Past exam of the mathematics course of the University of Cambridge / 2004 / ib / Paper 2 / 15E (Further Analysis)

For a sequence $ϵ$, let $C_n(ϵ)$ be its prefix cylinder set, consisting of sequences agreeing with $ϵ$ in the first $n$ coordinates. The specified collection consists exactly of unions of such cylinder sets.

The empty set meets the defining condition vacuously, and $X$ meets it because every cylinder set is contained in $X$. For a union ${\bigcup }_{\lambda }{Y}_{\lambda }$, a point lies in one $Y_{\lambda }$ and has a prefix cylinder set contained in that set, hence contained in the union. For a finite intersection, a point has a cylinder neighbourhood in every constituent. Taking the largest of the finitely many prefix lengths gives one cylinder set contained in all of them. Therefore the collection contains the empty and whole sets, is closed under arbitrary unions and finite intersections, and is a topology on $X$, the prefix topology of Cantor space.

The weighted binary-sequence metric is finite because $0\le d(ϵ,\eta )\le {\sum }_{n\ge 1}{2}^{-n}=1$. It is symmetric and nonnegative. If $d(ϵ,\eta )=0$, every nonnegative summand is zero, so every coordinate agrees; conversely agreement makes the sum zero. For three sequences, apply $|{ϵ}_n-{\zeta }_n|\le |{ϵ}_n-{\eta }_n|+|{\eta }_n-{\zeta }_n|$ in each coordinate and sum. This proves the triangle inequality, so $d$ is a metric.

If two sequences differ in one of their first $n$ coordinates, their distance is at least $2^{-n}$. Thus the open ball $B_d(\eta ,2^{-n})$ is contained in $C_n(\eta )$. For every $\eta \in C_n(ϵ)$, the latter cylinder equals $C_n(\eta )$, so each cylinder set is open in the metric topology. Every set in the specified topology, being a union of cylinder sets, is therefore metric-open.

Conversely, agreement in the first $n$ coordinates givesGiven a metric-open set containing $ϵ$, choose a ball $B_d(ϵ,r)$ inside it, then choose $n$ with $2^{-n}<r$. The bound gives $C_n(ϵ)\subseteq B_d(ϵ,r)$, establishing the required prefix-neighbourhood condition. The two topologies coincide. The strict inequality $2^{-n}<r$ avoids incorrectly treating the closed tail bound as a strict ball bound.