If $x,y$ are elements of $\mathbb{R}$ and $x>0$ then there is a positive integer $n$ s.t. $nx > y$

Question

Im reading a proof about this The proof is here.

Let $A$ be the set of all $nx$, where $n$ runs through the positive integers. If $nx \le y$, then $y$ would be an upper bound of $A$.

(start confused) But then $A$ has a least upper bound in $\mathbb{R}$.

(why? I know there is a theorem There exists an ordered field $\mathbb{R}$ which has the least-upper-bound property. What is this mean?)

Put $a = \sup A$. Since $x>0$, $a-x < a$, and $a-x$ is not an upper bound of $A$. Hence $a-x < mx$ for some positive integer $m$. (why $mx$ is greater then $a-x$? what property is this.) But then $a < (m+1)x$ where $(m+1)x$ is an element of $A$, which is impossible, since $a$ is an upper bound of $A$.

Answer 1

(I'm assuming we are talking about the real numbers). For your first confusion part (A has a Least Upper Bound), what does the Completeness Property of the Real Numbers state about sets that are bounded above?

For your second confusion part ($a-x$ not being an upper bound $\Longrightarrow a-x<mx$ for some $m$), what is the definition of an upper bound of a set? What is the negation of that definition?