proof of $2^{\aleph_0} = \mathfrak c$

Question

claim

Let $\aleph_0 = card(\Bbb N)$ and $\mathfrak c = card(\Bbb R)$ then

$2^{\aleph_0}=\mathfrak c$

proof

$2^{\aleph_0}=\mathfrak c \iff \exists \text{ bijection } f: \Bbb N \times \{0,1\}\rightarrow \Bbb R$

How to show the existence of bijection for the given sets? What is the most widely used approach?

Answer 1

$2^{\aleph_0}=\aleph_1$ is the Continuum Hypothesis which is famously independent of ZFC so you will find it difficult to prove.

Added following comments.

Look at the Beth Nummbers

Look at the link within that to Cardinality of the continuum where you will find the proof that you want.

$\mathfrak{c}$ is a common symbol for the cardinality of the continuum. Hence $\mathfrak{c} = \beth_1 = 2^{\aleph_0}$ but maybe not $\aleph_1$.