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What properties of "numbers" are used to assert that for given numbers $a$ and $b$, $a≠b$, there exists a number $x$ such that $a < x < b$ ?

In the texts I've read, this seems to be assumed without explanation in discussions that are otherwise quite careful about such things.

For example, in Spivak's Calculus (4E, p. 123) this fact is used to demonstrate that the function $f(x) = x^2$ does not take on its maximum on the interval (0,1) because for any $0 ≤ y < 1$ there is an $x$ such that $y < x < 1$. Up to that point, the only properties of "numbers" (whatever they may be) that have been defined are those of an ordered field, and it is not clear to me that this property can be derived from those.

I gather this amounts to the "numbers" in question having "dense order". (Correct me if I'm wrong.) But I'm unclear where that comes from.

Also note: up to the point where this question arises, there has been no mention of what "numbers" are (i.e., whether they are $\mathbb Q$ or $\mathbb R$), only that they have the properties of an ordered field.

2 Answers 2

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It is easy to show that in any ordered field if $a < b$, then $a < \frac{a + b}{2} < b$. In any ordered field 2 is invertible, so this intermediate value exists.

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    @Arturo: tha$n$ks agai$n$; very clear a$n$d helpful--as usual.2011-09-09
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If we just assume connectedness of the reals, if there were no numbers between y and 1, we would be lead to the contradiction that we could then express $\mathbb R= (-\infty,y]\cup[1,\infty)$, i.e., $\mathbb R$ is the union of two sets with disjoint closure, i.e., we get a disconnection of $\mathbb R$, as the union of two non-disjoint sets, or as the union of non-empty separated sets.

Alternatively,using field properties, if we assume $\frac {1}{2}$ is in $\mathbb R$ ($1$ is in $\mathbb R$, then so is $1+1$, then so is $(1+1)^{-1})$, then so is $\frac{1}{2^n}$ for all natural n. By the Archimedean property , there is some integer $m$ such that $\frac{1}{2^m}<1-y$ by field closure properties then, $y+\frac {1}{2^m}$ is a real number, but $y+\frac{1}{2^m}<1$.

EDIT: 3rd paragraph removed, since I have no good proof for it.

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    @Jonas: I had an explanation for this. I will delete this part until I have time to write a better explanation.2011-10-24