PCF Scales and ZFC Dowker Spaces: Examples and Counterexamples

In a discrete space complements of subsets are open, so every subset is also closed. Thus $G_n=F_n$ is open, contains $F_n$, and has $\overline{G_n}=F_n$. Consequently ${\bigcap }_n\overline{G_n}={\bigcap }_n{F}_n=\mathrm{\varnothing }$. The same equations hold when $F_n$ or $X$ is empty.

For the instance on $\omega$, $F_0=\omega$, $F_1=\omega \setminus \{0\}$, and $F_{n+1}\subseteq F_n$. For each $m\in \omega$, $m\mathrm{\notin }{F}_{m+1}$, proving the empty intersection despite every $F_n$ being nonempty. The singleton family $\{\{m\}:m\in \omega \}$ is an open refining cover of every open cover: a member containing $m$ also contains $\{m\}$. The neighborhood $\{m\}$ meets exactly one singleton, so the family is locally finite, verifying countable paracompactness directly. No simultaneous selection of cover members is involved. QED.

All coordinates of $h_k$ are in $[0,{\mathrm{\aleph }}_n]$: the prefix equals its tops and the tail has ${\omega }_1<{\mathrm{\aleph }}_n$ since $n\ge 2$. By F2 and A1 their cofinalities are respectively ${\mathrm{\aleph }}_n$ and ${\mathrm{\aleph }}_1$. Set $m=\max (k,1)+1$. These cofinalities are above $\omega$ and strictly below ${\mathrm{\aleph }}_m$, including for $k=0,1$ when the prefix is empty and $m=2$. F1 therefore gives $h_k\in X_R(B)$.

For every $n\le k$ in $B$ the defining value is ${\mathrm{\aleph }}_n$, so F3 gives $h_k\in F_k$. If $l>k$ and $l\in B$, its $l$-th coordinate is ${\omega }_1<{\mathrm{\aleph }}_l$, violating the equality required for $F_l$. At $k=3$, the first two coordinates are ${\omega }_2,{\omega }_3$, and the next is ${\omega }_1<{\omega }_4$, giving the asserted concrete instance. QED.

Given the earlier $g_{\eta }$ for $\eta <\xi <{\omega }_1$, define $t_{\xi }(n)=\sup (\{1\}\cup \{g_{\eta }(n)+1:\eta <\xi \})$. By F3–F4 each countable supremum is below ${\mathrm{\aleph }}_n$; thus $t_{\xi }$ is in the strict product. The earlier indices are countable and bounded below $\lambda$ by F3–F4. F1 supplies a scale term strictly eventually above $t_{\xi }$ with index above all earlier indices, by cofinality followed by a later scale term if necessary. Let ${\alpha }_{\xi }$ be the least eligible index and put $g_{\xi }(n)=\max (t_{\xi }(n),f_{{\alpha }_{\xi }}(n))$. This is in the product, is eventually equal to $f_{{\alpha }_{\xi }}$, and is above every earlier $g_{\eta }$ pointwise. F5 now supplies the sequence by the specified rule. At stage zero, $t_0(n)=1$ and $g_0(n)=\max (1,f_{{\alpha }_0}(n))\ge 1>0$. At stage one, $t_1(n)=g_0(n)+1$ and $g_1(n)\ge g_0(n)+1$. These are the first two calculations of the instance.

Set $h(n)={\sup }_{\xi <{\omega }_1}{g}_{\xi }(n)$ and $\delta ={\sup }_{\xi <{\omega }_1}{\alpha }_{\xi }$. The sequence in step 1.1 satisfies every hypothesis of F2, and $n>1$ for every $n\in B$, so its tail is the entire coordinate set. Therefore $h(n)<{\mathrm{\aleph }}_n$, $\mathrm{cf}(h(n))={\omega }_1$, $\delta <\lambda$, and $h{=}^{\ast }{f}_{\delta }$. The cofinalities have uniform strict bound ${\mathrm{\aleph }}_2$, so $h$ is a Rudin point and F1 gives $h\in X$. The calculation $h(n)\ge g_1(n)>g_0(n)\ge 1>0$ verifies strict domination of the chosen zero bound.

For a prescribed $b$ replace $1$ in step 1.1 by $b(n)+1$. This value is below the limit cardinal ${\mathrm{\aleph }}_n$, so the same countable-supremum and least-index arguments still apply. Then $g_0(n)\ge b(n)+1>b(n)$ and the resulting supremum satisfies $h(n)\ge g_0(n)>b(n)$ for every $n$. Thus the example exhibits the actual representative construction behind pointwise cofinality, with no choice of a member from a possibly empty scale class. QED.

Set $B=\{2,3,4,\dots \}$ and take the specific witness $X=X_R(B)$. Explicitly its points are functions $h(n)\le {\mathrm{\aleph }}_n$ whose coordinate cofinalities are all uncountable and strictly bounded by one finite aleph, with the relative ordinal box topology. The set $B$ is infinite and avoids zero and one, so F1 and A1 apply and verify that $X$ satisfies the asserted normality and $T_1$ hypotheses while failing countable paracompactness.

If this $X\times [0,1]$ were normal, F2 would imply that $X$ is countably paracompact, contradicting step 1.1. Thus the witness has a nonnormal interval product and refutes the universal assertion. The product in the conclusion is the ordinary product of the already defined space $X$ and the entire interval, including its endpoints. QED.