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I was reading this post: Cardinality of power set of $\mathbb N$ is equal to cardinality of $\mathbb R$. Here, bof considered the function $f:\mathcal{P}(\mathbb{N})\to\mathbb{R}$, where for all $S\subset \mathbb{N}$, $f(S)=\sum_{i\in S}10^{-i}$. ¿Why is $f$ inyective?

I tried to solve this. I suposse that there are $S,T\subset\mathbb{N}$ such that $\sum_{i\in S}10^{-i}=f(S)=f(T)=\sum_{i\in T}10^{-i}$. If $S\neq T$, there is $k\in S-T$ or $k\in T-S$. If $k\in S-T$, then \begin{eqnarray} f(S)&=&\sum_{i\in S}10^{-i}\\ &=&\sum_{i\in S-\{k\}}10^{-i}+10^{-k}\\ &>&\sum_{i\in S-\{k\}}10^{-i}\\ &=&\sum_{j\in T-\{k\}}10^{-i}\\ &=&\sum_{j\in T}10^{-i}=f(T) \end{eqnarray} Is true that $\sum_{i\in S-\{k\}}10^{-i}=\sum_{j\in T-\{k\}}10^{-i}$? I'm not sure. I really appreciate your help.

Asaf Karagila
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Jose Esparrago
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1 Answers1

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Hint Since $S \neq T$, there is a smallest $k \in \Bbb N$ such that $k$ belongs to one but not the other. WLOG, $k \in S \setminus T$. In that case, now we can write $$10^{-k} + \sum_{i \in S, i > k} 10^{-i} = \sum_{i \in T, i > k} 10^{-i}.$$ Check that the LHS is at least $10^{-k}$ whereas the maximum value that the RHS can take is...

Aryaman Maithani
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