Prove the existence of $n\in\mathbb{N}$ such that $0<1/n<\epsilon$

Question

I'm seeking to prove that for any $\epsilon>0$, there exists an $n\in\mathbb{N}$ such that $$ 0<\frac{1}{n}<\epsilon. $$

I think that—in some form—I'm going to need to use the fact that for any $x\in\mathbb{R}$, we can find $a\in\mathbb{N}$ such that $a\geq x$ (i.e. the Archimedean property).

Or perhaps I could construct a set with an infimum of zero and somehow use that to generate an element between zero and $\epsilon$.

What's the most sensible approach here?

Answer 1

Note: that $0 < \frac 1n < \epsilon \iff n > \frac 1\epsilon > 0$

So you have to show:

1. If $\epsilon > 0$ then $\frac 1{\epsilon}$ exists (that's easy; $\mathbb R$ is a field so multiplicative inverses exist).
2. If $\epsilon > 0$ then $\frac 1{\epsilon} > 0$. (You should have proven that already and can take that as a given without even acknowledging it is in question.)[The proof is If $\frac 1\epsilon \le 0$ and $\epsilon > 0$ we would have $1 =\frac 1\epsilon \cdot \epsilon \le 0\cdot \epsilon = 0$; a contradiction].
3. For any $M = \frac 1{\epsilon}\in \mathbb R$, there exists a natural number $n: n> M$.

And 3: (which says nothing more or less than "The Natural numbers are not bounded above") is the real heart of the issue.

Depending on the level of the book you may or may not be able to take that for granted.

But it is equivalent to the Archimedian principal. $1 > 0$ so for any $M =\frac 1{\epsilon} \in \mathbb R$ there is an $n\in \mathbb N$ so that $n*1 > M$.

.....

To sum up:

$\epsilon > 0$.

So $\frac 1{\epsilon} > 0$.

So there exists an $n\in \mathbb N$ so that $n > \frac 1\epsilon$.

So $0 < \frac 1n < \epsilon$.

That's all you have to do.

Answer 2

Consider $n$ equal to $\lceil\frac{1}{\epsilon} \rceil$ (that means the smallest natural number that is greater or equal to $\frac{1}{\epsilon}$). Then you can easily see that $$0<\frac{1}{n} = \frac{1}{\lceil\frac{1}{\epsilon}\rceil} <\frac{1}{\frac{1}{\epsilon}}=\epsilon,$$ so for any $\epsilon\in\mathbb{R}$, you can just pick $n\in\mathbb{N}$ that way and the condition is verified (for that $n$ and for any other natural number grater that $n$).

Answer 3

You have almost said it yourself if you accept the Archimedian property: Let $\epsilon > 0$. Then there exists $n \in \mathbb{N}$ such that

$$n > \frac{1}{\epsilon} \implies \epsilon > \frac{1}{n}.$$