Why these two sets are equinumerous?

$$[0,1]^\Bbb N\text{ and }\Bbb Q^\Bbb N$$

Here is my reason: The set of rational numbers $\Bbb Q$ is countably infinite. However, $[0, 1]$ is not countable and is infinite. So, they shouldn't be equinumerous.

Even, there is the power of $\Bbb N$, it shouldn't change anything.

But, I am wrong.

Can anybody tell me what is wrong please?

Thank you in advance!

  • 1
    $\begingroup$ This is one way that infinite cardinal operations are not the same as finite ones. One hint: $|[0,1]| = |\mathbb{Q}^\mathbb{N}|$, and $\mathbb{Q}^\mathbb{N} \cong \mathbb{Q}^{\mathbb{N}\times\mathbb{N}}$. $\endgroup$ – Carl Mummert Oct 15 '14 at 2:16
  • $\begingroup$ Why QN has the same cardinality as R?? $\endgroup$ – Blackgirl5 Oct 15 '14 at 2:50
  • $\begingroup$ It will have the same cardinality as $\mathbb{N}^\mathbb{N}$. $\endgroup$ – Carl Mummert Oct 15 '14 at 11:12

First of all, note that $\Bbb{Q^N}$ includes $\{0,1\}^\Bbb N$, so it too is uncountable. But just being uncountable doesn't mean much because there are uncountable sets of different cardinalities.

But note that $|[0,1]|=2^{\aleph_0}$ and $|\Bbb Q|=\aleph_0$. Therefore $[0,1]^\Bbb N$ has cardinality $(2^{\aleph_0})^{\aleph_0}$, and $\Bbb{Q^N}$ has cardinality $\aleph_0^{\aleph_0}$.

What do you know about these two cardinalities?

  • $\begingroup$ Well, I understand that (2ℵ0)ℵ0 is equinumerous to R. So, ℵℵ00 should also be equinumerous with R. But why? $\endgroup$ – Blackgirl5 Oct 15 '14 at 2:49
  • $\begingroup$ $2^{\aleph_0}\leq\aleph_0^{\aleph_0}\leq(2^{\aleph_0})^{\aleph_0}$. $\endgroup$ – Asaf Karagila Oct 15 '14 at 2:58

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

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