Past exam of the mathematics course of the University of Cambridge / 2013 / ia / Paper 1 / 10E (Analysis I)

Suppose continuity at $y$ failed. The negation of the relative-domain epsilon-delta definition gives an ${\epsilon }_0>0$ such that for every $\delta >0$ there is an $x\in [a,b]$ with $|x-y|<\delta$ but $|f(x)-f(y)|\ge {\epsilon }_0$. Choosing $\delta =1/n$ produces a sequence $x_n\to y$ whose image does not converge to $f(y)$, contradicting the hypothesis. Thus $f$ is continuous at $y$.

This proves the sequential characterization of continuity in metric spaces. At $a$ or $b$, the same argument uses points only in $[a,b]$, giving the appropriate one-sided continuity automatically.

The intermediate value theorem states that a real-valued continuous function on a closed interval takes every value between its endpoint values. Explicitly, if $v$ lies between $f(s)$ and $f(t)$, there exists $u\in [s,t]$ with $f(u)=v$. No monotonicity is assumed.

If $f$ is a strictly increasing function and $x\ne y$, one of $x<y$ or $y<x$ holds. The corresponding strict inequality of function values shows that $f$ is injective. Conversely, if $f(x)<f(y)$, then $x=y$ is impossible and $x>y$ would imply $f(x)>f(y)$. ThusTogether with strict increase, this says that such a function preserves and reflects the ordering of points.

Fix $a<x<b$. Injectivity excludes $f(x)=c$ or $d$. If $f(x)<c$, the intermediate value theorem on $[x,b]$ gives a point other than $a$ with value $c$, contradicting injective function. If $f(x)>d$, the same theorem on $[a,x]$ gives a point other than $b$ with value $d$. ThereforeTo prove strict increase, take $x<y$. If $x=a$, the displayed endpoint ordering already gives $f(x)<f(y)$. Otherwise $f(a)<f(y)$, and the same interior-point argument on the subinterval $[a,y]$ yields $f(a)<f(x)<f(y)$. Hence $f$ is a strictly increasing function. This is the interval order theorem for continuous injections; the assumed endpoint order rules out strict decrease.

First prove continuity at $b$. Given $\epsilon >0$, choose $v$ with $\max (c,d-\epsilon )<v<d$. Surjectivity gives $x_v\in [a,b]$ with $f(x_v)=v$. Strict increase implies $x_v<b$. For $b-(b-x_v)<x\le b$,Thus $\delta =b-x_v$ proves left continuity at $b$. At $a$, choose $c<v<\min (d,c+\epsilon )$ and use its preimage to prove right continuity by the same ordering argument.

For $a<x_0<b$, choose values $v_{-},v_{+}$ in the range such thatTheir preimages satisfy $x_{-}<x_0<x_{+}$. If $\delta =\min (x_0-x_{-},x_{+}-x_0)$ and $|x-x_0|<\delta$, strict increase traps $f(x)$ between $v_{-}$ and $v_{+}$, so $|f(x)-f(x_0)|<\epsilon$. Therefore $f$ is continuous on the entire closed interval. This is continuity of an increasing interval surjection: a jump would omit values, contradicting the stated surjectivity.