Obviously we can use the completeness of the real numbers (least upper bound axiom, or one of the equivalent principles) to prove the IVT. Can we go in the opposite direction?

This isn't a homework problem or something. I'm just wondering. If the answer is "yes", then I'm not really asking for much explanation. A reference, or a place to look if I'm stuck, will do.

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    $\begingroup$ Highly Highly recommend you the book The Real Numbers and Real Analysis by Ethan D. Bloch(published in 2011). It has the best introduction on these topic than others(Rudin, Wade, Apostol, ...etc). $\endgroup$
    – Eric
    Aug 10, 2017 at 4:51

2 Answers 2


Noah's answer is excellent but makes things a bit more difficult than necessary since he is proving that $F$ is isomorphic to $\mathbb{R}$, rather than merely that $F$ is complete (and so he is basically also reproducing the proof that every complete ordered field is isomorphic to $\mathbb{R}$). Here is a quick direct proof that if an ordered field $F$ satisfies the intermediate value theorem, then it is Dedekind-complete.

Suppose $X\subset F$ is a nonempty set that is bounded above but has no least upper bound. Define a function $f:F\to F$ by $f(x)=1$ if $x$ is an upper bound of $X$ and $f(x)=0$ if $x$ is not an upper bound of $X$. Let $a\in X$ and let $b$ be an upper bound for $X$. Then $a-1<b$, $f(a-1)=0$ and $f(b)=1$. But there is no $c$ between $a-1$ and $b$ such that $f(c)=1/2$. So, assuming $f$ is continuous, this violates the intermediate value theorem for $F$.

It thus remains only to show that $f$ is continuous. To show this, it suffices to show that for any $x$, there is an open interval $(c,d)$ containing $x$ such that $f(y)=f(x)$ for all $y\in (c,d)$. First suppose $f(x)=0$. Then $x$ is not an upper bound for $X$, so there is some $d\in X$ such that $x<d$. We then have $f(y)=0$ for all $y\in (-\infty,d)$. Now suppose $f(x)=1$. Since $x$ is not the least upper bound of $X$, there is some $c<x$ such that $c$ is also an upper bound of $X$. We then have $f(y)=1$ for all $y\in (c,\infty)$.

(Note that in fact this argument does not really use the field structure in any essential way, and with some minor modifications it proves that any totally ordered set with more than two points which satisfies the intermediate value theorem is Dedekind-complete.)

  • $\begingroup$ In proving IVT in the Field of Reals we use two properties: 1. LUB 2. Local Property of Continuous function: f continuous at x, f(x)>0 then f is positive in a nbd. of x. So in a general ordered field if one assumes LUB do we have property 2? Well in that case do we generally take order topology in general ordered fields to bring the notion of continuity to its functions. $\endgroup$
    – Saikat
    Jul 25, 2020 at 2:32
  • $\begingroup$ Yes, when I talk about $f$ being continuous, I mean with respect to the order topology on $F$. $\endgroup$ Jul 25, 2020 at 2:55

Yes, the intermediate value theorem implies completeness - in fact, the following is true:

Suppose $F$ is an ordered field satisfying IVT, in the sense that for any continuous (with respect to the order topology on $F$) function $s:F\rightarrow F$ and any $a<b$, if $s(a)<y<s(b)$ then we can find some $x\in(a, b)$ with $s(x)=y$ (there are other ways we might phrase IVT but they'll wind up being equivalent). Then $F$ is isomorphic to the field of real numbers.

One direction of course is immediate. For the other direction, we can argue as follows. I make no claim that this is the best argument, but it's the one that flows most naturally for me, and hopefully you find it helpful.

  • Suppose $F$ were not Archimedean. Consider the standard embedding $e: \mathbb{Q}\rightarrow F$. Let $A$ be the set of elements of $F$ less than some element in the range of $e$, and $B=F\setminus A$ (that is, $B$ consists of the "infinitely large" elements of $F$). Then the characteristic function of $A$ is continuous, but violates the IVT (it goes from $1$ to $0$ without ever crossing ${1\over 2}$).

  • Now, any Archimedean ordered field embeds into the reals, by building on the map $e$ above. Namely, to each Dedekind cut $C=(D, U)$, we may assign the set $C^F$ of elements of $F$ bigger than or equal to $e(d)$ for each $d\in D$ but smaller than $e(u)$ for each $u\in U$. If $a. b\in C^F$ are distinct, then ${1\over b-a}$ demonstrates that $F$ is non-Archimedean; so each $C^F$ has at most one element, and conversely since $F$ is Archimedean each element of $F$ lives in exactly one $C^F$, so we get an embedding $i$ of $F$ into $\mathbb{R}$.

  • Finally, we claim that $i$ is surjective (hence a bijection). This goes back to the first bullet: supposing $\alpha\in\mathbb{R}\setminus \operatorname{ran}(i)$, we can partition $F$ into $\{a\in F: i(a)<\alpha\}$ and $\{a\in F: i(a)>\alpha\}$; each is open, so the characteristic function of either contradicts the claim that $F$ satisfies IVT.

(I guess I've stated without proof that $i$ is in fact an embedding - that is, preserves the ordered field structure - but that's not hard to show so I'll leave it as an exercise.)

Here's a reference I think you'll find very cool - this article by James Propp. It's also worth gesturing at reverse mathematics in general as an approach to this kind of question (although for various reasons this is explicitly not the framework Propp looks at).

  • $\begingroup$ This asssumes you're talking about a field and not just a linearly ordered set. There's still the question of whether it's true of linearly ordered sets generally. $\endgroup$ Aug 10, 2017 at 4:08

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