0

Show that any upper bounded subset of $\mathbb{Z}$ has a unique maximal element

This is what I've managed to do so far.

Let $X \subseteq \mathbb{Z}$ be a nonempty upper bounded nonempty subset of $\mathbb{Z}$. Since $\mathbb{Z} \subseteq \mathbb{R}$ we have that $X$ is upper bounded in $\mathbb{R}$ hence $\sup X$ exists. Let $\alpha = \sup X$. We claim $\alpha = \max X$.

This is where I get stuck, ultimately I have to show that $\alpha \in X$, and I'm not exactly sure how to go about doing that.

Perturbative
  • 13,656
  • See also here. – Dietrich Burde Apr 01 '18 at 18:20
  • @DietrichBurde The second answer there uses compactness to prove this, don't you think that's really unecessary and overpowered machinery to prove this? You essentially have to develop some basic notions of topology and invoke metric spaces to do that – Perturbative Apr 01 '18 at 18:26
  • Compactness is a basic notion of topology, I would say, but I cannot find the word "compact" in the answers of the duplicate. So I hope this will be helpful for you. – Dietrich Burde Apr 01 '18 at 18:31
  • @DietrichBurde The answers in the linked duplicate aren't very satisfactory, I cannot follow the arguments made in the first answer for example – Perturbative Apr 01 '18 at 19:25

1 Answers1

1

Since $\alpha$ is a least upper bound, for any $\epsilon > 0$, there is an $n \in X$ such that $\alpha - \epsilon < n \leq \alpha$. This follows from the definition of a least upper bound.

If $\alpha \not\in X$, we have two cases. If $\alpha$ is an integer, then we immediately get a contradiction using the previous paragraph, since there is no integer between $\alpha - 1/2$ and $\alpha$. If $\alpha$ is not an integer, take $\lfloor \alpha\rfloor$ and consider $\epsilon = \alpha - \lfloor\alpha\rfloor > 0$, and get a similar contradiction, using the first paragraph.

Chris
  • 5,231