Is it true that $({\aleph_1})^{\aleph_0} = {\aleph_1}$?

Question

My scenario is as follows:

The cardinal number of $\mathbb R$ is $|\mathbb R|={\aleph_1}$ and the cardinal number of the Cartesian product of countable sets of real numbers $\mathbb R \times \mathbb R \times \mathbb R \times\ldots$ is $({\aleph_1})^{\aleph_0}$. I wonder whether $({\aleph_1})^{\aleph_0}$ is still ${\aleph_1}$？

Also, what is $({\aleph_0})^{\aleph_0}$, the cardinal number of the Cartesian product of countable sets of natural numbers $\mathbb N \times\mathbb N \times\mathbb N \times\ldots$?

score 7 · Accepted Answer · answered Jan 13 '15 at 21:47

7

No. It is not necessarily true. And it is also not necessarily true that $|\Bbb R|=\aleph_1$, this is known as the Continuum Hypothesis, and it is not provable nor refutable from the usual axioms of set theory.

Note that $2\leq\aleph_1\leq2^{\aleph_0}$, and therefore $2^{\aleph_0}\leq\aleph_1^{\aleph_0}\leq(2^{\aleph_0})^{\aleph_0}=2^{\aleph_0}$.

But if $\aleph_1<2^{\aleph_0}$, meaning if the Continuum Hypothesis fails, then $\aleph_1<\aleph_1^{\aleph_0}$.

answered Jan 13 '15 at 21:47

Asaf Karagila

393,674

Why $(2^{\aleph_0})^{\aleph_0}=2^{\aleph_0}$? – zxzx179 Jan 13 '15 at 21:50
Do you know how cardinal exponentiation is defined? – Asaf Karagila Jan 13 '15 at 21:51
I don't know the exact definition. I think $(2^{\aleph_0})$ should be the cardinal number of the power set of some set. – zxzx179 Jan 13 '15 at 21:55
This might be a good start, then, or this place. This might be a good place to continue, too. – Asaf Karagila Jan 13 '15 at 21:58
Thank you so much! I will study them. But how to calculate $({\aleph_0})^{\aleph_0}$, if we assume ${\aleph_1} = 2^{\aleph_0}$? – zxzx179 Jan 13 '15 at 22:03
1

Note that in that inequalities I wrote, you can put $\aleph_0$ between $2$ and $\aleph_1$. The result should be apparent now. – Asaf Karagila Jan 13 '15 at 22:10
Got it. Interesting:P – zxzx179 Jan 13 '15 at 22:15

Is it true that $({\aleph_1})^{\aleph_0} = {\aleph_1}$?

1 Answers1