Prove that the set of reciprocals of natural numbers has no positive infimum. Prove that for each $x>0$, there is an $ n \in \mathbb N $ such that $\frac{1}{n}$ $<$ $x$, WITHOUT using the Archimedean theorem.
Its simple enough with the theorem, but without the theorem I am having difficulties. Its clear that if we assume the contrary, then $x$ will be the infimum of the set {$\frac{1}{n}$:$ n \in \mathbb N $}. Then how do we derive a contradiction without the Archimedean theorem?
EDITING after thinking of something.
Since, after assuming the contrary, x is supposed to be the infimum, then there exists an $m \in \mathbb N $ such that $\frac{1}{m}$ $<$ $x+a$ for some $a > 0$. Thus, $\frac{1}{m}$ - $a$ $< x$. Now, if we choose $a$ as $\frac{1}{m}$ -$\frac{1}{l}$, for $l \in \mathbb N $, $l > m$, then we will have $\frac{1}{m}$ -$\frac{1}{l}$ $> 0$, and $\frac{1}{m}$ - $a$  =  $\frac{1}{m}$ -$\left(\frac{1}{m} -\frac{1}{l}\right)$ = $\frac{1}{m}$ $ < x$, thus arriving at a contradiction.
Is this a valid proof?
 A: I suppose we are allowed to use the following:

If $S$ is a non-empty subset of $\Bbb R$ and is bounded from below, then there exists a real number denoted as $\inf S$ such that

*

*$s\ge \inf S$ for all $s\in S$

*If $a>\inf S$, then there exists $s\in S$ with $s<a$.


The set $S:=\{\,\frac1n\mid n\in\Bbb N\,\}$ is non-empty and bounded from below by $0$, hence $\inf S$ exists.
Clearly, $\inf S\ge 0$ (as otherwise the second bullet point fails for $a=0$).
Suppose $\inf S>0$. Then with $a:=2\inf S$, the second bullet point tells us that there exists $n\in\Bbb N$ with $\frac1n<2\inf S$. But then $\frac1{2n}<\inf S$, contradicting the first bullet point.
A: The proof will depend on which model you take for the real numbers (e.g. Dedekind cuts, equivalence classes of Cauchy sequences, etc.). Perhaps the easiest model for this question is (despite its arbitrariness and its awkwardness in other respects) the traditional one of decimal expansions. Thus, any positive real number $x$ may be expressed in the form $x=m+\sum_{k=1}^\infty n_k10^{-k}$, where $m\in\Bbb N$ and $n_k\in\{0,...,9\}\,$ $(k=1,2,...)$, and where not all of $m$ and the $n_k$ are $0$. If $m>0$, then $n=2$ will do. Otherwise, $m=0$ and at least one of the $n_k$, say $n_l$, is nonzero. Then we can take $n=10^{l+1}$.
A: If you're allowed to use the ceiling, then you can use the following.
Let $x$ be a positive real.
Suppose $x > 1$, choose $n = 1$.
Suppose $x = 1$, choose $n = 2$.
Suppose $x < 1$, choose $n = \lceil x \rceil$. Since x is not an integer $\lceil x \rceil$ will be strictly greater than $x$ and hence $\frac{1}{\lceil x \rceil}$ will be strictly less than $x^{-1}$.
