Why $|\Bbb{R}|=2^{\aleph_0}$ and not $|\Bbb{R}|=10^{\aleph_0}$?

Question

Suppose we have a real number of this form:

$$...x_3x_2x_1,x_{-1}x_{-2}x_{-3}...$$

Since $x_i\in\{0,1,2,3,4,5,6,7,8,9\} \ \ \ \forall i$. We have 10 choices for every number, this means that:

$$|\Bbb{R}|=10^{|\Bbb{Z}|}$$

But since $|\Bbb{Z}|=|\Bbb{N}|=\aleph_0$:

$$|\Bbb{R}|=10^{\aleph_0}$$

But as far as i know $2^{\aleph_0}$. Where am I wrong?

score 6 · Accepted Answer · answered Apr 29 '19 at 09:48

6

Because in term of cardinal numbers, $2^{\aleph_0}= 10^{\aleph_0}$.

Another way to say it is that the set of functions from $\mathbb N$ to $\{0,1\}$ is in bijection with the set of functions from $\mathbb N$ to $\{0,1, \dots,9\}$.

answered Apr 29 '19 at 09:48

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Binary numbers ? – Kandinskij Apr 29 '19 at 09:56
@Eureka Yes, you can code a number in ${0,1, \dots, 9}$ as a sequence of $4$ numbers in ${0,1}$. – mathcounterexamples.net Apr 29 '19 at 09:58

Why $|\Bbb{R}|=2^{\aleph_0}$ and not $|\Bbb{R}|=10^{\aleph_0}$?

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