text:
\section{Introduction}
A \emph{Dowker space} is a normal topological space $\mathbb X$ whose product with the unit
interval $\mathbb X\times[0,1]$ is not normal. Whether such a space exists was asked by Dowker back in 1951 \cite{Dowker_C.H.}.
As of now, there are just three constructions of Dowker spaces in ZFC: Rudin's space of size $(\aleph_\omega)^{\aleph_0}$ \cite{Rudin_first_Dowker_ZFC_example}, Balogh's space of size continuum \cite{Balogh_space}, and the Kojman-Shelah space of size $\aleph_{\omega+1}$ \cite{MR1605988}.
Rudin's \emph{Conjecture~4} from \cite{MR1078646},
asserting that there exists a Dowker space of size $\aleph_1$ remains open.

\medskip

In \cite{paper48}, the guessing principle $\clubsuit_{\ad}$ was introduced,
and it was shown that
for every regular uncountable cardinal $\kappa$, each of the following two implies the existence of a Dowker space of size $\kappa$:
\begin{enumerate}
\item[(i)] $\clubsuit_{\ad}(\mathcal S,1,2)$ holds
for a partition $\mathcal S$ of some nonreflecting stationary subset of $\kappa$
into infinitely many stationary sets;
\item[(ii)] $\clubsuit_{\ad}(\{E^{\kappa}_{\lambda}\},\lambda,1)$ holds,
where $\kappa$ is the successor of a regular cardinal $\lambda$.
\end{enumerate}

It was also shown that in each of the scenarios of \cite{kappa_Dowker_from_souslin_Rudin,de_Caux_space,MR628595,Good_Dowker_large_cardinals} in which there exists a Dowker space of size $\kappa$,
either (i) or (ii) indeed hold.

The exact definition of the guessing principle under discussion reads as follows:

\begin{defn}\label{clubad}  Let $\mathcal S$ be a collection of stationary subsets of a regular uncountable cardinal $\kappa$,
and $\mu,\theta$ be nonzero cardinals $<\kappa$.
The principle
$\clubsuit_{\ad}(\mathcal S,\mu,\theta)$ asserts the existence of a sequence $\langle \mathcal A_\alpha\mid \alpha\in\bigcup\mathcal S\rangle$ such that:
\begin{enumerate}
\item For every $\alpha\in\acc(\kappa)\cap\bigcup\mathcal S$, $\mathcal A_\alpha$ is a pairwise disjoint family of $\mu$ many cofinal subsets of $\alpha$;
\item For every $\mathcal B\s[\kappa]^\kappa$ of size $\theta$, for every $S\in\mathcal S$, there are stationarily many $\alpha\in S$ such that $\sup(A\cap B)=\alpha$ for all $A\in\mathcal A_\alpha$ and $B\in\mathcal B$;
\item For all $A\neq A'$ from $\bigcup_{S\in\mathcal S}\bigcup_{\alpha\in S}\mathcal A_\alpha$, $\sup(A\cap A')<\sup(A)$.
\end{enumerate}
\end{defn}
\begin{remark}
The variation $\clubsuit_{\ad}(\mathcal S,\mu,{<}\theta)$ is defined in the obvious way.
\end{remark}

In \cite{two_more_S_spaces}, Juh\'asz, Kunen and Rudin constructed
a Dowker space of size $\aleph_1$ assuming the continuum hypothesis.
This was then improved by the third author \cite[p.~53]{TodPartTop} who got such a space from the existence of a Luzin set (cf.~\cite{MR1283587}).

The first main result of this paper shows that also in the above scenario, (i) and (ii)  hold.
This answers Question~2.35 of \cite{paper48} in the negative.

\begin{mainthm}\label{thma} Let $\lambda=\lambda^{<\lambda}$ be an infinite regular cardinal.

Then $(1)\implies(2)\implies(3)$:
\begin{enumerate}
\item There exists a $\lambda^+$-Luzin subset of ${}^{\lambda}\lambda$;
\item There exists a tight strongly unbounded coloring $c:\lambda\times\lambda^+\rightarrow\lambda$;
\item For every partition $\mathcal S$ of $E^{\lambda^+}_{\lambda}$ into stationary sets,
$\clubsuit_{\ad}(\mathcal S,\lambda,{<}\lambda)$ holds.
\end{enumerate}
\end{mainthm}

Note that as in the case $\lambda:=\aleph_0$, for every infinite cardinal $\lambda=\lambda^{<\lambda}$,
the existence of a $\lambda^+$-Luzin subset of ${}^{\lambda}\lambda$ follows from $2^\lambda=\lambda^+$.
Our second main result derives $\clubsuit_{\ad}$ from another consequence of $2^\lambda=\lambda^+$, namely,
from the \emph{stick} principle:

\begin{mainthm}\label{thmb} Let $\lambda$ be an infinite regular cardinal.

Then ${\ensuremath \mspace{2mu}\mid\mspace{-12mu} {\raise0.6em\hbox{$\bullet$}}}(\lambda^+)$ implies that
for every partition $\mathcal S$ of $E^{\lambda^+}_{\lambda}$ into stationary sets,
$\clubsuit_{\ad}(\mathcal S,\lambda,\lambda)$ holds.
\end{mainthm}

What's interesting about Theorem~B is that it uncovers
a scenario for the existence of a Dowker space that was not known before.
In particular, it yields the following contribution to the small Dowker space problem:

\begin{maincor} ${\ensuremath \mspace{2mu}\mid\mspace{-12mu} {\raise0.6em\hbox{$\bullet$}}}$ entails the existence of a Dowker spaces of size $\aleph_1$. In fact, it entails the existence of $2^{\aleph_1}$ many pairwise nonhomeomorphic such spaces.
\end{maincor}

\subsection{Organization of this paper}
In Section~\ref{Sec2}, we define \emph{unbounded}, \emph{strongly unbounded} and \emph{tight} colorings, and provide sufficient conditions for their existence. The proof of the implication $(1)\implies(2)$ of Theorem~\ref{thma} will be found there.

