$\textit{Define the Maximum}$ of $x, y \in \mathbb{R}$ by the following: $\max\{x,y\} = \begin{cases} x & \text{if } x \geq y \\ y & \text{if } y > x \end{cases}$
How can I generalize this definition so that it applies to any finite collection $\{x_1,...x_n\}\subset\mathbb{R}$ for $n\in\mathbb{N}$?
I guess I have to make sure I understand the definition first.
My attempt: I can choose a starting point $x_n$ and my max would be $x_{n+1}$?
Please help!
On
Let $\{x_1,...x_n\}\subset\mathbb{R}^{*}_{+}$ be a finite collection. And $m = \displaystyle\max \{x_{k} \, | \, {1\le k \le n}\} = x_{i} \, , \, i \in \{1,2,\cdots,n\}$
We can proove easly : $$\displaystyle\forall p \in\mathbb{N}^{*} \, , \quad m \leq \left(\sum_{k=1}^{n}x_{k}^{p}\right)^{\frac{1}{p}} \le n^{\frac{1}{p}} m$$
Then : $$\displaystyle m = \lim_{p\rightarrow +\infty} \left(\sum_{k=1}^{n}x_{k}^{p}\right)^{\frac{1}{p}}$$
For general case $\{x_1,...x_n\}\subset\mathbb{R}$ you can extract just strictly positive elements, and if $\{x_1,...x_n\}\subset\mathbb{R}^{*}_{-}$ we have : $\displaystyle\max \{x_{k} \, | \, {1\le k \le n}\} = - \min\{-x_{k} \, | \, {1\le k \le n}\}$
It might be useful to show that $\max$ is associative in the following sense: Define $x\vee y:=\max\{x,y\}$ and observe $$(x\vee y)\vee z=x\vee(y\vee z).$$
One can show via induction that $x_1\vee x_2\vee\cdots\vee x_n$ is well-defined in the following sense: Any placement of parentheses to make $x_1\vee x_2\vee\cdots\vee x_n$ well-defined has the same value in $\mathbb{R}$.
Then define $\max\{x_1,x_2,\ldots,x_n\}:=\bigvee_{i=1}^n x_i:=x_1\vee x_2\vee\cdots\vee x_n$.