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Let $m$ be the upper bound of $A$. Then, $a \leq m$ for all $a\in A $.

Now I should maybe find a way to show that that inequality is in fact an equality. Or maybe there's an easier way that I don't see.

Thank you in advance.

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Hint: what can you say about $m + m$?

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  • $\begingroup$ Since the rng has to be closed under addition, $m+m \in A$, but $m$ is an upper bound, so $m+m \leq m$. That implies $m \leq 0$. And that implies that 0 is the only element, maybe, because is $m$ is the upper bound and $m \leq 0$, any other element $a$ is such that $a \leq 0$, but in that case, $-a \geq 0$, which cannot be, since that means $a \geq m$ $\endgroup$ – dumb_undergrad Jun 6 '17 at 22:25
  • $\begingroup$ But wait, $m$ isn't necessarily in $A$. $\endgroup$ – dumb_undergrad Jun 7 '17 at 15:53
  • $\begingroup$ If it isn't in $A$, where is it? I think you'll find the question is about the ordered rng $A$ and not about $A$ extended with additional elements that aren't in the domains of the rng operations. (It is possible to set up theories of systems that could involve a rng and an ordered set containing the elements of the rng and other things as well, but I doubt that you are being asked to solve this problem in a context like that.) $\endgroup$ – Rob Arthan Jun 7 '17 at 19:30
  • $\begingroup$ Well, upper or lower bound in a ring means upper or lower bound in the set, right? So, a set $A$ can have a upper bound $m$ and $m \not\in A$. For example, $B = ]-x,x[$ has $x$ as a upper bound, but $x \not\in B$ and it doesn't have a biggest element. I know that set can't be a ring, but you get the idea. $\endgroup$ – dumb_undergrad Jun 7 '17 at 20:06
  • $\begingroup$ You are considering a relation between what is technically called the universe (or carrier set) of the ordered ring and a larger ordered structure that might contain an upper or lower bound for the unverse of the ring. We certainly can make the open interval $B=]-x, x[$ into a ring (by identifying it with $\Bbb{R} = ]-\infty, +\infty[$ under a suitable order-preserving mapping), but $x$ and $-x$ are not elements of the universe of that ring. (Aside: your comments have helped me to understand what "overthinking" means. Thank you.) $\endgroup$ – Rob Arthan Jun 7 '17 at 20:24

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