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I am not a mathematician, but rather a computer engineer with a curious mind.

The continuum hypothesis states (I believe) that there does not exist a set $S$ such that $\aleph_0 < |S| < 2^{\aleph_0}$

The cardinality $2^{\aleph_0}$ always seems to be taken for granted and not really defined in explanations that I run across.

My interpretation of this value is the following:

$2^{\aleph_0} = $ the number of ways of choosing between 2 possibilities at least 0 times and at most $\aleph_0$ number of times.

Is this an appropriate interpretation?

The main reason I thought this even might be true: if this is true, then $2^{\aleph_0}$ is like the cardinality of the binary representations of real numbers, whereas something like $10^{\aleph_0}$ would be the cardinality of the decimal representations of of real numbers. This led me to believe $n^{\aleph_0}$ are equal for ALL $n \in \mathbb{N}, n > 1$. This also showed to me that unary numbers cannot represent the real numbers (because my definition would lead to $1^{\aleph_0} = \aleph_0$) which I found interesting and (at least intuitively) consistent.

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  • $\begingroup$ The interpretation is reasonable. However, $1^{\aleph_0}=1$. $\endgroup$ Jan 2, 2016 at 21:03
  • $\begingroup$ @AndréNicolas thanks for the comment, and the clarification. So your comment has me with more questions! Any $\epsilon > 0$ will yield $(1+\epsilon)^{\aleph_0} = \aleph_1$ (by the CH), but $1^{\aleph_0} = 1$? $\endgroup$ Jan 2, 2016 at 21:17
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    $\begingroup$ Cardinals are discrete objects; it doesn't make sense to talk about $1.5^{\aleph_0}$ or anything like that. Just whole cardinal numbers for the base and exponent. $\endgroup$ Jan 2, 2016 at 21:18
  • $\begingroup$ @JohnCarpenter Other than what Carl said, it should be equal to $2^{\aleph_0}$, not $\aleph_1$ $\endgroup$
    – Kitegi
    Jan 2, 2016 at 21:19
  • $\begingroup$ Thanks for clarifications, this has helped me reconcile many things in my intuition. $\endgroup$ Jan 2, 2016 at 21:21

1 Answer 1

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$2^{\aleph_0}$ is defined to be the number of ways of deciding between two alternatives exactly $\aleph_0$ times.

This can be restated as: $2^{\aleph_0}$ is the number of infinite sequences of 0s and 1s, where the sequence has one entry for each natural numbers.

So you are correct that $n^{\aleph_0}$ is, when $n \geq 2$, the number of $n$-ary expansions of numbers in the interval $[0,1]$.

The number of ways of deciding between two alternatives a finite number $m$ times is $2^m$. We have $\bigcup_{m \geq 2} 2^m = \aleph_0$, so in your characterization using "at most $\aleph_0$ number of times", the real issue is with the sequences the make a choice exactly $\aleph_0$ times.

Regarding the equality at the end of the question, something stronger is true. We can show in ZFC that, for $n \geq 2$, $n^{\aleph_0} = 2^{\aleph_0} = \aleph_0^{\aleph_0}$.

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  • $\begingroup$ Thanks for this answer, this is exactly what I was looking for. I need to keep reading because the fact that $2^{\aleph_0} = \aleph_0^{\aleph_0}$ is mind boggling to me! $\endgroup$ Jan 2, 2016 at 21:20
  • $\begingroup$ The most direct way to prove it is with the Schroeder-Bernstein theorem. You construct an injection from $2^{\aleph_0}$ to $\aleph_0^{\aleph_0}$, and an injection going the other way. $\endgroup$ Jan 2, 2016 at 21:21
  • $\begingroup$ @JohnCarpenter Just like how the $n^{\aleph_0}$ correspond to $n$-ary expansions, you could think of $\aleph_0^{\aleph_0}$ as the continued fraction expression. $\endgroup$
    – Kitegi
    Jan 2, 2016 at 21:22

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