In Section~\ref{Sec3}, we get an instance of $\clubsuit_{\ad}(\ldots)$ from a tight strongly unbounded coloring. In particular, the proof of the implication $(2)\implies(3)$ of Theorem~\ref{thma} will be found there.

In Section~\ref{Sec4}, we get instances of $\clubsuit_{\ad}(\ldots)$ from the principles ${\ensuremath \mspace{2mu}\mid\mspace{-12mu} {\raise0.6em\hbox{$\bullet$}}}(\lambda^+)$ and $\diamondsuit(\mathfrak b)$.
In particular, the proof of Theorem~\ref{thmb} will be found there.

In the Appendix, we slightly extend Clause~(i) above, showing that for every regular uncountable cardinal $\kappa$, if $\clubsuit_{\ad}(\mathcal S,1,2)$ holds
for an infinite partition $\mathcal S$ of some nonreflecting stationary subset of $\kappa$
into $\mu$ many stationary sets,
then there are $2^\mu$ many pairwise nonhomeomorphic Dowker spaces of size $\kappa$.

\section{Unbounded colorings and generalized Luzin sets}\label{Sec2}
In this section, $\kappa$ denotes a regular uncountable cardinal, and $\nu$ and $\lambda$ denote infinite cardinals $<\kappa$.
\begin{defn}\label{deftrees} Suppose that $c:\nu\times\kappa\rightarrow\lambda$ is a coloring.
\begin{itemize}
\item For each $\beta<\kappa$, derive the fiber map $c_\beta:\nu\rightarrow\lambda$ via $c_\beta(\eta):=c(\eta,\beta)$;
\item $c$ is \emph{unbounded} iff for every cofinal $B\s\kappa$, there is an $\eta<\nu$ such that
$$\sup\{ c_\beta(\eta)\mid \beta\in B\}=\lambda;$$
\item $c$ is \emph{strongly unbounded} iff for every cofinal $B\s\kappa$, there are an $\eta<\nu$
and a map $t:\eta\rightarrow\lambda$ such that
$$\sup\{ c_\beta(\eta)\mid \beta\in B\  \&\ t\s c_\beta\}=\lambda;$$
\item For every $\mathbb T\s{}^{<\nu}\lambda$, let $[\mathbb T]_c:=\{ \beta<\kappa\mid \forall\eta<\nu\,(c_\beta\restriction\eta\in\mathbb  T)\}$;
\item Set $\mathcal T_c:=\{ \mathbb T\s{}^{<\nu}\lambda\mid \sup([\mathbb T]_c)=\kappa\}$;
\item $c$ is \emph{tight} iff $\cf(\mathcal T_c,{\supseteq})\le\kappa$.
\end{itemize}
\end{defn}
\begin{remark}
It is clear that ${}^{<\nu}\lambda\in\mathcal T_c$, so that $1\le|\mathcal T_c|\le2^{(\lambda^{<\nu})}$.
\end{remark}

\subsection{From unbounded to strongly unbounded}
\begin{lemma}\label{lemma23} Suppose that $\lambda=\aleph_0$ or $\lambda$ is strongly inaccessible.

Then any unbounded coloring $c:\lambda\times\kappa\rightarrow\lambda$
is strongly unbounded.
\end{lemma}
\begin{proof} Suppose that $c:\lambda\times\kappa\rightarrow\lambda$  is a given unbounded coloring. Let $B$
be some cofinal subset of $\kappa$.
By hypothesis, there exists an $\eta<\lambda$ such that
$\sup\{ c_\beta(\eta)\mid \beta\in B\}=\lambda$.
For the least such $\eta$, it follows that there exists some ordinal $\mu<\lambda$ such that
$$\{  c_\beta(i)\mid \beta\in B, i<\eta\}\s \mu.$$
As $|{}^\eta\mu|<\cf(\lambda)=\lambda$, there must exist some $t\in{}^\eta\mu$ such that
$$\sup\{ c_\beta(\eta)\mid \beta\in B\ \& \ t\s c_\beta\}=\lambda,$$
as sought.
\end{proof}

\begin{defn}[Shelah, {\cite[\S2]{Sh:107}}] For a regular uncountable cardinal $\lambda$:
\begin{enumerate}
\item $D\ell_\lambda$ asserts the existence of a sequence $\langle \mathcal P_\eta\mid\eta<\lambda\rangle$ such that:
\begin{itemize}
\item  for every $\eta<\lambda$, $\mathcal P_\eta\s\mathcal P(\eta)$ and $|\mathcal P_\eta|<\lambda$;
\item for every  $A\s\lambda$, for stationarily many $\eta<\lambda$,
$A\cap\eta\in\mathcal P_\eta$.
\end{itemize}
\item $D\ell^*_\lambda$ asserts the existence of a sequence $\langle \mathcal P_\eta\mid\eta<\lambda\rangle$ such that:
\begin{itemize}
\item  for every $\eta<\lambda$, $\mathcal P_\eta\s\mathcal P(\eta)$ and $|\mathcal P_\eta|<\lambda$;
\item for every  $A\s\lambda$, for club many $\eta<\lambda$,
$A\cap\eta\in\mathcal P_\eta$.
\end{itemize}
\end{enumerate}
\end{defn}

\begin{fact}[Shelah, {\cite[Claim 3.2]{sh:460} and \cite[Claim 2.5]{sh:922}}]\label{dlfacts} For a regular uncountable cardinal $\lambda$:
\begin{enumerate}
\item If $\lambda$ is strongly inaccessible, then $D\ell^*_\lambda$ holds;
\item $\diamondsuit_\lambda$ implies $D\ell_\lambda$,
and $\diamondsuit^*_\lambda$ implies $D\ell^*_\lambda$;
\item If $\lambda\ge\beth_\omega$ then $D\ell_\lambda$ iff $\lambda^{<\lambda}=\lambda$;
\item If $\lambda$ is a successor of an uncountable cardinal, then $D\ell_\lambda$ iff $\lambda^{<\lambda}=\lambda$.
\end{enumerate}
\end{fact}

