Prove that $2^\mathbb{N}$ is equinumerous to $\mathbb{R}$

Question

I was thinking using the fact that $[0,1]$ is equinumerous with $\mathbb{R}$, but i cant think of a bijection from there to $2^\mathbb{N}$.

https://en.wikipedia.org/wiki/Cardinality_of_the_continuum#Cardinal_equalities — StackTD, Jun 16 '16 at 17:15
I was tempted to close this as a duplicate of http://math.stackexchange.com/questions/1449619/cardinality-of-the-reals-and-the-power-set-of-naturals instead. — Asaf Karagila, Jun 16 '16 at 17:24

score 0 · Answer 1 · answered Jun 16 '16 at 17:21

For $2^{\mathbb{N}} \le \mathbb{R}$ consider the map $T:2^{\mathbb{N}}\to \mathbb{R}$ given by

$$T(f)=\sum_{n=0}^{\infty} \frac{f(n)}{10^n}$$

For $\mathbb{R}\le 2^{\mathbb{N}}$, note that $2^{\mathbb{N}}\sim \mathcal{P}(\mathbb{N})\sim \mathcal{P}(\mathbb{Q})$

and use the map

$$ S(x)=\{q\in\mathbb{Q}:q<x\} $$

for $x\in \mathbb{R}$

Now you just have to show that these maps are injective. I believe it's not very hard from here

1 Answers1