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I want to show $2^{ℵ_0}=\mathfrak c$.

I already showed $\mathfrak c \leq 2^{ℵ_0}$ as follows:

Each real number is constructed from an integer part and a decimal fraction. The decimal fraction is countable and has $ℵ_0$ digits. So we have

$\mathfrak c \leq ℵ_0 * 10^{ℵ_0} \leq 2^{ℵ_0} * (2^4)^{ℵ_0} = 2^{ℵ_0}$ since $ℵ_0 + 4ℵ_0=ℵ_0$

But how can I prove the other way $2^{ℵ_0} \leq \mathfrak c$?

user126154
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user8314628
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2 Answers2

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$2^{\aleph_0}$ is the cardinality of all reals (belonging to $(0,1)$ if you prefer) that you can write by using only $0,1$. Those numbers clearly form a subset of $\mathbb R$ which must therefore have cardinality at least $2^{\aleph_0}$.

user126154
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You can define a function $F$ from the set $\{ (x_n) | n\in \mathbb{N}, x_n\in \{ 0,1\} \}$ to $\mathbb {R}$ such that $$F[(x_1,x_2,x_3,\ldots )]=0\ .\ x_1x_2x_3\ldots $$
Then this function is injective. So $$ \text{Cardinal} \{ (x_n) | n\in \mathbb{N}, x_n\in \{ 0,1\} \} \leq \text{Cardinal}(\mathbb{R}) $$ So $2^{ℵ_0} \leq \mathfrak c$.

Darman
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    It’s not 1-1 because $F(0,1,1,\dots)=F(1,0,0,0\dots)$, if I’m not mistaken. But I think the idea is accurate. –  Nov 26 '18 at 13:28
  • No. It's One to One and your saying is false%% – Darman Nov 26 '18 at 13:32
  • Well, I'll let you know that the downvote is mine. –  Nov 26 '18 at 13:40
  • But why! was my solution false? – Darman Nov 26 '18 at 13:50
  • I take it that the sequence $0.x_1 x_2 x_3 \dots$ is a binary sequence, in which case $F$ is not 1-1. However, every binary sequence in $[0,1]$ has at most two distinct pre-images under $F$, and the cardinality of the set of points in $[0,1]$ that have exactly two pre-images is countable. From all of this one can still deduce that $2^{\aleph_0} \leq \mathfrak{c}$ is true. –  Nov 26 '18 at 14:14
  • If you meant something else by $0.x_1 x_2 x_3 \dots$ then please edit your answer to clarify. I'll be happy to change my vote if the answer is clear. –  Nov 26 '18 at 14:15
  • How is it that $F(0,1,1,\dots)=F(1,0,0,0\dots)$? I don't see this. – Git Gud Nov 26 '18 at 21:21
  • I don't like this answer at all because the codomain of $F$, when properly formalized, is $9^{\aleph_0}$, so this doesn't really add any value. – Git Gud Nov 26 '18 at 21:22
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    @GitGud For the same reason that 0.999...=1, assuming that $0.x_1 x_2 x_3 \dots$ is in binary. –  Nov 27 '18 at 15:36