, 2012

Let $\alpha = 3/2$ so $1 + \alpha > \alpha^2$.
Given $y$, we construct $Y$ algorithmically. Let $Y = \emptyset$ and of course $S_\emptyset = 0$. For $i = 0$ to $m$, perform the following operation:
If $S_Y + 2^i 3^{m-i} \le y$, then replace $Y$ by $Y \cup \{2^i 3^{m-i}\}$.
When this process is finished, we have a subset $Y$ of $P_m$ such that $S_Y \le y$.
Notice that the elements of $P_m$ are in ascending order of size as given, and may alternatively be described as $2^m, 2^m\alpha, 2^m\alpha^2, \dots, 2^m\alpha^m$. If any member of this list is not in $Y$, then no two consecutive members of the list to the left of the omitted member can both be in $Y$. This is because $1 + \alpha > \alpha^2$, and the greedy nature of the process used to construct $Y$.
Therefore either $Y = P_m$, in which case $y = 3^{m+1} - 2^{m+1}$ and all is well, or at least one of the two leftmost elements of the list is omitted from $Y$.
If $2^m$ is not omitted from $Y$, then the algorithmic process ensures that $(S_Y - 2^m) + 2^{m-1}3 > y$, and so $y - S_Y < 2^m$. On the other hand, if $2^m$ is omitted from $Y$, then $y - S_Y < 2^m$.

Note that $3^{m+1} - 2^{m+1} = (3-2)(3^m + 3^{m-1} \cdot 2 + \dots + 3 \cdot 2^{m-1} + 2^m) = S_{P_m}$. Dividing every element of $P_m$ by $2^m$ gives us the following equivalent problem:
Let $m$ be a positive integer, $a = 3/2$, and $Q_m = \{1, a, a^2, \dots, a^m\}$. Show that for any real number $x$ satisfying $0 \le x \le 1 + a + a^2 + \dots + a^m$, there exists a subset $X$ of $Q_m$ such that $0 \le x - S_X < 1$.
We will prove this problem by induction on $m$. When $m = 1$, $S_\emptyset = 0$, $S_{\{1\}} = 1$, $S_{\{a\}} = 3/2$, $S_{\{1,a\}} = 5/2$. Since the difference between any two consecutive of them is at most 1, the claim is true.
Suppose that the statement is true for positive integer $m$. Let $x$ be a real number with $0 \le x \le 1 + a + a^2 + \dots + a^{m+1}$. If $0 \le x \le 1 + a + a^2 + \dots + a^m$, then by the induction hypothesis there exists a subset $X$ of $Q_m \subset Q_{m+1}$ such that $0 \le x - S_X < 1$.
If $\frac{a^{m+1} - 1}{a - 1} = 1 + a + a^2 + \dots + a^m < x$, then $x > a^{m+1}$ as
$$\frac{a^{m+1} - 1}{a - 1} = 2(a^{m+1} - 1) = a^{m+1} + (a^{m+1} - 2) \ge a^{m+1} + a^2 - 2 = a^{m+1} + \frac{1}{4}.$$
Therefore $0 < (x - a^{m+1}) \le 1 + a + a^2 + \dots + a^m$. Again by the induction hypothesis, there exists a subset $X$ of $Q_m$ satisfying $0 \le (x - a^{m+1}) - S_X < 1$. Hence $0 \le x - S_{X'} < 1$ where $X' = X \cup \{a^{m+1}\} \subset Q_{m+1}$.