\begin{lemma}\label{lemma26} Suppose that $\lambda$ is a regular uncountable cardinal and $D\ell^*_\lambda$ holds.
Then there exists a strongly unbounded coloring $c:\lambda\times\kappa\rightarrow\lambda$,
for $\kappa:=\mathfrak b_\lambda$.
\end{lemma}
\begin{proof} We commence with verifying the following variation of weak diamond.
\begin{claim} For every function $F:{}^{<\lambda}\lambda\rightarrow\lambda$,
there exists a function $g:\lambda\rightarrow\lambda$ with the property that for every function $f:\lambda\rightarrow\lambda$,
the following set covers a club:
$$\{\eta<\lambda\mid F(f\restriction\eta)\le g(\eta)\}.$$
\end{claim}
\begin{proof}
Using $D\ell^*_\lambda$, we may fix a sequence $\langle \mathcal F_\eta\mid\eta<\lambda\rangle$ such that:
\begin{itemize}
\item  for every $\eta<\lambda$, $\mathcal F_\eta\s{}^\eta\eta$ and $|\mathcal F_\eta|<\lambda$;
\item for every function $f:\lambda\rightarrow\lambda$, for club many $\eta<\lambda$,
$f\restriction\eta\in\mathcal F_\eta$.
\end{itemize}
Now, given any function  $F:{}^{<\lambda}\lambda\rightarrow\lambda$,
define an oracle function $g:\lambda\rightarrow\lambda$ via
$$g(\eta):=\sup\{ F(t)\mid t\in\mathcal F_\eta\}.$$
Next, given any function $f:\lambda\rightarrow\lambda$, the set $C:=\{ \eta<\lambda\mid f\restriction\eta\in\mathcal F_\eta\}$ covers a club,
and it is clear that, for every $\eta\in C$, $g(\eta)\ge F( f\restriction\eta)$.
\end{proof}

For functions $f,g\in{}^\lambda\lambda$, let $f<_{cl}g$ iff $\{\alpha<\lambda\mid f(\alpha)<g(\alpha)\}$ covers a club.
By \cite[Theorem~6]{CuSh:541}, $\mathfrak b_\lambda$ coincides with the least size of a family of functions from $\lambda$ to $\lambda$
that is not bounded with respect to $<_{cl}$. It follows that we may construct a $<_{cl}$-increasing sequence of functions $\langle f_\beta\mid\beta<\kappa\rangle$
for which $\{ f_\beta\mid \beta<\kappa\}$ is not bounded with respect to $<_{cl}$.
Define $c:\lambda\times\kappa\rightarrow\lambda$ via $c(\eta,\beta):=f_\beta(\eta)$.
Then, for every function $g:\lambda\rightarrow\lambda$,
for a tail of $\beta<\kappa$, $S_\beta(g):=\{\eta<\lambda\mid g(\eta)\le c_\beta(\eta)\}$ is stationary.
We claim that $c$ is strongly unbounded.
Towards a contradiction, suppose that this is not the case, as witnessed by a cofinal set $B$.
Then, we may define a function $F:{}^{<\lambda}\lambda\rightarrow\lambda$ by letting for all $\eta<\lambda$ and $t:\eta\rightarrow\lambda$,
$$F(t):=\sup\{ c_\beta(\eta)\mid \beta\in B, t\s c_\beta\}+1.$$
Now, pick a corresponding oracle $g:\lambda\rightarrow\lambda$ such that for every function $f:\lambda\rightarrow\lambda$,
the following set covers a club:
$$C_f:=\{\eta<\lambda\mid F(f\restriction\eta)\le g(\eta)\}.$$
Pick $\beta\in B$ such that $S_\beta(g)$ is stationary.
Then, find $\eta\in S_\beta(g)\cap C_{c_\beta}$.
Altogether, $c_\beta(\eta)<F(c_\beta\restriction\eta)\le g(\eta)\le c_\beta(\eta)$.
This is a contradiction.
\end{proof}

\begin{lemma}\label{lemma27} Suppose that $\lambda$ is a regular uncountable cardinal and $D\ell_\lambda$ holds.

If $\kappa=\mathfrak b_\lambda=\mathfrak d_\lambda$,
then there exists a strongly unbounded coloring $c:\lambda\times\kappa\rightarrow\lambda$.
\end{lemma}
\begin{proof} As $\kappa=\mathfrak b_\lambda=\mathfrak d_\lambda$,
it is possible to construct a coloring $c:\lambda\times\kappa\rightarrow\lambda$ with the property that for every function $g:\lambda\rightarrow\lambda$,
for a tail of $\beta<\kappa$, $$C_\beta(g):=\{\eta<\lambda\mid g(\eta)\le c_\beta(\eta)\}$$ is co-bounded in $\kappa$.\footnote{See \cite[\S6]{paper53}: for $\lambda$ regular, $\mathfrak b_\lambda=\mathfrak d_\lambda=\kappa$ implies that
$\ubd([\lambda]^\lambda,J^{\bd}[\kappa],\lambda)$ holds.}
We claim that $c$ is strongly unbounded.
Towards a contradiction, suppose that this is not the case, as witnessed by a cofinal set $B$.
Then, we may define a function $F:{}^{<\lambda}\lambda\rightarrow\lambda$ by letting for all $\eta<\lambda$ and $t:\eta\rightarrow\lambda$,
$$F(t):=\sup\{ c_\beta(\eta)\mid \beta\in B, t\s c_\beta\}+1.$$

As $D\ell_\lambda$ holds, we may pick a corresponding oracle $g:\lambda\rightarrow\lambda$ such that for every function $f:\lambda\rightarrow\lambda$,
the following set is stationary:
$$S_f:=\{\eta<\lambda\mid F(f\restriction\eta)\le g(\eta)\}.$$
Pick $\beta\in B$ such that $C_\beta(g)$ covers a club.
Then, find $\eta\in C_\beta(g)\cap S_{c_\beta}$.
Altogether, $c_\beta(\eta)<F(c_\beta\restriction\eta)\le g(\eta)\le c_\beta(\eta)$.
This is a contradiction.
\end{proof}

