Is $\aleph_{1}^{\aleph_0}<\aleph_{2}^{\aleph_0}$ equivalent to the Continuum Hypothesis

cardinalsset-theory

While searching for exercises dealing with CH and GCH, I encountered an exercise with the following statement:

Study if:

$\aleph_{1}^{\aleph_0}<\aleph_{2}^{\aleph_0}$

Is equivalent to the Continuum Hypothesis

My first thought is that the statement has nothing to do with CH for it only gives a relation between $\aleph_{1}$ and $2^{\aleph_{0}}$.

I am getting started with this topic, so if you could give a clear and full answer, I would really appreciate it.

Thanks in advance.

EDIT: deleted part of the original question, as it was pretty ill-posed

Best Answer

The key is to observe that $\aleph_1^{\aleph_0}=(2^{\aleph_0})^{\aleph_0}$ regardless of whether CH holds. In particular, it's going to help in a bit to not simplify the right hand side.

Here's a proof of that observation:

  • $\le$: immediate, since we know without assumptions that $\aleph_1\le 2^{\aleph_0}$ and raising both sides to $\aleph_0$ doesn't break $\le$.

  • $\ge$: the right hand side simplifies to $2^{\aleph_0}$, and (by the same reasoning as the previous bullet) we know that $2^{\aleph_0}\le \aleph_1^{\aleph_0}$.

Now back to the main problem.

The comparison we're really trying to make is, by the observation above, $$(2^{\aleph_0})^{\aleph_0}\mbox{ versus }(\aleph_2)^{\aleph_0}.$$

We know that we get "$<$" if CH holds. Now suppose CH fails; what does that tell us about $2^{\aleph_0}$ versus $\aleph_2$ (and hence, what does that tell us about the comparison we want to make)?

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