Prove that $a \leq 0$ if $a \leq \frac 1 n$ for all $n$ The full question reads:
Suppose that $a$ is a number that has the property that for every $n \in \mathbb{N}$, $a \leq 1/n$. Prove $a \leq 0$.

Is there anyway to show this using Archimedean Property, or is it something related to the Completeness Axiom?  The problem using the Archimedean Property is that I get up to $a&lt; \epsilon$ but from there I am not able to conclude anything about whether $a \leq 0$ because $\epsilon &gt; 0$.
In other words, prove that $a\not&gt;0$. Assume $a&gt;0$, what does that tell you about some $n$?
$n$ is a natural number, therefore it can be $1,2,3, \ldots$. On the other hand, $1/n$ is a continuous function for every real $x$ distinct to zero.

Applying the following limit
$$
\mathop {\lim }\limits_{n \to \infty } a \le \mathop {\lim }\limits_{n \to \infty } \frac{1}{n} = 0
$$

$$
a \le 0
$$

$$Q.E.D.$$