Does the Intermediate Value Theorem for uniformly continuous functions imply the Least Upper Bound Property?

Question

Proofs that the Intermediate Value Theorem (IVF) implies the Least Upper Bound Property for an ordered field usually use a continuous function that is not uniformly continuous like here https://math.stackexchange.com/a/2388654/539499. Is IVF for uniformly continuous functions enough to prove the Least Upper Bound Property?

Let $F$ be an ordered field. $F_{>0}$ is the set of all positive elements of $F$. IVF for uniformly continuous functions is: "If $a,b,L \in F$, $a<b$, $f: [a,b] \to F$ is uniformly continuous and $f(a)<L<f(b)$, then there exists $c \in (a,b)$ such that $f(c)=L$"

A function $f: [a,b] \to F$ is uniformly continuous iff for all $\epsilon \in F_{>0}$, there exists $\delta\in F_{>0}$ such that for all $x,y \in [a,b]$, $|x-y| < \delta \implies |f(x)-f(y)|< \epsilon$.

We can't assume that a continuous function on $[a,b]$ is uniformly continuous as well because that that is not true for all ordered fields. It is false in $\mathbb{Q}$ for example.

Answer 1

This is a nice question! I believe the answer is yes, this property still implies the LUBP. I will refer to your property as the Uniform Intermediate Value Property.

The proof strategy is as follows. First, argue that UIVP implies $F$ is Archimedian. Therefore, $F$ is isomorphic to some subfield of $\Bbb R$, and it's complete if and only if that subfield is $\Bbb R$. We'll show this is the case, by building a uniformly continuous function without the IVP for all the proper subfields. (I think it's probably not essential to quote that Archimedian ordered fields embed in $\Bbb R$, but it makes this proof easier to write down)

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So firstly, suppose $F$ is not Archimedian. Let $i: \Bbb Q \to F$ be the unique embedding of $\Bbb Q$ into $F$. Let $X$ be the smallest interval in $F$ containing $i(\Bbb Q)$, which is to say $x \in X$ iff there are some $p, q \in \Bbb Q$ such that $i(p) \le x \le i(q)$. This set $X$ is the subset of $F$ of "finite-sized" elements. Since $F$ is non-Archimedian, $X$ is a proper subset of $F$.

Consider the function $g: F \to F$ defined by $g(x) = 1$ for $x \in X$ and $g(x) = 0$ for $x \notin X$. This function is uniformly continuous, for example by taking $\delta = 1$. This is a very strong form of uniform continuity, as $\delta$ is even independent of $\epsilon$!

But clearly $g$ does not have the IVP.

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So now let $F$ be some proper subfield of $\Bbb R$.

The following construction is inspired by a construction to prove that $\Bbb Q$ is order-isomorphic to $\Bbb Q \setminus \{0\}$. The construction is as follows. Firstly, pick some number $\alpha \in \Bbb R \setminus F$.

We'll define a continuous function $f: \Bbb R \to \Bbb R$, such that $f(F) \subseteq F$ and $f(\alpha) = 0$. Morally, we're building a function that sends the "hole" at $\alpha$ in $F$ to a hole at $0$.

Now let $a_n$ be an increasing sequence of elements of $F$ with $a_n \to \alpha$. Let $b_n$ be a decreasing sequence of elements of $F$ with $b_n \to \alpha$. Define $f: \Bbb R \to \Bbb R$ by $f(\alpha) = 0$, $f(a_n) = -\tfrac 1n$, $f(b_n) = \tfrac 1n$, and $f$ is piecewise linear on each interval $[a_n, a_{n + 1}]$ and $[b_{n + 1}, b_n]$. Take $f$ to be constant on $(-\infty, a_1]$ and $[b_1, \infty)$.

