Show that the set {1/6, 1/7 , 1/8,.....} does not have a least element

Question

Show that the set $\{\frac 16,\frac 17 ,\frac 18,\dots\}$ does not have a least element and conclude that no set containing this set is well ordered.

I am not sure how can I show this ... The set is infinite and strictly decreasing, and I know that a well-ordered set is a set that has the property that every nonempty subset has a least element...

Answer 1

Note that $$\left\{\frac16,\frac17,\frac18,\ldots\right\} = \{a_1,a_2,a_3,\ldots\}$$ where $a_1 > a_2 > a_3 > \cdots$. Can you show that for each $k$ it holds that $a_k$ is not the smallest element of the set? It would follow that the set does not have a least element.

Answer 2

Suppose the set had a least element. It would be of the form $\frac{1}{n_0}$ for some $n_0 \in \mathbb{N}$. Can you find another element in the set that is even smaller than $\frac{1}{n_0}$?

And what is the definition of a well-ordered set?