# Is the IVT equivalent to completeness?

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.

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, $f(a-1)=0$ 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. 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 such that $c$ is also an upper bound of $X$. We then have $f(y)=1$ for all $y\in (c,\infty)$.