Past exam of the mathematics course of the University of Cambridge / 2008 / ia / Paper 1 / 10D (Analysis I)

The intermediate value theorem states that, if $f$ is a continuous function on a closed interval $[a,b]$, every value between $f(a)$ and $f(b)$ equals $f(c)$ for some $c\in [a,b]$. If the value is strictly between the endpoint values, $c$ may be chosen in $(a,b)$.

Here is a proof from the supremum property of the real numbers. Suppose $f(a)<y<f(b)$; the reverse inequality is reduced to this case by replacing $f,y$ with $-f,-y$. LetThe set is nonempty because it contains $a$ and is bounded above by $b$. Continuity at $a$ supplies points of $E$ to the right of $a$, while continuity at $b$ excludes a neighborhood of $b$ from $E$. Hence $a<c<b$. By the definition of supremum, there are $t_n\in E$ with $t_n\to c$. Continuity gives $f(c)=\lim f(t_n)\le y$. If $f(c)<y$, continuity would give a point just to the right of $c$ still satisfying $f(t)<y$, contradicting that $c$ is an upper bound for $E$. Thus $f(c)=y$. Values equal to either endpoint are attained at that endpoint, completing the proof.

Let $u,v\in f(I)$ with $u<y<v$. Choose preimages $x_u,x_v\in I$, and let $a=\min (x_u,x_v)$, $b=\max (x_u,x_v)$. The defining property of a real interval ensures $[a,b]\subseteq I$. The restriction of $f$ to this closed interval is continuous, and its endpoint values are $u,v$ in one order or the other. The intermediate value theorem supplies $c\in [a,b]\subseteq I$ with $f(c)=y$. Thus every value between two image values lies in the image, provingThe continuous image of a real interval result does not require $I$ itself to be closed or bounded; only the segment between the chosen preimages is used. Empty and single-point images also satisfy the interval property.

No such continuous function exists. The extreme value theorem makes every continuous function on the closed interval $[0,1]$ bounded and gives it a finite maximum. Its image therefore cannot equal the unbounded real interval $[0,\infty )$.

The continuous functionhas the required image. Its values are nonnegative, and for every $y\ge 0$ the point $x=1/(y+1)\in (0,1]$ satisfies $f(x)=y$. Thus the image is exactly $[0,\infty )$. The missing left endpoint permits unboundedness without a failure of continuity on the domain.

A suitable continuous function isFor $t_k=\pi /2+2\pi k$ and $u_k=3\pi /2+2\pi k$,All these arguments lie in $(0,1]$. Its image is a real interval by the continuous image of a real interval result, and it is unbounded both above and below. Given any real $y$, choose two of the displayed points with function values on opposite sides of $y$, then apply the intermediate value theorem on the segment joining them. Hence the image is all of $\mathbb{R}$. A monotone example would not suffice here; the oscillation gives both directions of unboundedness near the same missing endpoint.