Dense domains and directed unions of specializing conditions

Fix $p\in P(T)$ and $t\in T$. If $t\in \mathrm{dom}(p)$ take $q=p$. Otherwise choose the explicit natural $N=0$ when $\mathrm{ran}(p)=\mathrm{\varnothing }$ and $N=1+\max \mathrm{ran}(p)$ when the finite range is nonempty. Put $q=p\cup \{(t,N)\}$. It is a finite function extending $p$, and $N$ differs from every old label. Pairs in the old domain satisfy F1 already; any new comparable pair involves $t$ and has unequal labels. Thus $q\in P(T)\cap D_t$ and $q\le p$, proving density.

Put $f=\bigcup G$ as a union of graphs. If $(t,a),(t,b)\in f$, take $p,q\in G$ containing the respective pairs. Directedness gives $r\in G$ extending both; as $r$ is a function, $a=r(t)=b$. Thus $f$ is a function with domain contained in $T$ and values in $\omega$. For each $t\in T$, meeting $D_t$ supplies a condition in $G$ with $t$ in its domain, so $t\in \mathrm{dom}(f)$. Hence $\mathrm{dom}(f)=T$. No simultaneous selection of the conditions is needed for this pointwise conclusion.

If $x{<}_{T}y$, totality gives $p,q\in G$ whose domains contain $x,y$ respectively. A common stronger $r\in G$ contains both nodes, so F1 gives $r(x)\mathrm{\ne }r(y)$. Since $r\subseteq f$ by its membership in $G$, these are $f(x)$ and $f(y)$. Therefore $f$ specializes $T$ by F2. This uses only the supplied directed family; density alone does not provide that family.