Successor-occurrence sets of a serial relation are open and dense

Each $U_n$ is a set by Separation in $Y$. The graph $\{(n,U)\in \omega \times \mathcal{P}(Y):(\mathrm{\forall }f\in Y)(f\in U\leftrightarrow (\mathrm{\exists }m\in \omega )f(n)Rf(m))\}$ is also a set by Separation. For each $n$ there is exactly one such $U$, so this graph defines an indexed family with domain $\omega$.

If $f\in U_n$, fix one witnessing $m$. Every $g$ in the cylinder $[f\upharpoonright (\max (n,m)+1)]$ has $g(n)=f(n)$ and $g(m)=f(m)$, hence $g(n)Rg(m)$ and $g\in U_n$. This is an open neighbourhood of $f$, so $U_n$ is open.

Fix one $a_{\ast }\in A$, and let $s:l\to A$ be any finite prefix; $l=0$ is allowed. Set $L=\max (l,n+1)$. Extend $s$ to $t:L\to A$ by assigning $a_{\ast }$ at each new coordinate. Thus $t(n)$ is defined and $L>n$. Seriality gives one $b\in A$ with $t(n)Rb$. Define $g(i)=t(i)$ for $i<L$, $g(L)=b$, and $g(i)=a_{\ast }$ for $i>L$. Then $g\in [s]$ and $g(n)Rg(L)$, so $g\in U_n$. This is one explicit extension for a fixed cylinder and fixed $n$, not a choice of extensions for a family of cylinders.

Every nonempty open subset of $Y$ contains a cylinder, and hence meets $U_n$ by the preceding construction. Equivalently every ball meets $U_n$, which is exactly density by the metric closure definition. Thus every $U_n$ is open dense. For singleton $A=\{a_{\ast }\}$ seriality forces $a_{\ast }R{a}_{\ast }$ and the same construction gives $U_n=Y$. No infinite relation-path was assumed in proving nonemptiness of a cylinder.