The Archimedean Property

# The Archimedean Property

We will now look at a very important property known as the Archimedean property which tells us that for any real number $x$ there exists a natural number $n_x$ that is greater or equal to $x$. This is formalized in the following theorem:

 Theorem 1 (The Archimedean Property): For every element $x \in \mathbb{R}$ there exists an element $n_x \in \mathbb{N}$ such that $x ≤ n_x$. • Proof of Theorem: Consider the case where $x ≤ 0$. Then let $n_x = 1$. We've proven that $0 < 1$ already so any $x$ such that $x ≤ 0$ also has that $x ≤ 1$.
• Now for the case where $x > 0$ we will do proof by contradiction. Suppose that there exists an element $x > 0$ such that for all $n_x \in \mathbb{N}$, $x > n_x$. In other words, $x$ is an upper bound for the natural numbers and hence $\mathbb{N}$ would be bounded above. Since $\mathbb{N} \subset \mathbb{R}$ it follows by the axiom of completeness that there exists an element $s \in \mathbb{R}$ such that $s = \sup (\mathbb{N})$, that is, $\forall n \in \mathbb{N}$, $n ≤ s$.
• Now we note that $n + 1 \in \mathbb{N}$, and so since $s$ is the supremum of the natural numbers then $n + 1 ≤ s$ which implies that $n ≤ s - 1$ $\forall n \in \mathbb{N}$. Therefore $s - 1$ is an upper bound for $\mathbb{N}$. But $s - 1 < s = \sup \mathbb{N}$ which is a contradiction since the supremum is defined to be the least upper bound.
• So $\forall x > 0$ there exists a natural number $n_x \in \mathbb{N}$ such that $x ≤ n_x$. $\blacksquare$

We will now look at some important corollaries regarding the Archimedean property.

 Corollary 1: If $S := \left \{ \frac{1}{n} : n \in \mathbb{N} \right \}$ then $\inf S = 0$.
• Proof: We note that $S$ is a nonempty set since $1 \in S$. Furthermore we note that $S$ is bounded below by $0$ since as $n$ gets very large, $\frac{1}{n}$ approaches 0 but does not equal 0. Therefore this set has an infimum in $\mathbb{R}$.
• Let $w = \inf S$. We note that $w ≥ 0$ since we have already deduced that $0$ is a lower bound for $S$.
• Now by the Archimedean property, for any $\epsilon > 0$ there exists a natural number $n$ such that $\frac{1}{n} < \epsilon$, and so we have that $0 ≤ w ≤ \frac{1}{n} < \epsilon$. Now recall that if $0 ≤ w < \epsilon$ for any $\epsilon > 0$ then $w = 0$. And therefore $\inf S = w = 0$. $\blacksquare$
 Corollary 2: If $a$ is a real number such that $a > 0$ then there exists a natural number $n_a \in \mathbb{N}$ such that $0 < \frac{1}{n_a} < a$.
• Proof: Let $S := \left \{ \frac{1}{n} : n \in \mathbb{N} \right \}$. We note that $\inf S = 0$ and since $a > 0$ then $a$ is not a lower bound for $S$ and so there exists a natural number $n_a \in \mathbb{N}$ where $0 < \frac{1}{n_a} < a$. $\blacksquare$
 Corollary 3: If $a$ is a real number such that $a > 0$ then there exists a natural number $n_a \in \mathbb{N}$ such that $n_a - 1 ≤ a < n_a$.
• Proof: Consider the set $S := \{ y \in \mathbb{N} : a < y \}$. This subset of $N$ is nonempty by the Archimedean property. Furthermore, recall that the Well Ordering Principle says that any nonempty subset of the natural numbers has a least element. which we will denote as $n_a$. Therefore $(n_a - 1) \not \in S$ and so $n_a - 1 ≤ a < n_a$. $\blacksquare$