\subsection{Tightness}
\begin{lemma} Suppose that $\mathfrak d=\mathfrak c=\kappa$.
Then there exists a tight strongly unbounded coloring $c:\omega\times\kappa\rightarrow\omega$.
\end{lemma}
\begin{proof} It is easy to construct an unbounded coloring $c:\omega\times \mathfrak d\rightarrow\omega$ (see \cite[\S6]{paper53}).
By Lemma~\ref{lemma23}, $c$ is moreover strongly unbounded.
As $\mathcal T_c\s\mathcal P({}^{<\omega}\omega)$, it follows that $|T_c|\le\mathfrak c$.
So, if $\mathfrak d=\mathfrak c=\kappa$, then $c$ is tight.
\end{proof}

\begin{cor} Suppose that $\lambda=\lambda^{<\lambda}$ is an infinite cardinal satisfying any of the following:
\begin{itemize}
\item $\lambda=\aleph_0$, or
\item $\lambda=\aleph_1$ and $\diamondsuit_\lambda$ holds, or
\item $\lambda>\aleph_1$ is a successor cardinal, or
\item $\lambda\ge\beth_\omega$, or
\item $\lambda$ is strongly inaccessible.
\end{itemize}

If $\kappa=\mathfrak b_\lambda=2^\lambda$,
then there exists a tight strongly unbounded coloring $c:\lambda\times\kappa\rightarrow\lambda$.
\end{cor}
\begin{proof} As in the proof of the previous lemma, the fact that $\kappa=2^{\lambda^{<\lambda}}$
implies that any strongly unbounded coloring $c:\lambda\times\kappa\rightarrow\lambda$ is tight.
Assuming $\kappa=\mathfrak b_\lambda$, it is also easy to obtain an unbounded coloring $c:\lambda\times\kappa\rightarrow\lambda$.
So the heart of the matter is to get a strongly unbounded one.
Lemma~\ref{lemma23} takes care of the first and last bullet.
The remaining bullets follow from Lemma~\ref{lemma27} together with Fact~\ref{dlfacts}.
\end{proof}

\begin{defn} A \emph{$\kappa$-Luzin subset of ${}^\lambda\lambda$}
is a subset $L\s{}^\lambda\lambda$ of size $\kappa$ having the property that for every $B\in[L]^\kappa$,
there exists $t\in{}^{<\lambda}\lambda$ such that, for every $t'\in{}^{<\lambda}\lambda$ extending $t$,
there exists an element of $B$ extending $t'$.
\end{defn}

It is well-known that $\ma$ implies the existence of a  $\mathfrak c$-Luzin subset of ${}^\omega\omega$. More generally, $\cov(\mathcal M)=\cof(\mathcal M)=\kappa$ entails the existence of a $\kappa$-Luzin subset of ${}^\omega\omega$.
Also, the following fact is standard:

\begin{fact}[Luzin] For every infinite cardinal $\lambda=\lambda^{<\lambda}$,
if $2^\lambda=\lambda^+$, then there exists a $\lambda^+$-Luzin subset of ${}^\lambda\lambda$.
\end{fact}

\begin{lemma} Suppose that there exists a $\kappa$-Luzin subset of ${}^{\lambda}\lambda$, with $\kappa$ regular.

If $\lambda^{<\lambda}<\kappa$, then there exists a tight strongly unbounded coloring $c:\lambda\times\kappa\rightarrow\lambda$.
\end{lemma}
\begin{proof} Fix an injective enumeration $\vec g=\langle g_\beta\mid \beta<\kappa\rangle$ of a $\kappa$-Luzin subset of ${}^\lambda\lambda$.
Let $c:\lambda\times\kappa\rightarrow\lambda$ denote the unique coloring such that $c_\beta=g_\beta$ for all $\beta<\kappa$.
\begin{claim} $c$ is strongly unbounded.
\end{claim}
\begin{proof} Let $B\in [\kappa]^\kappa$; we need to find
$\eta<\lambda$ and a map $t:\eta\rightarrow\lambda$ such that
$\sup\{ c_\beta(\eta)\mid \beta\in B\  \&\ t\s c_\beta\}=\lambda$.
As $\im(\vec g)$ is a $\kappa$-Luzin subset of ${}^\lambda\lambda$,
fix some $t\in {}^{<\lambda}\lambda$ such that, for every $t'\in{}^{<\lambda}\lambda$ extending $t$,
there exists $l\in \{g_\beta \mid \beta \in B\}$ extending $t'$. Set $\eta:=\dom(t)$.
Then for every $\gamma<\lambda$, we can find $\beta \in B$ such that $g_\beta$ extends $t{}^\smallfrown\langle\gamma\rangle$.
Altogether, $\{ c_\beta(\eta)\mid \beta\in B\  \&\ t\s c_\beta\}=\lambda$.
\end{proof}
For every $s\in{}^{<\lambda}\lambda$, denote $\mathbb T_s:=\{ t\in{}^{<\lambda}\lambda\mid s\s t\text{ or }t\s s\}$.
\begin{claim} For every $\mathbb T\in\mathcal T_c$,
there exists  $s\in{}^{<\lambda}\lambda$ such that $\mathbb T\supseteq\mathbb T_s\in\mathcal T_c$.
\end{claim}
\begin{proof} Let $\mathbb T\in\mathcal T_c$,
so that $[\mathbb T]_c=\{ \beta<\kappa\mid \forall\eta<\lambda\,(c_\beta\restriction\eta\in\mathbb  T)\}$
is cofinal in $\kappa$. For every $t\in{}^{<\lambda}\lambda$, write $A_t:=\{ \beta\in [\mathbb T]_c\mid t\s c_\beta\}$.
As  $\lambda^{<\lambda}<\cf(\kappa)=\kappa$, the set $N:=\bigcup\{ A_t\mid t\in{}^{<\lambda}\lambda, |A_t|<\kappa\}$ has size $<\kappa$.
In particular, $B:= [\mathbb T]_c\setminus N$ has size $\kappa$.
As $\im(\vec g)$ is a $\kappa$-Luzin subset of ${}^\lambda\lambda$,
fix some $s\in {}^{<\lambda}\lambda$ such that, for every $s'\in{}^{<\lambda}\lambda$ extending $s$,
there exists $\beta \in B$ such that $s'\s g_\beta$. As $B\s[\mathbb T]_c$, it follows that $\mathbb T_s\s\mathbb T$.