Clearly $f: \Bbb R \to \Bbb R$ is continuous. Since $[a_1, b_1]$ is compact, $f$ is uniformly continuous on that interval. But outside that interval, $f$ is constant, so $f$ is uniformly continuous even as a function $\Bbb R \to \Bbb R$, which certainly implies that its restriction to $F$ is uniformly continuous.

We have $f(F) \subseteq F$, because $1/n \in F$ for all $n$, and "linear interpolation in an interval with endpoints in $F$, between values in $F$" requires only field operations.

Hence, $f|_F: F \to F$ is uniformly continuous. But clearly $0$ is not in the image of $f|_F$, so $f$ does not have the IVP. So we are done.

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I believe that if you take a bit more care in the construction of $a_n$ and $b_n$, you can even make $f$ be Lipschitz, for example. I think for instance asking that $a_n \in (\alpha - \tfrac 1n - \tfrac 1{n^3}, \alpha - \tfrac 1n + \tfrac 1{n^3})$ and something similar for $b_n$ should do the job.

Answer 2

Yes, the IVT for uniformly continuous functions (hereafter abbreviated UIVT) is enough to imply the least upper bound property.

To see this, first observe that the same example function used at your link can be used to show that the UIVT implies the Archimedean property. That is, if the Archimedean property fails, then let $X=\{x\in F\mid \exists n\in \mathbb N (x\leq n)\}$. Then the indicator function $\chi_X$ is uniformly continuous, yet takes on values $0$ and $1$ and nothing else, contradicting UIVT.

Having established the Archimedean property, suppose $F$ has UIVT but fails the LUB property. We lose no generality supposing it fails the LUB property for an initial segment $X\subset F$, i.e., $x\in X$ and $y<x$ implies $y\in X$, and $X$ has no least upper bound, i.e., $Y:=F\backslash X$ has no least element, and $X$ has no greatest element. Take $0\in X$ with no loss of generality.

From the Archimedean property for each $n$ there are points $x_n,y_n$, with $0\leq x_n\in X$, $y_n\in Y$, and $|y_n-x_n|<\frac{1}{n}$, (otherwise starting from $0$ and increasing by $\frac{1}{n}$ gives a contradiction), and we may take $x_n<x_{n+1}$, $y_n>y_{n+1}$ (we may do this by forcing $|y_n-x_n|$ to be smaller than $\frac{1}{m}$ for $m$ large enough so that $y_{n-1}-\frac{1}{m}\in Y$ and $x_{n-1}+\frac{1}{m}\in X$).

Now define $f(x_n)=1-\frac{1}{n}$, $f(y_n)=1+\frac{1}{n}$, and extend $f$ linearly between consecutive $x_n$'s, and likewise between consecutive $y_n$'s, and let $f(x)=f(x_1)=0$ for $x<x_1$, and $f(x)=f(y_1)=2$ for $x>y_1$.

Then $f$ is uniformly continuous - to see this, note that it is certainly uniformly continuous on $F\backslash [x_n,y_n]$ for each $n$ (this is an easy exercise using the maximum slope to determine a modulus of continuity).

Therefore if $f$ fails to be uniformly continuous in general, then we have sequences $p_n$, $q_n$ with $|p_n-q_n|\to 0$, and $$|f(p_n)-f(q_n)|>\epsilon,\tag{1}$$ which can only happen if $p_n$ and $q_n$ are eventually in each interval $[x_N,y_N]$*, but then we have $|f(p_n)-f(q_n)|\leq \frac{2}{N}$ eventually for each $N$, which for large enough $N$ contradicts (1) (by the Archimedean property).

Thus $f$ is indeed uniformly continuous. Since $f$ never takes the value of $1$, the UIVT property is contradicted.

*Remark

An additional clarification of the argument here: The precise argument is that for a given $N$, eventually, we must always have either $p_n$ or $q_n$ in $[x_{N+1},y_{N+1}]$, which in turn means that when $|p_n-q_n|$ is sufficiently small, both $p_n$ and $q_n$ lie in $[x_N,y_N]$.