice we define the cardinals differently. With the axiom of choice we can prove that the definition amounts to that of initial ordinals. – Asaf Karagila Dec 13 '12 at 20:33
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@AsafKaragila A cardinal is by definition an ordinal that is not in bijection with a smaller ordinal. Why do we have to define them differently without AC and what's the definition you are talking about? – Rudy the Reindeer Dec 13 '12 at 21:19
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Got it. Ignore previous comments. – Rudy the Reindeer Dec 13 '12 at 21:28
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Good. :-)${}{}{}$ – Asaf Karagila Dec 13 '12 at 22:00
