How to prove that $C\cdot\aleph_0=C$

Question

How can I prove that $C\cdot\aleph_0=C$?

I tried this:

Given that $k\cdot 1=k$ and $C\cdot C=C$
if $C\cdot C = C \wedge C\cdot 1 = C \wedge C>|\mathbb N|>1$ then
$C\cdot |\mathbb N|= C$

c is the size of the continuum and k is any cardinal.

Is this correct?

Answer 1

Observe that $$ C\le C.\aleph_0\le C.C=C $$ Now use Cantor-Bernstein theorem to deduce the statement.

For proving that $C.C=C$ you can use decimal expansion of real numbers and the function $$ f:[0,1]\times[0,1]\to [0,1]\\ f(.r_1r_2r_3\dots,r'_1r'_2r'_3\dots)=.r_1r'_1r_2r'_2r_3r'_3\dots $$

Answer 2

If $C$ is the cardinality of the continuum, that is, $C=2^{\aleph_0}$, you can use a direct method, by knowing that $C=|\mathbb{R}|$: write $$ \mathbb{R}=\bigcup_{n\in\mathbb{Z}}(n,n+1] $$ Since $|(n,n+1]|=C$, you have the thesis.