Universal Borel sets and strictness on Cantor space

If $X$ is nonempty, fix a countable dense sequence. Balls at its centres of positive rational radii form a countable basis: for $x\in O$ choose $ϵ>0$ with $B(x,ϵ)\subseteq O$, a centre within $ϵ\mathrm{/}4$ of $x$ and a rational radius between that distance and $ϵ\mathrm{/}2$. This ball contains $x$ and is inside $O$ by the triangle inequality. Enumerate this basis as $W_n$; if $X=\mathrm{\varnothing }$ take all $W_n=\mathrm{\varnothing }$. Set $U_1={\bigcup }_n\{c:c(n)=1\}\times W_n$. This is open by F4. For open $O$, the parameter $c(n)=1$ iff $W_n\subseteq O$ has section exactly $O$ by the basis property. Put $V_1=(\mathcal{C}\times X)\setminus U_1$; its sections exhaust the closed sets.

For each countable $\alpha >1$ choose a nondecreasing positive sequence ${\beta }_n^{\alpha }<\alpha$ with ${\sup }_n({\beta }_n^{\alpha }+1)=\alpha$. At successor $\alpha =\gamma +1$ take constant $\gamma$. At a limit enumerate its ordinals and take the maximum of $1$ and the first $n+1$ listed ordinals; finite maxima stay below the limit and are cofinal. These choices form a set-indexed family, so A1 applies. Recursively, using F5, set

Suppose $B={\bigcup }_j{B}_j$ with $B_j\in {\mathrm{\Pi }}_{{\gamma }_j}^0(X)$ and $1\le {\gamma }_j<\alpha$. Recursively choose $n_j>n_{j-1}$ least with ${\beta }_{n_j}^{\alpha }\ge {\gamma }_j$. Such indices occur arbitrarily late: otherwise the nondecreasing sequence would be bounded below ${\gamma }_j$, contradicting its cofinal property. By F1 place $B_j$ in rank ${\beta }_{n_j}^{\alpha }$, and place the empty set at every unused index. Inductive universality and A1 select a parameter for each of these sets; F3 codes this parameter sequence as $c$. Then $(U_{\alpha }{)}_c={\bigcup }_j{B}_j=B$. Complementation proves universality of $V_{\alpha }$. This proves the recursive universality assertion, including the empty set, without matching the original jth rank to the jth cofinal rank.

Apply this to $X=\mathcal{C}$ and put $D=\{c:(c,c)\mathrm{\notin }{U}_{\alpha }\}$. The diagonal is continuous since the preimage of a basic product $O\times P$ is $O\cap P$, so F2 gives $D\in {\mathrm{\Pi }}_{\alpha }^0$. If $D\in {\mathrm{\Sigma }}_{\alpha }^0$, universality gives $d$ with $D=(U_{\alpha }{)}_d$. At $d$ this says $d\in D$ iff $(d,d)\in U_{\alpha }$ iff $d\mathrm{\notin }D$, impossible. The complement of $D$ lies in the opposite difference.

Let $Y\subseteq Z$ be homeomorphic to $\mathcal{C}$ and $Z$ metrizable. By F2 the homeomorphism and its inverse preserve both classes, so step 3.1 supplies $D_Y\in {\mathrm{\Pi }}_{\alpha }^0(Y)\setminus {\mathrm{\Sigma }}_{\alpha }^0(Y)$. F2 lifts $D_Y$ to a ${\mathrm{\Pi }}_{\alpha }^0(Z)$ set $D_Z$. Were $D_Z$ also ${\mathrm{\Sigma }}_{\alpha }^0(Z)$, its trace would contradict the choice of $D_Y$. Thus $D_Z$ lies in the required difference, and its complement proves the dual difference. QED.