Past exam of the mathematics course of the University of Cambridge / 2015 / ia / Paper 1 / 11D (Analysis I)

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

Endpoint values are already attained. Suppose $f(a)<y<f(b)$ and define $E=\{x\in [a,b]:f(x)<y\}$. It is nonempty and bounded above, so $c=\sup E$ exists. Continuity at $a$ ensures that $E$ contains points to the right of $a$, and continuity at $b$ ensures that no point sufficiently close to $b$ belongs to $E$. Hence $a<c<b$.

By the definition of the supremum, there are points of $E$ approaching $c$. If $f(c)>y$, continuity would exclude all such nearby points from $E$, a contradiction. If $f(c)<y$, continuity would put some point to the right of $c$ in $E$, again a contradiction. Therefore $f(c)=y$. When $f(a)>f(b)$, apply this argument to $-f$. This proves the intermediate value theorem in all cases.

For a chord of length $1/2$, consider the continuous function $h(x)=f(x+1/2)-f(x)$ on $[0,1/2]$. Since $f(0)=f(1)$,An endpoint zero already supplies the chord. Otherwise the endpoint values have opposite signs, so the intermediate value theorem supplies $h(\alpha )=0$ at some $\alpha \in (0,1/2)$. Thus a horizontal chord of length $1/2$ exists.

For each integer $n>1$, define $h_n(x)=f(x+1/n)-f(x)$ on $[0,1-1/n]$. The sampled differences telescope:If any sampled value is zero, it gives the required chord. Otherwise there must be both a positive and a negative sampled value. The intermediate value theorem, applied between their sample points, then gives $h_n(\alpha )=0$. Hence a horizontal chord of length $1/n$ exists for every positive integer $n$; for $n=1$, the endpoint chord itself suffices. This proves the horizontal chord of reciprocal-integer length result, without restricting $n$ to powers of two.