Borel subspaces admit polish presentations

If $P$ is empty there is only the empty subset and the assertion holds. On nonempty $P$, fix a compatible complete metric and a countable basis using [F1], [F4] and [F5]. An open $U$ is $G_{\delta }$ (repeat $U$), so [F2] completely metrizes it; its closed complement is complete in the restricted original metric since a limit of a sequence in a closed set stays there. Both subspaces have countable trace bases and hence are Polish by [F4].

Bound each component metric by replacing $d$ with $\min (d,1)$; this preserves its topology and Cauchy sequences, hence completeness. On the disjoint union of $U$ and $P\setminus U$, retain those metrics within components and set cross-component distance equal to two. The triangle inequality holds within a component and across components (any cross-component path includes an edge of length two). A Cauchy sequence is eventually in one component and converges there. A union of the two countable bases is countable. This Polish topology is finer than $\tau$, makes $U$ clopen, and has the same Borel sets: each new open is the union of two old trace-open sets, hence old Borel. Empty components simply contribute no points.

Let $\mathcal{R}$ be the old Borel subsets that can be made clopen by such a refinement. Step 2.1 puts every open set in $\mathcal{R}$; closure under complements uses the same topology. Given $B_n\in \mathcal{R}$, [F5] selects a witnessing Polish topology ${\tau }_{n+1}$, complete bounded metric and countable basis for each $n$. Include ${\tau }_0=\tau$. In ${\prod }_{n\ge 0}(P,{\tau }_n)$ let $\mathrm{\Delta }=\{(x,x,\dots ):x\in P\}$.

The diagonal $\mathrm{\Delta }$ is closed. If two coordinates differ, disjoint neighbourhoods in the original metric topology pull back to open neighbourhoods in both refined coordinates; their product cylinder misses $\mathrm{\Delta }$. The product is completely metrized by [F3], so its closed subspace $\mathrm{\Delta }$ is complete. It has a countable basis of finite cylinders restricted to $\mathrm{\Delta }$; For each finite length the coordinate-index and basis-index lists form a countable set by [F8]; [F7] makes the union over lengths countable, with its countable-choice hypothesis supplied by [F5]. By [F4] it is Polish. Pull its topology back to $P$ along $x\mapsto (x,x,\dots )$. This refines every ${\tau }_n$. Each basic open is a finite intersection of old Borel sets, and every open is a union of a subfamily of the countable basis. Thus every new open is old Borel; the two Borel sigma-algebras coincide.

Each $B_n$ is clopen in the common refinement, so ${\bigcup }_n{B}_n$ is open. Apply the splitting construction of step 2.1 to that Polish topology, making the union clopen while preserving its Borel sets, hence the original Borel sets. Consequently $\mathcal{R}$ is a sigma-algebra containing $\tau$, and contains every old Borel set. For the specified $B$, restrict the resulting complete metric and countable basis to the closed set $B$. This is a Polish topology on $B$, finer than its original subspace topology; its Borel sets are precisely the old traces, since relative opens generate traces of Borel sets in either topology. The identity is the Polish presentation required by [F6].