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

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.

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

• If CH is true, then $\aleph_1^{\aleph_0} = (2^{\aleph_0})^{\aleph_0} = 2^{\aleph_0\aleph_0} = 2^{\aleph_0} = \aleph_1\lt \aleph_2 \leq \aleph_2^{\aleph_0}$. So CH certainly implies the inequality. – Arturo Magidin Apr 26 at 20:54
• Yes... but that's just the contrapositive of what I just said! CH implies the inequality, so Not(inequality) implies Not(CH). That is not the question you should ask next however; the question that remains is whether the inequality implies CH, or if the negation of CH implies that the inequality is false. – Arturo Magidin Apr 26 at 21:10
• We can have $\aleph_1^{\aleph_0} = \aleph_2^{\aleph_0}$ in models of MA and non-CH, say with continuum equal to $\aleph_3$. – Henno Brandsma Apr 26 at 21:21
• @Arthur There is a binary operation on the ordinals called exponentiation. Under this operation, $2^\omega = \omega$. – Alex Kruckman Apr 26 at 21:21
• @Akerbeltz No, that's not true: we could have CH$_{\aleph_1}$ hold even if CH itself fails. (Here "CH$_\kappa$" means "$2^\kappa=\kappa^+$," so e.g. CH$_{\aleph_1}$ means $2^{\aleph_1}=\aleph_2$; and usual CH is CH$_{\aleph_0}$ in this notation.) – Noah Schweber Apr 26 at 21:57

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)?

• So if HC doesn't hold, $\aleph_{2}\le 2^{\aleph_0}$ – Akerbeltz Apr 26 at 22:15
• @Akerbeltz CH, but yes. – Noah Schweber Apr 26 at 22:57
• In Spanish it's all the way around, it's lame that comments can't be edited... – Akerbeltz Apr 26 at 22:58

Using Hausdorff formula we have $$\aleph_2^{\aleph_0}=\aleph_1^{\aleph_0}\cdot\aleph_2,$$ Now using the fact that $$\aleph_0^{\aleph_0}\leq\aleph_1^{\aleph_0}\leq (2^{\aleph_0})^{\aleph_0}=2^{\aleph_0}$$, we get that $$\aleph_2^{\aleph_0}=2^{\aleph_0}\cdot\aleph_2.$$

If $$2^{\aleph_0}\geq\aleph_2$$, then $$\aleph_1^{\aleph_0}=\aleph_2^{\aleph_0}$$. Therefore, the assumption that $$\aleph_1^{\aleph_0}<\aleph_2^{\aleph_0}$$ implies that CH holds.