Past exam of the mathematics course of the University of Cambridge / 2009 / ia / Paper 1 / 10D (Analysis I)

The intermediate value theorem says that if $f:[a,b]\to \mathbb{R}$ is a continuous function, every value between $f(a)$ and $f(b)$ is attained at some point of $[a,b]$.

For a proof, endpoint values are immediate, and replacing $f$ by $-f$ reduces the remaining case to $f(a)<y<f(b)$. The set $S=\{x\in [a,b]:f(x)\le y\}$ is nonempty and bounded above; let $c=\sup S$ by the least-upper-bound property. Continuity and the strict endpoint inequalities ensure $a<c<b$: near $a$ there are points in $S$, and near $b$ there are none. If $f(c)>y$, continuity gives a neighborhood of $c$ containing no points of $S$, contradicting the defining approximation property of the supremum. If $f(c)<y$, continuity gives points to the right of $c$ still in $S$, contradicting its being an upper bound. Hence $f(c)=y$, proving the intermediate value theorem.

For the distance problem, define the continuous functionIt satisfies $d(x)\ge |x-a|$, so it is a coercive function: $d(x)\to \infty$ as $|x|\to \infty$. Choose $R>|a|+d(0)+1$. Outside $[-R,R]$, the distance is greater than $d(0)$. On $[-R,R]$, the extreme value theorem gives a minimizer $x_0$, with $A=d(x_0)\le d(0)$. It is therefore a global minimizer and $A\ge 0$.

For any $y>A$, choose $X>x_0$ so large that $d(X)>y$. Apply the intermediate value theorem to $d$ on $[x_0,X]$ to obtain a point whose distance is $y$. The distance $A$ itself is attained at $x_0$, and no smaller value is possible. This is the range of a continuous coercive real function. Thus the distance set is exactlyIn particular $A=0$ precisely when the fixed point lies on the graph.