I know this sounds like an incredibly dumb question, but why is a single nonnegative number smaller than a sum of nonnegative numbers in a vector? I know it's true, but I want to know why it's true. Would it have to be due to the order topology or metric topology on $\mathbb{R}$?
Example: $x = (1, 0, 2, 3, 4, 0, 5)$.
$1+0+2+3+4+0+5 = 15$.
One can see that any $x_i \in x$ is smaller than the sum of the entries of $x$. And there would be a $\leq$ if say, $x = (1, 0, 0, 0, 0, 0, 0)$ or some other vector where the $x_i$ you chose was the only element in the vector. Hopefully what I'm saying here is clear.
But why specifically is it that any $x_i$ is less than the sum of all the elements in the vector?
edit: Ultimately, what I want to know is why is this the case.
For any $x \in \mathbb{R}^p, |x_i| \leq |\sum{x_i}|$.
I originally started my question by just positing that every number had to be nonnegative for the sake of simplicity, but in reality, what I want to know why is the case is this right here.
edit: I'm sorry for anybody that comes across this ridiculously stupid question. You know how sometimes you're working on a problem and it causes you to leap down rabbit holes and question things that you shouldn't because like, you already know them or something? Yeah. this was that. I'm just stupid. Sorry for wasting your time with this.
I tried to write a proof for this, wish it helps.
let $x_i$ be arbitrary nonegative integers such that $x_i \ge 0$ for all $i = 1,2,\ldots,n$.
let $s = x_1 +x_2 + \ldots +x_n$ be the sum of them.
$s - x_i = x_1 + \ldots + x_{i-1} + x_{i+1} + \ldots + x_n \ge 0$, as $x_1, x_2,\ldots, x_n \ge 0$.
$s - x_i \ge 0$, which means $s \ge x_i$.