 Solution

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.