Archimedean Property and Real Numbers

Solution 1:

The Archimedean Property of $\mathbb{R}$ comes into two visually different, but mathematically equivalent versions:

Version 1: $\mathbb{N}$ is not bounded above in $\mathbb{R}$.

This essentialy means that there are no infinite elements in the real line.

Version 2: $$\forall \epsilon>0\ \exists n\in \mathbb{N}:\frac1n<\epsilon$$ This essentially means that there are no infinitesimally small elements in the real line, no matter how small $\epsilon$ gets we will always be able to find an even smaller positive real number of the form $\frac1n$.

Note that $0$ is not infinitesimally small as it is not positive (remember that we take $\epsilon>0$) and $\infty$ doesn't belong in the real line. The extended real line $\overline{\mathbb{R}}$ is in fact not Archimedean, not only because it has infinite elements, but because it is not a field! ($+\infty$ has no inverse element for example).

You may want to note that the Archimedean Property of $\mathbb{R}$ is one of the most important consequences of its completeness (Least Upper Bound Property). In particular, it is essential in proving that $a_n=\frac1n$ converges to $0$, an elementary but fundumental fact.

The notion of Archimedean property can easily be generalised to ordered fields, hence the name Archimedean Fields.

Now, surreal numbers are not exactly $\pm \infty$ and I suggest you read this Wikipedia entry. You might also want to read the Wikipedia page for Non-standard Analysis. In non standard analysis, a field extension $\mathbb{R}^*$ is defined with infinitesimal elements! (of course that's a non Archimedean Field but interesting enough to study)

Solution 2:

Similar to the other answers. The Archimedean property for an ordered field $F$ states: if $x,y>0$, then there is $n \in \mathbb N$ so that $$ x+x+\dots+x \ge y $$ where we have added $n$ terms all equal to $x$.

Consequences: There are no infinite elements $u \in F$, that is, there is no $u$ so that $1+1+\dots+1 \lt u$ (with $n$ terms) for all $n$.

There are no infinitesimal elements $v \in F$, that is, there is no $v$ so that $v>0$ and $v+v+\dots+v < 1$ (with $n$ terms) for all $n$.

There is no real number called $\infty$, so we say the real numbers satisfy the Archimedean property. The "extended real numbers" do not form a field, but may be useful for certain computations in analysis. Instead of saying $\infty$ is defined or undefined maybe it is better to say whether $\infty$ is an element of the set you are talking about.

The surreal numbers No does form an ordered Field, but has infinite and infinitesimal elements, so the Archimedian property fails in No.

Solution 3:

The Archimedean property states that if $x$ and $y$ are positive numbers, there is some integer $n$ so that $y < nx$. This is a property of the real number field. It can be shown that any Archimedean ordered complete fields is isomorphic to the reals.

In an ordered field in which the Archimedean property does not apply, there are numbers $\epsilon > 0$ so that $n\epsilon$ will not eventually exceed every element in the field. These are the so-called infinitesimals.

The extended real numbers are not a field. The Archimedean property applies specifically to the reals.

Fields with infinitesimals are studied in non-standard analysis. See this Wikipedia article.