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 $2^{ℵ_0} \leq \mathfrak c$?

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