Converse Of Intermediate Value Theorem

A function $f: [a,b] \to \mathbb{R}$ is said to have the intermediate value property (IVP) if, for all $$x\in \Big[\min\big(f(a), f(b)\big), ~\max\big(f(a), f(b)\big)\Big],$$ there exists $c\in [a,b]$ such that $f(c) = x$. In other words, $f$ has the IVP if it attains every value between $f(a)$ and $f(b)$ at some point in the interval $[a,b]$.

The intermediate value theorem states that if $f$ is continuous, then $f$ has the IVP. This fact leads to the question of whether or not there exist noncontinuous functions that have the IVP. How could one exhibit such a function? This article will settle the question by constructing a collection of noncontinuous functions $\mathbb{R} \to \mathbb{R}$ that have the IVP.

Let $f: \mathbb{R} \to \mathbb{R}$ be a function. One calls $f$ strongly surjective if for any interval $[a,b] \subset \mathbb{R}$, the image of this interval under $f$ is $\mathbb{R}$. In symbols, $f([a,b]) = \mathbb{R}$. $_\square$

Any strongly surjective function certainly satisfies the IVP on any subinterval of the real line. Thus, it suffices to construct a strongly surjective function. These functions are always, by the following argument, not continuous at any point in $\mathbb{R}$.

Let $f: \mathbb{R} \to \mathbb{R}$ be a strongly surjective function. Suppose $f$ is continuous at $x\in \mathbb{R}$. By the epsilon-delta formulation of continuity, this means that for any $\epsilon > 0$, there exists $\delta > 0$ such that $y \in (x-\delta, x+\delta)$ implies $f(y) \in (f(x) - \epsilon, f(x) + \epsilon)$. But by strong surjectivity, $$(f(x) - \epsilon, f(x) + \epsilon) \supset f((x-\delta, x+\delta)) \supset f([x-\delta/2, x+\delta/2]) = \mathbb{R}.$$ This is impossible unless $\epsilon = \infty$, which is absurd, as $\epsilon$ is a finite real number. By contradiction, we conclude $f$ is not continuous at $x$. $_\square$

In modular arithmetic, two integers $x$ and $y$ are considered to be equivalent modulo $n$ if $x$ and $y$ give the same remainder upon division by $n$. This is the same as requiring $x-y$ is a multiple of $n$. If $n\mathbb{Z}$ denotes the set of multiples of $n$, then this may be again reformulated as the condition that $x-y\in n\mathbb{Z}$.

One can do something similar in a more general setting. Let $S$ be an ideal of $\mathbb{R}$. Declare $x, y \in \mathbb{R}$ to be equivalent if $x-y\in S$. This is an equivalence relation, and the set of equivalence classes obtained in this way is denoted $\mathbb{R}/S$.

If $S = \mathbb{Q}$, then one obtains the set $\mathbb{R}/\mathbb{Q}$. This set comes equipped with a natural surjective map $p: \mathbb{R} \to \mathbb{R}/\mathbb{Q}$ which sends each real number to its equivalence class in $\mathbb{R}/\mathbb{Q}$. In fact, $p$ has a property stronger than surjectivity; in fact $p(I) = \mathbb{R}/\mathbb{Q}$ for any interval $I \subset \mathbb{R}$.

Let $I = [a,b]$ (if $I$ is not a closed interval, replace it with a closed subinterval of itself). For any $x\in \mathbb{R}$, let $q$ be a rational number such that $a-x \le q \le b-x$. Then $x+q \in [a,b]$ and $p(x+q) = p(x)$, so any element of $\mathbb{R}/\mathbb{Q}$ is mapped to by some number in $[a,b]$. $_\square$

Since the projection $p: \mathbb{R} \to \mathbb{R}/\mathbb{Q}$ is surjective when restricted to any subinterval of $\mathbb{R}$, to construct a strongly surjective function, it would suffice to compose $p$ with a surjection $g: \mathbb{R}/\mathbb{Q} \to \mathbb{R}$. Does such a surjection exist?

For a set $S$, denote by $|S|$ its cardinal number. By Lagrange's theorem, one knows $$|\mathbb{R}| = |\mathbb{Q}| \cdot |\mathbb{R}/\mathbb{Q}| = \max( |\mathbb{Q}| , |\mathbb{R}/\mathbb{Q}|).$$ There is an injection $\mathbb{N} \to \mathbb{R}/\mathbb{Q}$ given by $n \mapsto \{\text{equivalence class of } n \pi\}$. Thus, it follows that $|\mathbb{R}/\mathbb{Q}| \ge |\mathbb{N}| = |\mathbb{Q}|$, and hence $|\mathbb{R}| = | \mathbb{R}/\mathbb{Q}|$. Let $g: \mathbb{R}/\mathbb{Q} \to \mathbb{R}$ be a bijection.

Then, $g\circ p: \mathbb{R} \to \mathbb{R}$ is a strongly surjective function. In fact, since this will hold for any bijection $g$, the construction yields a family of strongly surjective functions, one for each such bijection.