Balogh hereditary normality

For each $A\subseteq \kappa$ and $n<\omega$, we prove by finite induction that $A\times \{n\}$ and $(\kappa \setminus A)\times \{n\}$ have disjoint open neighborhoods contained in $U_n$. For $n=0$, use the two sets themselves, open by F1. At the successor stage let $c_A\in C$ be the characteristic function of $A$ and put $A^{\mathrm{\prime }}=\{\beta :d_{c_A}(\beta )=1\}$. The induction assertion gives disjoint open $O_1,O_0\subseteq U_n$ containing $A^{\mathrm{\prime }}\times \{n\}$ and its complement on $L_n$. Adjoin $A\times \{n+1\}$ to $O_1$ and its complement on $L_{n+1}$ to $O_0$. At a new top point $\alpha$, the witness $F(\alpha ,\{c_A\},\mathrm{\varnothing })$ lies in $A^{\mathrm{\prime }}$ if $\alpha \in A$, and in its complement otherwise. F1 therefore makes the enlarged sets open; lower points already lie in the old opens. Disjointness holds on the new level and on the lower levels separately. This proves the assertion at every height.

We will use the following open-extension calculation. Suppose $C_0$ is closed, $O\subseteq U_m$ is open, and $L_m\setminus C_0\subseteq O$. For $t>m$, put $O^{\ast }=O\cup (U_t\setminus (U_m\cup C_0))$. At points of height at most $m$, openness follows from $O$. At a new point of height $m+1$, the open set $X\setminus C_0$ supplies by F1 a witness whose height-$m$ slice avoids $C_0$, and this slice is contained in $O$. At greater heights the same kind of witness lies above height $m$ and outside $C_0$, hence inside the newly added part. Thus $O^{\ast }$ is open. The identical check at every finite height proves that $O\cup (X\setminus (U_m\cup C_0))$ is open as well. No claim that $U_m$ is closed was used.

Let $H,K$ be separated, and fix $m<n$. Apply step 1.1 at level $m$ to the partition $L_m\cap \bar{K}$, $L_m\setminus \bar{K}$. Obtain disjoint open $O_K,O_H\subseteq U_m$ containing those sets. Then $O_K$ contains $K\cap L_m$. By step 1.2, $P=O_H\cup (U_n\setminus (U_m\cup \bar{K}))$ is open; it contains $H\cap L_n$ because those points have height $n>m$ and avoid $\bar{K}$. Its newly added portion is outside $U_m$, while $O_K\subseteq U_m$; the old portions are disjoint. Hence $P,O_K$ are disjoint open neighborhoods of these different-level subsets, both lying in $U_n$.

Fix $n$. For every $m<n$ choose such disjoint open $P_m,Q_m\subseteq U_n$ about $H\cap L_n$ and $K\cap L_m$. At level $n$, step 1.1 supplies disjoint open $P_n,Q_n\subseteq U_n$ about $L_n\cap \bar{H}$ and $L_n\setminus \bar{H}$. In particular these contain $H\cap L_n$ and $K\cap L_n$. Put $V_n={\bigcap }_{m\le n}{P}_m$ and $O^{\mathrm{\prime }}={\bigcup }_{m\le n}{Q}_m$. Both are open, disjoint, and respectively contain $H\cap L_n$ and $K\cap U_n$. In addition $L_n\setminus \bar{H}\subseteq O^{\mathrm{\prime }}$ by the choice of $Q_n$. Step 1.2 therefore makes $O=O^{\mathrm{\prime }}\cup (X\setminus (U_n\cup \bar{H}))$ open. It contains all of $K$, because $K$ avoids $\bar{H}$, and remains disjoint from $V_n\subseteq U_n$. Thus $\overline{V_n}\cap K=\mathrm{\varnothing }$: each point of $K$ has the neighborhood $O$ avoiding $V_n$. Repeating with $H,K$ interchanged yields open $W_n$ containing $K\cap L_n$ with $\overline{W_n}\cap H=\mathrm{\varnothing }$. A1 permits fixing these choices for all $n$. At $n=0$ the finite intersection contains only the same-level choice; no different-level choices are needed.

Define $V={\bigcup }_{n<\omega }(V_n\setminus {\bigcup }_{i\le n}\overline{W_i})$ and $W={\bigcup }_{n<\omega }(W_n\setminus {\bigcup }_{i\le n}\overline{V_i})$. Each summand is open because only finitely many closed sets are removed. Every point of $H$ lies in some $V_n$ and in none of the $\overline{W_i}$, so $H\subseteq V$; likewise $K\subseteq W$. If a point belonged to a $V$-summand indexed by $n$ and a $W$-summand indexed by $m$, then $m\le n$ would make the first omit $W_m$, while $n\le m$ would make the second omit $V_n$. At least one comparison holds, including equality, so no such point exists. These are disjoint open neighborhoods of $H,K$.

For any subspace $Y\subseteq X$ and disjoint relatively closed $H,K\subseteq Y$, one has $Y\cap \bar{H}=H$ and $Y\cap \bar{K}=K$, by the subspace closure definition. Thus $H,K$ are separated in $X$. Step 4.1 gives disjoint open neighborhoods in $X$, whose intersections with $Y$ prove normality of $Y$. This also handles empty subsets and the empty subspace. F2 gives $T_1$ for $X$; apply its normality to two distinct closed singletons to get Hausdorff separation. Therefore all the asserted hereditary and separation properties hold. QED.