Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

How to simplify the following:


Thank you for every help.

share|cite|improve this question
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 $\kappa^\lambda = 2^\lambda$


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

share|cite|improve this answer
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). – Rudy the Reindeer 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 $\alpha = \beta + n$. – Rudy the Reindeer Dec 13 '12 at 20:09

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

share|cite|improve this answer

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.