	Finally, to show that $[\mathbb T_s]_c=\{ \beta<\kappa\mid \forall\eta<\lambda\,(c_\beta\restriction\eta\in\mathbb  T_s)\}$  is in $\mathcal T_c$,
	we need to prove that $\sup([\mathbb T_s]_c)=\kappa$.
Recalling that $s$ extends $g_\beta$ for some $\beta\in B\s(\kappa\setminus N)$, we infer that $|A_s|=\kappa$.
As $[\mathbb T_s]_c$ clearly covers $A_s$, we infer that $\sup([\mathbb T_s]_c)=\kappa$.
\end{proof}
In particular, $\cf(\mathcal T_c,{\supseteq})\le\lambda^{<\lambda}$. So, we are done.
\end{proof}

\section{Theorem~A}\label{Sec3}
\begin{defn}[{\cite[\S3.3]{paper36}}] Let $\lambda<\kappa$ be a pair of infinite cardinals, $e:[\kappa]^2\rightarrow\lambda$ be a coloring, and $S$ be a subset of $\kappa$.
\begin{enumerate}
\item $e$ is \emph{$S$-coherent} iff for all $\beta\le\gamma<\delta<\kappa$ with $\beta\in S$,
$$\sup\{\xi<\beta\mid e(\xi,\gamma)\neq e(\xi,\delta)\}<\beta;$$
\item $\partial(e):=\{ \alpha\in\acc(\kappa)\mid \forall \gamma<\kappa\forall \nu<\lambda\,\sup\{ \xi<\alpha\mid e(\xi,\gamma)\le\nu\}<\alpha\}$.
\end{enumerate}
\end{defn}
\begin{fact}[{\cite[Lemma~3.31]{paper36}}]\label{nonreflecting} Let $\lambda<\kappa$ be a pair of infinite regular cardinals.

For a stationary subset $S\s E^\kappa_{\lambda}$, the following are equivalent:
\begin{itemize}
\item $S$ is nonreflecting;
\item There exists an $S$-coherent coloring $e:[\kappa]^2\rightarrow\lambda$  such that $\partial(e)\supseteq S$.
\end{itemize}
\end{fact}

\begin{theorem}\label{maintheorem} Suppose:
\begin{enumerate}
\item $\theta\le\lambda<\kappa$ are infinite cardinals, with $\lambda,\kappa$ regular,
\item $c:\lambda\times\kappa\rightarrow\lambda$ is a strongly unbounded coloring,
\item $c$ is tight. Furthermore, $(\cf(\mathcal T_c,{\supseteq}))^{<\theta}\le\kappa$, and
\item $\mathcal S$ is a partition of some nonreflecting stationary subset of $E^{\kappa}_{\lambda}$ into stationary sets.
\end{enumerate}

Then there exists a club $C\s\kappa$ such that $\clubsuit_{\ad}(\{ S\cap C\mid S\in \mathcal S\},\allowbreak\lambda,{<}\theta)$ holds.
If either $\kappa=\lambda^+$ or $\lambda^{<\lambda}=\lambda$, then $C$ can moreover be taken to be whole of $\kappa$.
\end{theorem}
\begin{proof}  The proof is an elaboration of a construction from \cite[\S2]{TodPartTop}.
Let $\mathcal T$ be a dense subfamily of $\mathcal T_c$ of minimal size.
Let $\seq_{<\theta}(\mathcal T)$ denote the collection of all nonempty sequences of elements of $\mathcal T$ of length $<\theta$.
By Clause~(3) above, $|\seq_{<\theta}(\mathcal T)|\le\kappa$.
Let $\mathcal S$ be a given partition of some nonreflecting stationary subset $\mathbb{S}$ of $E^{\kappa}_{\lambda}$ into stationary sets.
Then, for every $S\in\mathcal S$, let $\langle S_\sigma\mid \sigma\in\seq_{<\theta}(\mathcal T)\rangle$ be a partition of $S$ into stationary sets.

\begin{claim}\label{claim3331} The set $\Sigma:=\{c_\beta\restriction\Lambda\mid \beta<\kappa, \Lambda<\lambda\}$ has size $<\kappa$.
\end{claim}
\begin{proof} Suppose not. Since $\kappa$ is a regular cardinal greater than $\lambda$, it follows that there exist $B\in[\kappa]^\kappa$ and $\Lambda<\lambda$ on which the map $\beta\mapsto c_\beta\restriction\Lambda$
is injective. By possibly shrinking $B$ further, we may also assume the existence of some $\epsilon<\lambda$ such that $c_\beta[\Lambda]\s\epsilon$ for all $\beta\in B$.
But then $c$ cannot be strongly unbounded. Indeed, for every $t:\eta\rightarrow\lambda$, if $\eta<\Lambda$, then $\{ c_\beta(\eta)\mid \beta\in B, t\s c_\beta\}\s \epsilon$,
and if $\eta\in[\Lambda,\lambda)$, then $|\{ c_\beta(\eta)\mid \beta\in B, t\s c_\beta\}|\le1$.
\end{proof}
Set $T^*:=\{ t\in \Sigma\mid |\{\beta<\kappa\mid t\s c_\beta\}|=\kappa\}$.
By Claim~\ref{claim3331}, $|T^*|<\kappa$, so we may fix a surjection $f:\kappa\rightarrow T^*$ with the property that for every $\epsilon<\kappa$,
$\{f(\xi+1)\mid \epsilon<\xi<\epsilon+|T^*|\}=T^*$.
By Claim~\ref{claim3331}, we may also fix a large enough ordinal $\varrho<\kappa$ such that $\bigcup\{\{ \beta<\kappa \mid t\s c_\beta\}\mid  t\in \Sigma\setminus T^*\}\s\varrho$.

