# Is $f: \mathcal{P}(\mathbb N) \to [0 ,+\infty]$ bijection?

Consider the set-function

$f: \mathcal{P}(\mathbb N) \to [0 ,+\infty]$ with $\displaystyle{ f(A)= \sum_{n \in A } \frac{1}{3^n}}$ where $A \subset \mathbb N$

(a) Is $f$ one-to-one ?

(b) Is $f$ bijective ?

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There’s an obvious, natural way to try to prove that $f$ is one-to-one; have you tried it? As for (b), is there any $A$ such that $f(A)=2/3$? – Brian M. Scott Apr 17 '12 at 21:39
Never mind $2/3$, ask about $f(A) = 10$. – GEdgar Apr 17 '12 at 21:41
@GEdgar: Oh, my; I didn’t even notice the silly codomain! – Brian M. Scott Apr 17 '12 at 21:42
Have you tried to represent fractions in binary base, e.g. $1/3 = 0.010101\overline{01}_{bin}$? What about base $3$? – dtldarek Apr 17 '12 at 21:43
@GEdgar: O.K I see that we can't find such $A$ so $f$ is not bijective. – passenger Apr 17 '12 at 21:45

HINT for (a): Suppose that $f(A)=f(B)$, but $A\ne B$. Let $m$ be the smallest integer that is in exactly one of $A$ and $B$; without loss of generality suppose that $m\in A\setminus B$. Then $$\sum_{k\in A\atop{k<m}}\frac1{3^k}=\sum_{k\in B\atop{k<m}}\frac1{3^k}\;.$$ Call this sum $s$. Then $$f(A)\ge s+\frac1{3^m}\;,$$ and $$f(B)\le s+\sum_{k>m}\frac1{3^k}\;;$$ can you take it from there?
From these two inequalities and since $f(A)=f(B)$ we get that $\displaystyle{\frac{1}{3^m} \leq \sum_{k>m} \frac{1}{3^k}}$ why is this absurd? – passenger Apr 17 '12 at 22:19
@passenger: Sum the geometric progression $\sum_{k>m} \frac{1}{3^k}$. – Brian M. Scott Apr 17 '12 at 23:06
@passenger: Yes, you are missing that: $$\sum_{k>m, k\in B}\frac1{3^k}\le\sum_{k>m}\frac1{3^k}=\frac1{2\cdot 3^m}<\frac1{3^m}$$ – Asaf Karagila Apr 18 '12 at 0:23