Past exam of the mathematics course of the University of Oxford / 2021 / Prelims / Mathematics II / 5 / b / ii

We give the explicit point $\boxed{x=1/4}$. Repeatedly choose the left remaining third and then the right remaining third. After $2m$ choices, the resulting closed interval is $[L_m,R_m]$, whereThese formulas follow from a finite geometric series. They show $L_m<1/4<R_m$, and that both endpoints tend to $1/4$. The point $1/4$ is also strictly inside each intermediate left third: its distance from $L_m$ is one quarter of the current length. It therefore never enters any assigned closed middle third or its endpoints, so $1/4\in C$.

Endpoint values in the Cantor function construction obey the same left/right choices, with halving in place of thirding. ThusAll these endpoints are in the domain of the partial function: $L_m$ is the right endpoint of a middle third assigned at round $2m$, and $R_m$ is the left endpoint of a middle third assigned at round $2m-1$.

Given $\epsilon >0$, choose $m$ so that $(2/3)4^{-m}<\epsilon$, and put $\delta =9^{-m}/4$. If $t\notin C$ and $|t-1/4|<\delta$, then $L_m<t<R_m$. The assumed monotonicity givesConsequently $|f(t)-1/3|<\epsilon$. This proves directly, on the stated partial domain,There are defined points arbitrarily close to $1/4$, for example $L_m$ and $R_m$, so the limit of a function is not vacuous. No continuous extension has been assumed. The residual $C$ here omits the endpoints that the question assigns values to; it is the usual Cantor set with those assigned endpoints removed.