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I noticed a factor about Set Theory that has somewhat confused me.

Assuming CH, $2^{\aleph_0} = \aleph_1$.

Taking into account Combinatorics, you can treat a number as a set of numbers; 0, 1, 2, 3, 4, 5, 6, 7, 8, and 9. Along with this, the maximum number of digits possible in the Set of ℕ is $\aleph_0$. This would, assuming all combinations, where order is important and repetition is allowed, would be $10^{\aleph_0}$, which would satisfy $a^{\aleph_0}$ where $a \geq 2$. Therefore should be $\aleph_1$ and Biject to the Reals.

I am aware that the Cardinality of the Naturals (and the Rationals) is $\aleph_0$. What aspect of Set Theory prevents it from coming out as $\aleph_1$ due to Combinatorics?

EDIT: Question was answered, I was just being dumb.

Mark Saving
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Zoey
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    I don't entirely understand your question. But I think your question might be cleared up by the fact that $2^{\aleph_0} = 10^{\aleph_0}$. – HallaSurvivor Nov 09 '21 at 20:33
  • The fact that 2^ℵ(0) = 10^ℵ(0) is kinda the entire crux of my question. I know that it does. And that is why I have confusion at how the amount of numbers that result from ℵ(0) Digits is ℵ(1), when it should only be ℵ(0) – Zoey Nov 09 '21 at 20:35
  • Does it help if I point out that "numbers" are all of finite length? So really we have $10^{< \omega}$ (maybe you would rather $10^{< \aleph_0}$, though this notation is nonstandard), which is countable? That is, there are no numbers which require all infinitely many digits to write down, so really it's $10^0 \cup 10^1 \cup 10^2 \cup 10^3 \cup \ldots$ – HallaSurvivor Nov 09 '21 at 20:38
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    Wait, yeah, that is obvious. ℕ doesn't contain ℵ(0), it is just the Cardinality of it, not a number itself. I was just being dumb. Thanks for pointing that out. – Zoey Nov 09 '21 at 20:39
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    Trust me, everybody has been (and sometimes still is) confused by infinite cardinalities—it's definitely not a sign of being dumb! – Greg Martin Nov 09 '21 at 21:00
  • I know quite a bit about them, and the various levels of Transfinite Cardinal, and with Ordinals, I just had a lapse in thought for it. It is quite obvious now that I think more about it. – Zoey Nov 09 '21 at 21:37
  • There are $\beth_1:=2^{\aleph_0}$ strings, of which only $\aleph_0$ are of finite length. – J.G. Nov 09 '21 at 23:26

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