How to show $2^{ℵ_0} \leq \mathfrak c$ 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 $2^{ℵ_0} \leq \mathfrak c$?

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 $2^{ℵ_0} \leq \mathfrak c$.
