Is this function one to one? Why?

Question

Is the function $f:P(\mathbb{N})\to\mathbb{R}$ defined by $$f(A)=\sum_{n\in A}\dfrac{2}{3^{n+1}},\quad\forall\,\,A\in P (\mathbb{N}),$$ an one to one function? Please help understand why. For me, $\mathbb{N}=\{n\in\mathbb{Z}:n\geq0\}$, if that is important.

Answer 1

Let $A_1$ and $A_2$ be two distinct subsets.

Trivially, if $A_1 \subsetneq A_2$ then $f(A_1) < f(A_2)$ and if $A_2 \subsetneq A_1$, then $f(A_2) < f(A_1)$.

Else, $A_1 \setminus A_2$ and $A_2 \setminus A_1$ are nonempty. Let $n_1 = \min(A_1 \setminus A_2)$ and $n_2 = \min(A_2 \setminus A_1)$.

Now, note that $f(A_1) - f(A_2) = \displaystyle\sum_{n \in A_1}\dfrac{2}{3^{n+1}} - \displaystyle\sum_{n \in A_2}\dfrac{2}{3^{n+1}}$ $= \displaystyle\sum_{n \in A_1 \setminus A_2}\dfrac{2}{3^{n+1}} - \displaystyle\sum_{n \in A_2 \setminus A_1}\dfrac{2}{3^{n+1}}$.

If $n_1 < n_2$ then $\{n_1\} \subseteq A_1 \setminus A_2$ and $A_2 \setminus A_1 \subseteq \{n_2,n_2+1,n_2+2,\ldots\}$.

Hence, $f(A_1)-f(A_2) = \displaystyle\sum_{n \in A_1 \setminus A_2}\dfrac{2}{3^{n+1}} - \displaystyle\sum_{n \in A_2 \setminus A_1}\dfrac{2}{3^{n+1}} \ge \dfrac{2}{3^{n_1+1}} - \displaystyle\sum_{n = n_2}^{\infty}\dfrac{2}{3^{n+1}} = \dfrac{2}{3^{n_1+1}} - \dfrac{1}{3^{n_2}}$.

Can you show that this is greater than $0$? The case where $n_1 > n_2$ can be handled similarly.