Blair's sequence space for a serial relation

The relation is serial because for each $i\in \omega$, $i+1\in \omega$ and $iR(i+1)$. Thus the sequence-space and open-dense conclusions apply. Let $g=(2,0,2,3,0,0,\dots )$ and $h=(2,0,2,0,0,0,\dots )$. They extend the prefix, their first disagreement is at coordinate $3$, and $d(g,h)=1\mathrm{/}(3+1)=1\mathrm{/}4$. Since $g(3)=3=g(0)+1$, $g\in U_0$.

For $f$ the successor witnesses are explicit: at $n=0$ or $n=2$, use $m=4$ since $f(4)=3=2+1$; at $n=1$, use $m=3$ since $f(3)=1=0+1$; at $n=3$, use $m=0$ since $f(0)=2=1+1$; and at every $n\ge 4$, use $m=n+1$ since $f(n+1)=n=f(n)+1$. These cases cover all naturals, proving $f\in {\bigcap }_n{U}_n$. But $f(1)=0\mathrm{\ne }3=f(0)+1$, so $f(0)Rf(1)$ fails.

Let $q(i)=\min \{j:f(j)=f(i)+1\}$. The displayed values give $q(0)=q(2)=4$, $q(1)=3$, and $q(3)=0$: the first occurrences of $3,1,2$ are at $4,3,0$, respectively. For $i\ge 4$, $f(i)+1=i\ge 4$ cannot occur among coordinates $0,1,2,3$, whose values are $2,0,2,1$, and in the tail $f(j)=j-1$ it occurs only at $j=i+1$. Thus $q(i)=i+1$ for $i\ge 4$. Recursing from $k(0)=0$ gives $k(1)=4$ and $k(t)=t+3$ for every $t\ge 1$, because $q(t+3)=t+4$. Hence $f(k(0))=2$ and $f(k(t))=t+2$ for $t\ge 1$, so successive extracted values differ by exactly one.

For arbitrary serial $R$, membership in every $U_n$ says exactly that each coordinate value has an $R$-successor somewhere among the sequence's values. It does not specify the next coordinate. The explicit failure at coordinates zero and one demonstrates that distinction, and the computed least-index extraction demonstrates how to obtain a chain. The backward witness $q(3)=0$ also shows why unrestricted witness indices need not increase. These computations required no CM-Baire assumption to produce this particular $f$.