How is the property of having no infinitely large or small elements derived from the original statement of the archimedean property? There is already a question asked related to this.
Here's the link:Does this definition of archimedean property guarantee that a set has no infinitely large or infinitely small elements?
That question asks that, does his/her statement guarantees the property of having no infinitely large or small elements. But there is no proper answer and I  also don't know is his/her derived statement truly guarantees the property mentioned in the wikipedia page of the Archimedean property:

Roughly speaking, it is the property of having no infinitely large or
infinitely small elements

My question is kind of a generalized version of that question.
From the original statement
$\forall x,y \in \mathbb{R}_{+}$,
$\exists n \in \mathbb{N}:nx>y$
How to derive the property of having no infinitely large or infinitely small element?
(I am not a mathematician so please don't used too advanced mathematics in the answers)
 A: My statement is based on a simpler version of the Archimedean property, which states $\forall y \in \mathbb{R} \exists n \in \mathbb{N} (n > y)$. This is easy to prove from OP's definition: we note that $|y| + 1 \in \mathbb{R}_+$ and $1 \in \mathbb{R}_+$. Then, we take some $n \in \mathbb{N}$ such that $n \cdot 1 > |y| + 1$. Then $n \cdot 1 > |y| + 1 > |y| \geq y$. Then $n > y$.
What is an "infinitely large number?" We need to define this term precisely before we can answer your question.
A sensible definition of "$x$ is an infinitely large number" is $\forall n \in \mathbb{N} (|x| > n)$. Obviously, the Archimedean property immediately refutes the existence of an infinitely large $x$,  since it states that there exists $n \in \mathbb{N}$ such that $|x| < n$.
What about an "infinitely small number"? I would define "x is infinitely small" to be the statement $|x| > 0 \land \forall n \in \mathbb{N} (|x| < \frac{1}{n})$.
There can be no infinitely small $x$. For if there were such an $x$, take some $n$ such that $n > \frac{1}{|x|}$. In that case, we would have $\frac{1}{n} < |x|$.
