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Note: For the purpose of this question, $\Bbb N$ does not include $0$.

I have a function $f:\mathscr P(\Bbb N) \to \Bbb R$ defined by: $$f(I) = \sum_{n \in I} \frac 1 {n!}$$

This is essentially a transformation from binary sequences indexed by $\Bbb N$ to a number in $\Bbb R$.

I would like to prove that this function is injective.

Travis Willse
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DHMO
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    Do you mean injective, rather than bijective? It is certainly not bijective onto $\Bbb R$, as $f(I) \geq 0$ for all $I$, nor is it bijective onto $[\min f, \max f] = [f(\emptyset), f(\Bbb N)] = [0, e - 1]$, as it does not contain any element in $(f(\Bbb N - {1}), f({1})) = (e - 2, 1)$. – Travis Willse Mar 14 '17 at 14:05
  • @Travis Yes, I meant injective. – DHMO Mar 14 '17 at 14:05
  • BTW this is a special case of factorial expansion (an, in turn, a special case of Cantor series) mentioned - for example - in this answer. – Martin Sleziak Mar 14 '17 at 14:17

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Hint First prove that $f(\{k\}) > f(\{k + 1, k + 2, \ldots\})$ for all $k \in \Bbb N$.

Then, consider distinct elements $I, J \in \mathscr{P}(\Bbb N)$ and the smallest $k \in \Bbb N$ which is in one and not the other. (You'll probably also want to use the apparent fact that $f$ is monotonic under inclusion, that is, that if $I \subseteq J$ then $f(I) \leq f(J)$.)

Travis Willse
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