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 $a< \epsilon$ but from there I am not able to conclude anything about whether $a \leq 0$ because $\epsilon > 0$.  
 A: In other words, prove that $a\not>0$. Assume $a>0$, what does that tell you about some $n$?
A: I'll not give the complete answer to this (on purpose).
But here is some intuition:  if you have a positive number, what can you say about that positive number as compared to the numbers $1/n$ for each n?
Start with something concrete, like 0.1 for example.  It can't have the property you mention... why not?  
The same reason works for all positive numbers.
A: If $a \leq 1/n$ for all $n \in \mathbb{N}$, then $a \le \inf \{ 1/n : n \in \mathbb{N} \}=0$. Now, to prove that the inf is $0$, you need the Archimedean Property. (The inf being 0 is actually equivalent to the Archimedean Property.)
A: Suppose that $a>0$. Then by the Archimedean Property, there exists a natural number $n$ so that $1/n<a$. But this is a contradiction. Then $a\leq0$ if $a\leq 1/n$ for all $n$.
A: Assume $a>0$,then there exists some $\epsilon$ such that $0<\epsilon<a$.
Let $N=[\epsilon]+1$,then $\frac{1}{N}<\epsilon<a$, which contradicts to that for every $n∈N, a≤1/n$.
So, $a\leq0$.[Q.E.D]
Above is implied in @Daniel's and @lhf's answers.
A: $n$ 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
$$
\mathop {\lim }\limits_{n \to \infty } a \le \mathop {\lim }\limits_{n \to \infty } \frac{1}{n} = 0
$$
$$
a \le 0
$$
$$Q.E.D.$$
