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.