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

Question

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 $2^{ℵ_0} \leq \mathfrak c$?

Answer 1

$2^{\aleph_0}$ 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 $2^{\aleph_0}$.

Answer 2

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 $2^{ℵ_0} \leq \mathfrak c$.