Definition (Summation over finite sets) Let $X$ be a finite set with $n$ elements let $f:X \rightarrow\mathbb R$ be a function. We define $\sum_{x \in X} f(x)$ as follows. Select any bijection $g$ from {${i \in N : 1 \leq i \leq n}$} to $X$ we then define:
$\sum_{x \in X} f(x)$ $:=$ $\sum_{i=1}^{n} f(g(i))$
I've been stuck on this problem, its really been bugging me here it is:
Let $n\leq m$ and let $X$ be the set $X$ $:=$ {${i \in Z : n \leq i \leq m}$} if $a_i$ is a real number assigned to each integer $x$ $\in$ $X$, then we have
$\sum_{i=n}^{m} a_i$ = $\sum_{i \in X} a_i$
I've tried induction but keep getting strange results, I've tried finding some bijection for example:
let $h(i)$ $=$ $n$ + (i $-$ 1) if $n \leq i <k$ or $h(i) = m$ if $i=k$
where $h$ is a bijection from {$i \in N : 1 \leq\ i \leq k$} to $X$
any help or even solution for that matter would be greatly appreciated. Any thoughts?
The best way to think about this exercise for me was to realize the following: A finite sequence is a function $a: \{i \in \mathbb{Z} \mid m \leq i \leq n \} \to \mathbb{R}$. The domain of the sequence is a finite set. Thus, it should be possible to perform the sum $\sum_{i \in X} a_i$. Furthermore, this should just be a finite sequence as we have defined it at the beginning of the section; i.e., as a sum of finite sequence elemnts $\sum_{i=m}^n a_i$.
Proof: First note that $X$ is a finite set with $n - m +1$ elements. Thus, there exists a bijection $g: \{j \in \mathbb{N} \mid 1 \leq j \leq n-m+1\} \to X$. Consider now the specific bijection defined by $g(j) = j+m -1$. Note that $g$ is injective. Indeed, $g(j_1) = g(j_2)$ implies $j_2 + m -1 = j_1 + m -1$ and $j_2 = j_1$. Note that $g$ is surjective. Indeed, for every $i \in X$ one can choose $j = i - m +1$. Then, $g(i-m+1) = i$. Thus, $g$ is a bijection. Therefore, \begin{align*} \sum_{i \in X}a_i &= \sum_{i = 1}^{n-m+1}a_{g(i)} &\text{By summation over finite sets}\\ &= \sum_{i=m}^{n}a_{g(i-m+1)} &\text{by properties of finite series}\\ &= \sum_{i = m}^n a_i &\text{by construction of $g$}. \end{align*}