# Proving these sets are well-ordered with the order inherited from $\mathbb{R}$.

These questions arise in my Set Theory module but I don't understand how to go about deciding if the following sets, with order inherited from the reals, are well-ordered or not. $$\textbf{I would only like a hint or some intuition to help me along.}$$

$$A=\{\frac{1}{m}- \frac{1}{n} : n,m \in\mathbb{N}\}$$ $$B=\{-2^{-m}-3^{-n} : n,m \in\mathbb{N}\}$$

I argue that $$A$$ is not well-ordered. The generic element of $$A$$, $$\frac{1}{m}- \frac{1}{n}$$ for some $$n,m \in\mathbb{N}$$. We obtain that $$-1 < \frac{1}{m}- \frac{1}{n} < 1$$. In particular, since the value of this element will "converge" to -1 as m tends to infinity, we cannot pick a smallest element of this set. Hence we cannot pick a smallest element for every subset of it, and so it cannot be well-ordered.

For $$B$$, I am stuck: The whole set has a least element. However this does not tell us about the least element of any subset of it, so this does not help us. We could check if there is an infinite strictly decreasing sequence in this set. I tried taking the subsequence of the even or odd terms in this, but this leads to increasing sequences. I need some help with this one.

Thanks

Your argument for $$A$$ is good, but you could make your proof a little more rigorous and clear. Explicitly show there is no smallest element.
For $$B$$, suppose there were an infinite decreasing sequence, and prove a contradiction, this will give you well-ordered. Since $$\mathbb{N}$$ is well ordered and $$-2^{-n_j}-3^{-m_j}$$ decreasing has something to do with $$n_j, m_j$$ decreasing, you should be able to do this.
• Hi again, I've had a look at this and I sort of understand. Because it must be that $n_j$ and $m_j$ are non-decreasing, then the sequence $(-2^{-n_j}-3^{-m_j})$ cannot be decreasing, hence there cannot exist an infinite strictly decreasing sequence. But why can't $n_j$ and $m_j$ be decreasing? Commented May 3, 2020 at 15:24