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We know that the set of all natural numbers $N$ is countable. We know that the set of all real numbers $R$ is uncountable .

Define $A= N×N×N×... $ Cartesian product of $N$ in uncountable times.

Define $B=R×R×R×... $ Cartesian product of $R$ in countable times.

I think that the sets $A$ and $B$ are uncountable sets. Is my think correct?

Then whether the sets $A$ and $B$ have the same cardinality?

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    $\begingroup$ "Uncountable" does not specify sufficiently what $A$ is. For example, it could have $|\mathbb{R}|$ copies of $\mathbb{N}$, which gives you $|\mathbb{N}|^{\mathbb{R}} = (\aleph_0)^{2^{\aleph_0}} = 2^{2^{\aleph_0}}$ as cardinality; but if you take $2^{2^\aleph_0}$ copies, then you get $2^{2^{2^{\aleph_0}}}$ as the cardinality of $A$, which is strictly bigger. "Uncountable" just means "larger than $\aleph_0$". $\endgroup$ – Arturo Magidin Oct 24 '18 at 21:29
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    $\begingroup$ It is easy to exhibit uncountably many elements of $B$, which shows $B$ is uncountable. It is also easy to exhibit uncountably many elements of $A$, which shows that $A$ is uncountable. Whether they have the same cardinality depends on exactly how uncountably many times you take the product of $\mathbb{N}$ to define $A$. $\endgroup$ – Arturo Magidin Oct 24 '18 at 21:30
  • $\begingroup$ @ArturoMagidin yes sir. Your last line is correct. Uncountable means just Greater than aleph 0. $\endgroup$ – Avinash N Oct 25 '18 at 7:35
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If $A$ is infinite, then $\left\lvert \Bbb N^A\right\rvert=2^{\lvert A\rvert}$. This is due to the inequalities $$ 2^{\lvert A\rvert}= \left\lvert\{0,1\}^A\right\rvert\le \left\lvert \Bbb N^A\right\rvert\le \left\lvert \mathcal P(\Bbb N)^A\right\rvert=\left\lvert \left(\{0,1\}^{\Bbb N}\right)^A\right\rvert=\\=\left\lvert \{0,1\}^{\Bbb N\times A}\right\rvert\le\left\lvert \{0,1\}^{A\times A}\right\rvert=2^{\lvert A\rvert^2}=2^{\lvert A\rvert}$$

On the other hand $\left\lvert\Bbb R^{\Bbb N}\right\rvert=2^{\aleph_0^2}=2^{\aleph_0}$.

Therefore, it is true that $\Bbb N^A$ has larger cardinality than $\Bbb R^{\Bbb N}$ for all uncountable $A$ of cardinality at least $2^{\aleph_0}$. On the other hand, the assertion that $2^{\aleph_0}$ is the least uncountable cardinal and the assertion that $2^{\aleph_0}$ is the cardinality of the power set of some uncountable set are both consistent with ZFC.

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  • $\begingroup$ Isn’t $2^{\aleph_0}$ the cardinality, by definition of cardinal exponentiation, of the set of subsets of $\aleph_0$, and hence demonstrably the cardinality of some set, and by Cantor’s Theorem a demonstrably uncountable set? I don’t see why you single this assertion out: it seems like a theorem of ZFC (whereas the assertion that it is the least uncountable cardinality is of course the Continuum Hypothesis, which is a horse of a different color) $\endgroup$ – Arturo Magidin Oct 25 '18 at 3:58
  • $\begingroup$ @ArturoMagidin Thank you, I meant to write cardinality of the power set of some uncountable set. $\endgroup$ – Saucy O'Path Oct 25 '18 at 9:24

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