Prove that the set $X=\{1/n : n \in \mathbb{N}\}$ has no minimum.

Question

I am not sure how to go about the proof. It makes sense that the sense has no minimum because each element in the set is becoming smaller and smaller than the one preceding it. However, I am not sure how to write it is as proof. I know I am supposed to use a proof by contradiction and refer to the definition of minimum, but I am lost. Please help.

Answer 1

Let $1/n_0$ be the smallest element of $X$, then $1/(n_0+1)$ is in $X$.

Answer 2

A minimum is a number $x\in X$ that is less than or equal to any every element of $X$. So we must prove there is no such number. The way we do that is to consider an arbitrary $a\in X$ and prove that it is not a minimum. To prove $a$ is not a minimum, it suffices to find an element $b\in X$ such that $b<a.$ This means we must find an $a\in\mathbb N$ such that $1/n <a.$ This is the same as $n > 1/a$ (since $a\in X$ and all elements of $X$ are positive) so we must show there exists a natural number $n>1/a.$ Can you take it from here?

Answer 3

The important point in this problem is that the minimum must belong to the set and no other element of the set is smaller than the minimum.

Assume that the set $\{ 1/n\}_{n=1}^{\infty} $ has a minimum say $1/k$ for some$ k\in \mathbb N$.

Note that $1/(k+1)$ is in the set and it is less than $ 1/k$.

That is not possible because $1/k$ is the smallest element.

Thus the set does not have a minimum.