1
$\begingroup$

Consider an ordered field $S$ with $+,.,<,>,=$ not necessarily having their customary definitions. Let $ E \subset S, E \neq \varnothing$ be bounded above. Then $ \forall \epsilon > 0$, does $\exists \alpha \in E$ such that $\alpha + \epsilon \notin E $ ?

If the answer is in affirmative, how does one prove it? Otherwise, can a counter example be provided?

If the statement is indeed untrue, does it hold true for the specific cases of $S \equiv R$ and $S \equiv Q$ with $+,.,<,>,=$ having their customary meaning?

  • 0
    By $\phi$ do you mean, $\varnothing$, the empty set? – 2017-01-23
  • 0
    Yes. I have edited the post. – 2017-01-23

1 Answers 1

1

This fails in non-Archimedean ordered fields. Indeed, suppose $S$ is non-Archimedean, and let $x,y \in S$ be such that $x>0$ is infinitesimal with respect to $y$. This means that $nx < y$ for all $n \in \mathbb{N}$, where $$nx = \underbrace{x+x+\cdots+x}_{n \text{ times}}$$ Then the set $E = \{ nx \mid n \in \mathbb{N} \} \subseteq S$ is bounded above, since $y$ is an upper bound for $E$, but it does not satisfy the desired property since taking $\varepsilon = x$ we see that $\alpha + \varepsilon \in E$ for all $\alpha \in E$.

The property does hold in Archimedean ordered fields such as $\mathbb{R}$ or $\mathbb{Q}$, since if there were to exist $\varepsilon > 0$ such that $\alpha + \varepsilon \in E$ for all $\alpha \in E$, then we'd have $\alpha + n\varepsilon \in E$ for all $n \in \mathbb{N}$; the Archimedean property would then imply that $E$ is unbounded, contrary to assumption.

So in summary, the condition on $S$ that you state is equivalent to $S$ being Archimedean.

  • 0
    Thank you so much. You have been great help. – 2017-01-24