How can I show that the set of reals and the set of pairs of reals have the same cardinality?
I know that since reals are uncountable infinite, I can't create a list of reals and talk about the $i^{th}$ real mapping to the $i^{th}$ real pair. So how can I construct a one-to-one and onto mapping $f: \mathbb R \to \mathbb R^2$?
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Let the binary expansions of $(x,y)\in[0,1)\times[0,1)$ be $$ \begin{array}{} x=\sum_{k=1}^\infty x_k2^{-k}&\text{and}&y=\sum_{k=1}^\infty y_k2^{-k} \end{array} $$ (finite where possible) where $(x_k,y_k)\in\{0,1\}\times\{0,1\}$. Define $f:[0,1)\times[0,1)\mapsto[0,1)$ by $$ f(x,y)=\sum_{k=1}^\infty x_k2^{1-2k}+y_k2^{-2k} $$ that is, $f(x,y)$ interleaves the bits of $x$ and $y$. It is easy to see that $f$ is injective, which means the cardinality of $[0,1)\times[0,1)$ is less than or equal to that of $[0,1)$.
Define $g:[0,1)\mapsto[0,1)\times[0,1)$ by $g(x)=(x,0)$. $g$ is injective.
Use the Cantor-Bernstein-Schroeder Theorem to get the existence of a bijection between $[0,1)\times[0,1)$ and $[0,1)$.
If $a$ is the cardinality of $\mathbb N$, then we have $$2^a\cdot2^a=2^{a+a}=2^a.$$