The Archimedean Property

However small a step, repeating it eventually passes any target. That is the Archimedean property.

The reals contain no infinitely small numbers. However tiny a positive number, adding it to itself enough times outgrows any given large number. Without this fact,

\varepsilon

N

arguments could never pick a "large enough

N

". The proof uses the least-upper-bound axiom.

For example, one step of an 🐜 ant is tiny. But an ant that never stops crosses a soccer field, and eventually circles the Earth. It is the count of steps, not their size, that wins.

Examples & Exercises

Prove the Archimedean property completely from the least-upper-bound axiom: for every $a > 0$ and real $b$ there is a natural $n$ with $na > b$ .

\forall a > 0, \ \forall b \in \mathbb{R}, \ \exists n \in \mathbb{N}: \ na > b

Step 0.5, target 5 — 11 steps get past.

Show proof

Assume the contrary: $na \le b$ for every natural $n$ . Then the set $S = \{na : n \in \mathbb{N}\}$ is non-empty and bounded above by $b$ .
By the least-upper-bound axiom, $s = \sup S$ exists. Since $a > 0$ , $s - a < s$ , and $s - a$ , being below the supremum, is not an upper bound for $S$ . So for some $m$ , $ma > s - a$ .
Adding $a$ to both sides gives $(m + 1)a > s$ . But $(m + 1)a \in S$ , contradicting that $s$ is an upper bound for $S$ . So the assumption fails, and some $n$ has $na > b$ .

Prove completely, from the Archimedean property, that for every $\varepsilon > 0$ there is a natural $n$ with $\frac{1}{n} < \varepsilon$ .

\forall \varepsilon > 0, \ \exists n \in \mathbb{N}: \ \frac{1}{n} < \varepsilon

Step 0.25, target 1 — 5 steps get past, matching

\frac{1}{5} < 0.25

Show proof

Apply the Archimedean property with step $\varepsilon > 0$ and target $b = 1$ : there is a natural $n$ with $n\varepsilon > 1$ .
Both $n$ and $\varepsilon$ are positive, so dividing $n\varepsilon > 1$ by $n$ gives $\varepsilon > \frac{1}{n}$ . The proof is complete.

Prove completely that if $x \ge 0$ and $x < \frac{1}{n}$ for every natural $n$ , then $x = 0$ .

x \ge 0, \ (\forall n) \ x < \frac{1}{n} \ \Longrightarrow \ x = 0

However small the step, the march still passes the target — no positive number fits in between.

Show proof

Suppose for contradiction that $x > 0$ . Applying the corollary of the previous problem with $\varepsilon = x$ gives a natural $n$ with $\frac{1}{n} < x$ .
But the hypothesis gives $x < \frac{1}{n}$ for the same $n$ . A number cannot satisfy both $\frac{1}{n} < x$ and $x < \frac{1}{n}$ . The contradiction shows $x > 0$ is false, and with $x \ge 0$ this forces $x = 0$ .