0
$\begingroup$

How to simplify the following:

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

Thank you for every help.

2 Answers 2

3

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

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

  • 0
    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.2012-12-13
  • 0
    And now all three require the axiom of choice.2012-12-13
  • 0
    @AsafKaragila Actually: no. Dependent choice is enough to prove (i).2012-12-13
  • 0
    No. It's not. I have a counterexample, although it's not trivial enough for me to type from an iPhone. Soon...2012-12-13
  • 0
    @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 $\alpha = \beta + n$.2012-12-13
  • 0
    But DC won't let you to climb beyond $\omega_1$.2012-12-13
  • 0
    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 $|A|+|A|\cdot2=|A|\cdot3\neq|A|\cdot2$.2012-12-13
  • 0
    @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.2012-12-13
  • 0
    Of course I am already trying to figure it out on my own (while at the same time doubtful that it really needs AC)2012-12-13
  • 0
    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**.2012-12-13
  • 0
    @AsafKaragila But a cardinal is an ordinal.2012-12-13
  • 0
    ...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...?2012-12-13
  • 0
    @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.2012-12-13
  • 0
    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.2012-12-13
  • 0
    @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?2012-12-13
  • 0
    Got it. Ignore previous comments.2012-12-13
  • 0
    Good. :-)${}{}{}$2012-12-13
1

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