The concepts of monotonicity and continuity are very different. However, if we consider a monotonic function whose domain is an **interval**, we can deduce that right and left limits exist at each point in the domain (even though they may be different, of course). This is a consequence of the **Completeness** principle, which implies that every upper (lower) bounded set has a supremum (infimum).

We can therefore demonstrate a Theorem which is somehow close to the Intermediate Value Theorem, albeit “weaker”.

**Theorem. Let \(f: [0,1] \to [0,1]\) a monotonically increasing function. Then there exists a point \(\alpha \in [0,1]\) such that \(f(\alpha) = \alpha\).**

The proof, too, follows the same approach that can be used to demonstrate the I.V.T.

If \(f(0) = 0\) or \(f(1) = 1\), then we have found \(\alpha\). Conversely we can define a procedure that either will stop once the required point is found, or will allow us to construct two *sequences converging* to it. Here’s how.

Let \(a_0=0\) and \(b_0=1\). Then for \(n=1,2,3,\dots\) proceed as follows.

- Determine \(x_n = \frac{a_{n-1}+b_{n-1}}{2}\);
- If \(f(x_n) = x_n\) then \(\alpha = x_n\) and we are done;
- If \(f(x_n) > x_n\) then let \(a_n = x_n\) and \(b_n = b_{n-1}\);
- If \(f(x_n) < x_n\) then let \(a_n = a_{n-1}\) and \(b_n = x_n\).

Let us verify that if the search does not stop, then the sequences \((a_n)\) and \((b_n)\) converge to the desired value. First of all, the sequences do converge, since they are monotonic and bounded (here Completeness comes into play). Secondarily, since

\[a_n-b_n = \frac{1}{2^n},\]

they converge to the same value, say \(\overline x\).

Furthermore we have

\begin{equation}f(a_n)>a_n\tag{1}\label{eq656:a}\end{equation}

and

\begin{equation}f(b_n)< b_n\tag{2}\label{eq656:b}\end{equation}

for all \(n\).

Suppose now \(f(\overline{x}) < \overline x\). Then, because \((a_n)\) converges to \(\overline x\), for large enough \(n\) we should have

\[f(\overline x) < a_n < \overline x,\]

But, this, together with equation \eqref{eq656:a} gives \(f(a_n) > f(\overline x)\), contradicting monotonicity of \(f\). Analogously, if we assume \(f(\overline{x}) > \overline x\), then we get, for large enough \(n\),

\[\overline x < b_n < f(\overline x),\]

yielding, together with \eqref{eq656:b}, \(f(b_n) < f(\overline x)\), i.e. again a contradiction. Thus it must be \(f(\overline x) = \overline x\) and \(\overline x = \alpha\).

It shoud be observed that Completeness is in fact *necessary* for the Theorem to hold. As a counterexample consider the function \(f: [0,1]\cap \Bbb Q \to [0,1]\cap \Bbb Q\) defined as

\[f(x) = \frac{1}{3-x},\]

which is monotonic and even continuous in its domain. However there is no \(\alpha\) such that

\[\frac{1}{3-\alpha} = \alpha,\]

because then \(\alpha\) would be a solution to the quadratic equation

\[\alpha^2-3\alpha + 1=0\]

which we know does not have solutions in \(\Bbb Q\).

As a final exercise you can try to demonstrate again the Theorem after *replacing “monotonically increasing” with “continuous”* in the hypothesis. *(Hint: consider using the auxiliary function \(F(x) = f(x)-x\).)*