Next, we turn to recursively define an injective matrix $\langle \beta_{\alpha, j}\mid \alpha<\kappa,  j<\zeta(\alpha)\rangle$ of ordinals in $\kappa$, as follows.
Suppose that $\alpha<\kappa$ and that
$\langle \beta_{\bar\alpha,\bar j}\mid\bar\alpha<\alpha,\bar j<\zeta(\bar\alpha)\rangle$ has already been defined.
\begin{itemize}
\item[$\blacktriangleright$] If $\{ (S,\sigma)\in\mathcal S\times\seq_{<\theta}(\mathcal T)\mid \alpha\in S_\sigma\}$ is nonempty,
then it is a singleton, so let $(S,\sigma)$ denote its unique element.
Write $\sigma=\langle \mathbb T_ j\mid  j<\zeta\rangle$. Set $\zeta(\alpha):=\zeta$,
and then, by recursion on $ j<\zeta(\alpha)$, set
\begin{equation}\label{defofbetaalpha}
\tag{$\star$}\beta_{\alpha,j}:=\min([\mathbb T_ j]_c\setminus(\varrho\cup\{ \beta_{\bar\alpha,\bar j},\beta_{\alpha,j'}\mid\bar\alpha<\alpha , \bar  j\le \zeta(\bar\alpha),  j'< j\})).
\end{equation}
\item[$\blacktriangleright$] Otherwise, set $\zeta(\alpha):=1$ and
\begin{equation}\label{defofbetaalpha2}
\tag{$\star\star$}\beta_{\alpha,0}:=\min(\{\beta<\kappa\mid f(\alpha)\s c_\beta\}\setminus(\varrho\cup\{\beta_{\bar\alpha,\bar j}\mid \bar\alpha<\alpha, \bar j<\zeta(\bar\alpha)\})).
\end{equation}
\end{itemize}

Having constructed the above injective matrix of ordinals in $\kappa$,
we derive a corresponding injective matrix
$\vec d=\langle d_{\alpha, j}\mid\alpha<\kappa, j<\zeta(\alpha)\rangle$  by setting $d_{\alpha, j}:=c_{\beta_{\alpha, j}}$.

For all $x\neq y$ in ${}^\lambda\lambda$, denote $\Delta(x,y):=\min\{\eta<\lambda\mid x(\eta)\neq y(\eta)\}$.
As $\lambda$ is regular, for every $x\in {}^\lambda\lambda$, we may attach a strictly increasing function $\hat x:\lambda\rightarrow\lambda$
satisfying $\hat x(\eta)\ge x(\eta)$ for all $\eta<\lambda$.

Next, as $\mathbb{S}$ is a nonreflecting stationary subset of $E^{\kappa}_{\lambda}$,
by Fact~\ref{nonreflecting}, we may fix a coloring $e:[\kappa]^2\rightarrow\lambda$ that is $\mathbb{S}$-coherent
and such that $\mathbb{S}\s \partial(e)$.

Fix a surjection $\varsigma:\kappa\rightarrow\lambda$ such that $\partial(e)\cap \varsigma^{-1}\{i\}$ is stationary for every $i<\lambda$.
Next, for every $\alpha<\kappa$ and $ j<\zeta(\alpha)$, define a map $h_{\alpha, j}:\alpha\rightarrow\lambda$ via:
$$h_{\alpha, j}(\xi):=\varsigma(\min\{ \gamma\in(\xi,\alpha]\mid \gamma=\alpha\text{ or }e(\gamma,\alpha)\le{{\Delta(d_{\xi,0},d_{\alpha, j})}}\}).$$

For all $\alpha<\kappa$ and $i<\lambda$,
let $A^i_{\alpha,*}:=\bigcup_{ j<\zeta(\alpha)}A^i_{\alpha, j}$,
where for every $ j<\zeta(\alpha)$:
$$A^i_{\alpha, j}:=\{\xi<\alpha\mid  h_{\alpha, j}(\xi)=i\ \&\ e(\xi,\alpha)\le \hat{d}_{\alpha, j}(\Delta(d_{\xi,0},d_{\alpha, j}))\}.$$

\begin{claim}\label{claim331} Suppose that $\alpha\in\partial(e)$, $\beta<\kappa$, $j<\zeta(\alpha), j'<\zeta(\beta)$, and $i,i'<\lambda$.

If $(\alpha,j)\neq(\beta,j')$. Then $\sup(A_{\alpha, j}^i\cap A^{i'}_{\beta, j'})<\alpha$.
\end{claim}
\begin{proof} Suppose that $(\alpha,j)\neq(\beta,j')$,
and then let $\eta:=\Delta(d_{\alpha, j},d_{\beta, j'})$.
Towards a contradiction, suppose that $A^i_{\alpha, j}\cap A^{i'}_{\beta, j'}$ is cofinal in $\alpha$.
Set  $\nu:=\hat d_{\alpha, j}(\eta)$.
As $\alpha\in\partial(e)$, the following set is cofinal in $\alpha$:
$$Y:=\{ \xi\in A^i_{\alpha, j}\cap A^{i'}_{\beta, j'}\mid e(\xi,\alpha)>\nu\}.$$

