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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!

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    $\begingroup$ 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
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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?

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  • $\begingroup$ 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

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