Cardinality of $\mathbb{R}^n$ and $\mathbb{R}$ is equal How can we prove that cardinality of set of all real numbers $\mathbb{R}$ and $\mathbb{R}^n$ is equal for every $n$ in the set of all natural numbers?
 A: Consider z=(x,y) with x=0.x1x2x3… and y=0.y1y2y3… their decimal expansions (the standard ones without an infinite series of nines as a suffix). Then the number f(x,y)=0.x1y1x2y2x3y3… is a bijection between $(0,1)$ and $(0,1) \times (0,1)$. Since $card(0,1)=card(\mathbb{R})$ we have  $card(\mathbb{R}) = card(\mathbb{R^2})$. 
The rest is done by simple induction:
$card(\mathbb{R^n})=card(\mathbb{R^{n-1}} \times \mathbb{R})=card(\mathbb{R^{n-1}}) + card(\mathbb{R})=card(\mathbb{R}) + card(\mathbb{R})= card(\mathbb{R^{2}})=card(\mathbb{R})$
A: for any $m,n \gt 0$ look at the half-open open unit hypercubes. represent the reals in (say) binary notation, avoiding infinite sequences of 1's. then you can set up a 1-1 correspondence between the points of the hypercubes in $\mathbb{R^n}$ and $\mathbb{R^m}$ by interleaving the digits of $(r_1,r_2,...,r_n)$. clearly this is reversible.
since the set of such hypercubes in any $\mathbb{R^n}$ are countable, this bijection can be extended to include the whole spaces.