For every $\xi\in Y$, $\hat d_{\alpha, j}(\eta)=\nu<e(\xi,\alpha)\le \hat d_{\alpha, j}(\Delta( d_{\xi,0},d_{\alpha, j}))$,
so since $\hat d_{\alpha, j}$ is strictly increasing, $\Delta( d_{\xi,0},d_{\alpha, j})>\eta=\Delta(d_{\alpha, j},d_{\beta, j'})$,
and hence $\Delta(d_{\xi,0},d_{\beta, j'})=\eta$.
Set $\tau:=\hat d_{\beta, j'}(\eta)$. As $Y\s \alpha\cap A^{i'}_{\beta, j'}$, altogether $Y\s \{\xi<\alpha\mid e(\xi,\beta)\le \tau\}$.
As $\alpha\in\partial(e)$, $Y$ is bounded in $\alpha$. This is a contradiction.
\end{proof}

\begin{claim}\label{claim332} Let $\alpha\in\mathbb S$ and $i\neq i'$ in $\lambda$. Then $\sup(A^{i}_{\alpha,*}\cap A^{i'}_{\alpha,*})<\alpha$.
\end{claim}
\begin{proof}  Suppose not. As
$\zeta(\alpha)<\theta\le \lambda=\cf(\alpha)$, there must exist $ j,j'<\zeta(\alpha)$ such that
$\sup(A^i_{\alpha, j}\cap A^{i'}_{\alpha, j'})=\alpha$. As $\alpha\in\mathbb S\s\partial(e)$, Claim~\ref{claim331} implies that $j=j'$.
But it is evident that $A^i_{\alpha, j}$ and $A^{i'}_{\alpha, j}$ are disjoint.
\end{proof}

For all $\alpha\in\mathbb S$ and $i<\lambda$, let $$A^i_\alpha:=A^i_{\alpha,*}\setminus\bigcup_{i'<i} A^{i'}_{\alpha,*}.$$
Clearly, $\langle A^i_\alpha\mid i<\lambda\rangle$ consists of pairwise disjoint subsets of $\alpha$.
\begin{claim} Let $(\alpha,\beta)\in[\mathbb S]^2$ and $i,i'<\lambda$. Then $\sup(A^i_\alpha\cap A^{i'}_\beta)<\alpha$.
\end{claim}
\begin{proof} Suppose not. In particular, $\sup(A^i_{\alpha, *}\cap A^{i'}_{\beta, *})=\alpha$.
However, $\zeta(\alpha),\zeta(\beta)<\theta\le \lambda=\cf(\alpha)$, so there must exist $ j<\zeta(\alpha)$ and $ j'<\zeta(\beta)$ such that
$\sup(A^i_{\alpha, j}\cap A^{i'}_{\beta, j'})=\alpha$, contradicting Claim~\ref{claim331}.
\end{proof}

Now, we turn to inspect the guessing features of the matrix $\langle A_\alpha^i\mid \alpha\in\mathbb S, i<\lambda\rangle$.

\begin{claim}\label{claim312} Let $S\in\mathcal S$, and let $\langle X_ j\mid  j<\zeta\rangle$ be any sequence of cofinal subsets of $\kappa$ with $0<\zeta<\theta$.
Then $\{ \alpha\in S\mid \forall i<\lambda\forall j<\zeta\,\sup(A^i_\alpha\cap X_ j)=\alpha\}$ is stationary.
\end{claim}
\begin{proof} For all $ j<\zeta$ and $t\in \Sigma$, denote $X_ j^t:=\{ \xi\in X_ j\mid t\s d_{\xi,0}\}$.
Set $$T_ j:=\{ t\in \Sigma\mid |X_ j^t|=\kappa\}.$$
By Claim~\ref{claim3331}, $N_ j:=\bigcup\{ X_ j^t\mid t\in \Sigma\setminus T_ j\}$ is the small union of sets of size $<\kappa$, so that $|X_ j\setminus N_ j|=\kappa$.
For all $\xi\in X_ j\setminus N_ j$ and $\eta<\lambda$,
$c_{\beta_{\xi,0}}\restriction\eta=d_{\xi,0}\restriction\eta\in T_ j$,
so that $[T_ j]_c$ covers $\{ \beta_{\xi,0}\mid \xi\in X_ j\setminus N_ j\}$, and
hence $T_ j\in\mathcal T_c$.
Recalling that $\mathcal T$ is dense in $\mathcal T_c$,
we may now pick $\mathbb T_ j\in\mathcal T$ with $\mathbb T_j\s T_j$.
In particular, $\sigma:=\langle \mathbb T_ j\mid j<\zeta\rangle$ is in $\seq_{<\theta}(\mathcal T)$,
and $S_\sigma$ is stationary.

Towards a contradiction, suppose that $\{ \alpha\in S\mid \forall i<\lambda\forall j<\zeta\,\sup(A^i_\alpha\cap X_ j)=\alpha\}$ is nonstationary.
As $S_\sigma$ is a stationary subset of $S$, we may fix $i<\lambda$ and $j<\zeta$
for which the following set is stationary:
$$S^0:=\{\alpha\in S_\sigma\mid \sup(A^i_\alpha\cap X_ j)<\alpha\}.$$
For every $\alpha\in S^0$, since $i<\lambda=\cf(\alpha)$ and since $A^i_\alpha=A^i_{\alpha,*}\setminus\bigcup_{i'<i} A^{i'}_{\alpha,*}$,
Claim~\ref{claim332} implies that $\sup(A^i_{\alpha,*}\cap X_ j)<\alpha$.
In particular, $\sup(A^i_{\alpha,j}\cap X_ j)<\alpha$.
So by Fodor's lemma, we may fix an $\epsilon<\kappa$ such that the following set is stationary:
$$S^1:=\{\alpha\in S_\sigma\mid \sup(A^i_{\alpha,j}\cap X_ j)=\epsilon<\alpha\}.$$

By the choice of the map $\varsigma$, the set $\Gamma$ of all $\gamma\in\partial(e)\cap\varsigma^{-1}\{i\}$
for which there exists an elementary submodel $M_\gamma\prec H_{\kappa^+}$
containing $\{\vec d,X_ j,\Sigma\}$ and satisfying $\gamma=M_\gamma\cap\kappa$ is stationary.
Fix $\delta\in\mathbb S\cap\acc^+(\Gamma\setminus\epsilon)$.
As $\delta\in\mathbb S$ and $e$ is $\mathbb{S}$-coherent, we may fix $S^2\in[S^1\setminus\delta]^{\kappa}$ along with some $\varepsilon<\delta$ such that, for every $\alpha\in S^2$,
$$\{ \xi<\delta\mid e(\xi,\alpha)\neq e(\xi,\delta)\}\s\varepsilon.$$
Pick $\gamma\in\Gamma\cap\delta$ above $\max\{\epsilon,\varepsilon\}$,
and then fix a model $M_\gamma$ witnessing that $\gamma\in\Gamma$.

