# the cardinality of all finite subsets of real numbers

I do realise this question has already been answered here, but I want a more direct answer in the form of defining an injection. The first injection is trivial, from real numbers to the set of all finite subsets of real numbers, it assigns each real number a set that contains only that number. But how to form the injection the other way around? I tried classifying the finite sets by their cardinality but can't find an injection to the set of real numbers.

• This might help: math.stackexchange.com/a/611645/389703. And consider the injection from the set of finite substet of $\mathbb{R}$ to $\mathbb{R}^\mathbb{N}$ by mapping every subset to an ordered sequence containing its elements, with trailing zeros. – j3M May 31 '18 at 17:24

Map first to finite subsets of $(0,1)$ by mapping each member $x$ of the set to $(\tanh(x)+1)/2$. Then map $\{t_1, \ldots, t_n\}$ to $n + 0.t_{11}t_{21}\ldots t_{n1} t_{12} t_{22}\ldots t_{n2} \ldots$ where $t_{ij}$ is the $j$'th decimal digit of $t_i$.
• Assuming that $\{j\in \Bbb N: t_{ij}\ne 9\}$ is unbounded (infinite) for any $t_i$ so that $\{t_{ij}:j\in \Bbb N\}$ is uniquely defined................+1 – DanielWainfleet May 31 '18 at 18:29