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Proposition. Complete ordered fields are Archimedean.

Proof. Let $\mathbb{F}$ be a complete ordered field and consider $x\in\mathbb{F}$. We must show that there is an integer $N$ such that $x<N.$ If not, the monotone sequence $1,2,3,\ldots$ is bounded above by $x$ and converges by the monotone sequence property. We assert that this sequence cannot converge to a number, say $y$. If it did, for any $\varepsilon >0$ there would be an $N$ such that for $n\geq N$, $$1=|n+1-n|\leq |n+1-y|+|y-n|\leq 2\varepsilon$$ by the triangle inequality.This gives a contradiction if $\varepsilon <\frac{1}{2}.\square$

This proof for the proposition stated is from Marsden's Elementary Classical Analysis. My question here is that in this proof, we assume $\mathbb{F}$ to be a complete ordered field (which also means that $\mathbb{F}$ follows Monotone Sequence Property. Marsden's book defines completeness through MSP).

However, it seems to me that this proof additionally puts another implicit assumption that the integer number field $\mathbb{Z}$ also obeys Monotone Sequence Property.

Am I right about this implicit assumption, or did the author make no additional assumptions in regards to $\mathbb{Z}$?

Wooyong Park
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1 Answers1

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Since $\Bbb Z\subset\Bbb R$ and since it is being assumed that every bounded monotonic of real numbers converges, then, in particular, every bounded monotonic of integer numbers converges (to a real number). No extra assumption is needed.

José Carlos Santos
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  • I'm sorry but in which part is this part assuming the convergence of bounded monotonic real number sequence? I mean, the proof is assuming 's completeness, but it didn't say anything about ℝ. Would it be that since ℝ's completeness is an obvious fact, the author didn't mention it? – Wooyong Park Mar 05 '24 at 12:31
  • And if that's true, in which part is 's completeness being utilized? – Wooyong Park Mar 05 '24 at 12:32
  • I missed the fact that you are working here with an arbitrary complete ordered field, instead of the reals. But that changes nothing, since you can identify each $n\in\Bbb N$ with the element $1+1+\cdots+1$ ($n$ times) of $\Bbb F$. So, the sequence $1,2,3,\cdots$ can be seen as a monotonic sequence of elements of $\Bbb F$. – José Carlos Santos Mar 05 '24 at 13:33
  • Oh, thanks. That makes whole lot sense. – Wooyong Park Mar 09 '24 at 14:26