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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 ?

Thanks in advance!

  • 1
    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 $f(A)=2/3$?2012-04-17
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    Never mind $2/3$, ask about $f(A) = 10$.2012-04-17
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    @GEdgar: Oh, my; I didn’t even notice the silly codomain!2012-04-17
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    Have you tried to represent fractions in binary base, e.g. $1/3 = 0.010101\overline{01}_{bin}$? What about base $3$?2012-04-17
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    @GEdgar: O.K I see that we can't find such $A$ so $f$ is not bijective.2012-04-17
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    @BrianM.Scott: Can you explain me how $f$ is one-to-one because I can't see it. Thank's!2012-04-17
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    (Assuming it's homework) hint: think about a somewhat simpler case with a couple of features changed, for example $ \displaystyle{ f(A)= \sum_{n \in A } {2^n}}$. Try it out for some small subsets $A$. Does it look familiar? If so, then proceed back toward the original question, feature-by-feature.2012-04-17
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    If is$A$ proper subset of $B$ then obviously $f(A) \neq f(B)$. The same if is $B$ proper subset of $A$. But what happens if $A \cap B \neq \emptyset$ ?2012-04-17

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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?

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    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?2012-04-17
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    @passenger: Sum the geometric progression $\sum_{k>m} \frac{1}{3^k}$.2012-04-17
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    How can I sum this geometric progression? I mean that we can't know the ratio because it might missing some number's it may be k=m+1, k=m+2, k=m+4 the next.2012-04-17
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    Am I missing something obvious?2012-04-17
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    @passenger: Yes, you are missing that: $$\sum_{k>m, k\in B}\frac1{3^k}\le\sum_{k>m}\frac1{3^k}=\frac1{2\cdot 3^m}<\frac1{3^m}$$2012-04-18
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    O.K Thank you for you time both!2012-04-18