Small Dowker ladder normality

Every singleton $\{x\}$ is closed: for active $\alpha \mathrm{\ne }x$, its intersection with $L_{\alpha }$ is empty or a singleton $\{x\}$ with $x<\alpha$, bounded by $x+1<\alpha$ since $\alpha$ is a limit. Every initial segment $\xi <{\omega }_1$ is open, since $\alpha <\xi$ implies $L_{\alpha }\subseteq \alpha \subseteq \xi$. If $\xi \mathrm{\notin }\bar{S}$, the initial segment is also closed: for active $\alpha \mathrm{\notin }\xi$, equality $\alpha =\xi$ is excluded and $\xi <\alpha$ bounds $\xi \cap L_{\alpha }$. Points outside $\bar{S}$ are isolated, and every $W_n$ is open because all ladders at its points lie in lower levels.

If disjoint closed $K_0,K_1$ were both uncountable, F1 would give active $\alpha$ with both $K_i\cap L_{\alpha }$ cofinal. The closedness test would force $\alpha \in K_0\cap K_1$, impossible. Consequently one member is countable and is bounded by F2, whose choice hypothesis is supplied by A1.

Fix $\xi <{\omega }_1$ and an injective enumeration $({\alpha }_j{)}_{j\in J}$ of $\bar{S}\cap \xi$, where $J$ is a finite initial segment of $\omega$ or all of $\omega$. Such an enumeration exists by F2; if the domain is empty use the empty list. For each $j$, the finitely many intersections $L_{{\alpha }_j}\cap L_{{\alpha }_k}$ with $k<j$ are bounded in ${\alpha }_j$ by F1. Choose the least ${\epsilon }_j<{\alpha }_j$ whose initial segment contains their union, and put $T_j=L_{{\alpha }_j}\setminus {\epsilon }_j$. The empty union permits ${\epsilon }_0=0$. Each $T_j$ is a cofinal tail; for $k<j$, it misses $L_{{\alpha }_k}$ and hence $T_k$. For $c:\bar{S}\cap \xi \to \omega$, define $c^{\ast }(\gamma )=c({\alpha }_j)$ on $T_j$, and $0$ off their disjoint union. This is a well-defined function on $\xi$ and equals $c({\alpha }_j)$ on a tail of every $L_{{\alpha }_j}$. It proves almost $P_0$, with arbitrary natural-number colors.

Given disjoint closed sets, interchange their names if necessary so that $K_0$ is bounded, by step 1.2. Choose a countable successor ordinal $\xi$ with $K_0\subseteq \xi$; if $K_0=\mathrm{\varnothing }$, take $\xi =1$. Successors are inactive, so $\xi$ is clopen by step 1.1. Fix the disjoint tails $T_j$ from step 1.3 for this one $\xi$ once and for all. They give a specified extension operation $c\mapsto c^{\ast }$ by the formula there, valid for every later coloring on this same domain.

Suppose $H_0,H_1$ are disjoint closed sets with $H_0\subseteq \xi$. On $\bar{S}\cap \xi$ define $c_i(\alpha )=1$ for $\alpha \in H_i$ and $0$ otherwise, and extend each by the fixed operation of step 2.1. Set $H_i^{\mathrm{\prime }}=H_i\cup P_i$, where $P_i=\{\gamma \in \xi \setminus H_{1-i}:c_i^{\ast }(\gamma )=1\text{ and }{c}_{1-i}^{\ast }(\gamma )=0\}$. The old sets are retained and $H_0^{\mathrm{\prime }}\subseteq \xi$. New points of one set avoid the other old set by definition; new points of both sets would require both $c_0^{\ast }=1,c_1^{\ast }=0$ and $c_1^{\ast }=1,c_0^{\ast }=0$. Thus $H_0^{\mathrm{\prime }},H_1^{\mathrm{\prime }}$ are disjoint.

Each $H_i^{\mathrm{\prime }}$ in step 3.1 is closed. Indeed, let active $\alpha \mathrm{\notin }{H}_i^{\mathrm{\prime }}$ have $L_{\alpha }\cap H_i^{\mathrm{\prime }}$ cofinal. Since $\alpha \mathrm{\notin }{H}_i$ and $H_i$ is closed, its old intersection is bounded; hence $L_{\alpha }\cap P_i$ must be cofinal. All new points lie below $\xi$, so $\alpha \le \xi$. Equality is impossible since $\xi$ is inactive. Thus $\alpha \in \bar{S}\cap \xi \setminus H_i$, giving $c_i(\alpha )=0$. Its extension is eventually zero on $L_{\alpha }$, whereas every point of $P_i$ has extension value one, a contradiction. The closedness test of F1 now gives the assertion.

If active $\alpha \in H_i\cap \xi$, disjointness gives $c_i(\alpha )=1$ and $c_{1-i}(\alpha )=0$. On a common tail of $L_{\alpha }$ the extensions therefore have these two values. Also $H_{1-i}\cap L_{\alpha }$ is bounded because $H_{1-i}$ is closed and does not contain $\alpha$. Removing this additional bounded set puts the entire remaining tail in $P_i\subseteq H_i^{\mathrm{\prime }}$. This explicitly checks the exclusion of the other old set, as well as the two color requirements. Inactive indices have empty ladders and need no tail assertion.

Start $K_i^0=K_i$ and iterate the explicit operation of step 3.1 to obtain $K_i^{n+1}=(K_i^n{)}^{\mathrm{\prime }}$. Steps 4.1 and 4.2 preserve closedness, disjointness, $K_0^n\subseteq \xi$, and containment of a ladder tail at each active point of $K_i^n\cap \xi$ in $K_i^{n+1}$. The assignment is a fixed function of a pair, so finite iteration defines each stage; equivalently its graph consists of the unique endpoints of finite sequences of iterates. No countable selection of unspecified extensions is being assumed. Put $U_0={\bigcup }_n{K}_0^n$ and $U_1=({\omega }_1\setminus \xi )\cup {\bigcup }_n{K}_1^n$.

For any active $\alpha \in U_i\cap \xi$, membership in some $K_i^n$ puts a tail of $L_{\alpha }$ in $K_i^{n+1}\subseteq U_i$. The exterior ${\omega }_1\setminus \xi$ is open because $\xi$ is clopen; it supplies the open-set test at the remaining points of $U_1$. There are no points of $U_0$ outside $\xi$. Thus both $U_i$ are open. If $x\in U_0\cap U_1$, it lies below $\xi$ and in $K_0^r\cap K_1^s$ for some $r,s$. Monotonicity puts it in the disjoint pair at stage $\max (r,s)$, impossible. Each $U_i$ contains $K_i$. This proves normality, including empty closed members. Separating the closed singletons of two distinct points proves Hausdorffness. QED.