Question

Exercise 2. If x > 0, show that there exists n ∈ N such that 1/n < x.

Answer 1

If 1/n >= x > 0 for all n ∈ N, then for all n ∈ N 1/x > n.

Let's prove that such real number doesn't exist.

Let A be a set of real numbers $A = \{a\in \mathbb{R}: \exist n\in \mathbb{N}\,\, a<n\}$ ,

$B=\mathbb{R}\setminus A = \{b\in \mathbb{R}: \forall n\in \mathbb{N}\,\, b\geq n\}$ .

$\mathbb{N}\subset A$ implies $A\ne \empty$. If, in addition, $B\ne \empty$, then $\forall a\in A \,\forall b\in B\, a< b$ .

Indeed, there exists $n\in N$, such that a<n and, by the definition, $n\leq b$ , therefore a<b.

So we have Dedekind's section of the set of real numbers.

By the Dedekind's axiom of completeness there exists $c\in \mathbb{R}$ , such that $\forall a\in A \,\forall b\in B\, a\leq c\leq b$ .

We have $c-1<c$, therefore, $c-1\in A$.

We have $c+1>c$, therefore, $c+1\in B$.

By the definition of the set A we have $\exist n\in \mathbb{N} c-1<n$ and, hence, c+1 < n+2. The last inequality means that c+1 must belong to A. This contradiction proves that $B=\empty$ .