Put $\nu:=e(\gamma,\delta)$.
By Claim~\ref{claim3331}, we may find a cofinal subset of $S^3\s S^2$ on which the map $\alpha\mapsto d_{\alpha, j}\restriction\nu$ is constant.

Next, as $c$ is strongly unbounded and $\{ \beta_{\alpha, j}\mid \alpha\in S^3\}$ is cofinal in $\kappa$,
we may find an ordinal $\eta<\lambda$ and a map $t:\eta\rightarrow\lambda$ such that
$$\sup\{ c_{\beta_{\alpha,j}}(\eta)\mid \alpha\in S^3, t\s c_{\beta_{\alpha,j}}\}=\lambda.$$
Equivalently, for every $\tau<\lambda$, for some $\alpha\in S^3$, $d_{\alpha, j}\restriction\eta=t$ and $d_{\alpha, j}(\eta)>\tau$.
Clearly, $\eta\ge\nu$.

Pick for a moment $\alpha^*\in S^3$ such that $t\s d_{\alpha^*, j}$.
Since $\alpha^*\in S^3\s S_\sigma$,
Equation~\eqref{defofbetaalpha} and the definition of $\sigma$ implies that $\beta_{\alpha^*, j}$ is in $[\mathbb T_ j]_c$.
Recalling Definition~\ref{deftrees}, from $c_{\beta_{\alpha^*, j}}\restriction\eta=d_{\alpha^*, j}\restriction\eta=t$, we infer that $t\in \mathbb T_ j$.
As $\mathbb T_ j\s T_ j$, this means that $|X_ j^t|=\kappa$.
It thus follows from $\{\vec d,X_ j,\Sigma\}\in M_\gamma$ that $\sup(X_ j^t\cap\gamma)=\gamma$.
Now, as $\gamma\in\partial(e)$, $G:=\{\bar\gamma<\gamma\mid e(\bar\gamma,\delta)\le{{\eta}}\}$ is bounded below $\gamma$.
Altogether, we may find $\xi\in X_ j^t\cap \gamma$ above $\max\{\epsilon,\varepsilon,\sup(G)\}$.

Set $\tau:=\max\{e(\xi,\delta),d_{\xi,0}(\eta)\}$,
and then pick $\alpha\in S^3$ such that $d_{\alpha, j}\restriction\eta=t$ and $d_{\alpha, j}(\eta)>\tau$.
As $d_{\alpha, j}\restriction\eta=t=d_{\xi,0}\restriction\eta$ and $d_{\alpha, j}(\eta)>d_{\xi,0}(\eta)$, we infer that $\Delta(d_{\xi,0},d_{\alpha, j})=\eta$.
As $\varepsilon<\xi<\delta$, altogether, $$\hat d_{\alpha, j}(\Delta(d_{\xi,0},d_{\alpha, j}))=\hat d_{\alpha, j}(\eta)\ge d_{\alpha, j}(\eta)>\tau\ge e(\xi,\delta)=e(\xi,\alpha).$$
Next, from $\Delta(d_{\xi,0},d_{\alpha, j})=\eta$ and the fact that $\xi>\varepsilon$,
we also infer that $$h_{\alpha, j}(\xi)=\varsigma(\min\{ \bar\gamma\in(\xi,\alpha]\mid \bar\gamma=\alpha\text{ or }e(\bar\gamma,\delta)\le{{\eta}}\}).$$
Since $e(\gamma,\delta)=\nu\le{{\eta}}$ and $\varsigma(\gamma)=i$,
it follows that if $h_{\alpha, j}(\xi)\neq i$, then there exists $\bar\gamma\in(\xi,\gamma)$ such that $e(\bar\gamma,\delta)\le{{\eta}}$,
contradicting the fact that $\xi>\sup(G)$.
So, it is the case that $h_{\alpha, j}(\xi)=i$. Consequently, $\xi\in A^i_{\alpha,j}$.

Altogether, we established that $\xi$ is an element of $A^i_{\alpha,j}\cap X_j$ above $\epsilon$, contradicting the fact that $\alpha\in S^3\s S^1$.
\end{proof}

The next claim implies that there exists a club $C\s\kappa$ such that,
for every $\alpha\in C$,
for every $i<\lambda$, $\sup(A^i_\alpha)=\alpha$.

\begin{claim} Let $S\s\mathbb S$ be stationary.
Then $\{ \alpha\in S\mid \forall i<\lambda\,\sup(A^i_\alpha\cap S)=\alpha\}$ is stationary.
\end{claim}
\begin{proof} Suppose not, and fix $i<\lambda$
for which the following set is stationary:
$$S^0:=\{\alpha\in S\mid \sup(A^i_\alpha\cap S)<\alpha\}.$$
It follows that there exists an $\epsilon<\kappa$ such that
$$S^1:=\{\alpha\in S\mid \sup(A^i_{\alpha,0}\cap S)=\epsilon<\alpha\}.$$
Similarly to the proof of the previous claim,
find ordinals $\varepsilon<\gamma<\delta$ and a set $S^2\in[S^1\setminus\delta]^\kappa$ such that:
\begin{itemize}
\item $\gamma,\delta\in\partial(e)$;
\item $\delta=M_\delta\cap\kappa$ for some elementary submodel $M_\delta\prec H_{\kappa^+}$ containing $\{\vec d,S,\Sigma\}$;
\item $\gamma=M_\gamma\cap\kappa$ for some elementary submodel $M_\gamma\prec H_{\kappa^+}$ containing $\{\vec d,S,\Sigma\}$;
\item for every $\alpha\in S^2$, $\{ \xi<\delta\mid e(\xi,\alpha)\neq e(\xi,\delta)\}\s\varepsilon$.
\end{itemize}

Put $\nu:=e(\gamma,\delta)$.
Then find a cofinal subset of $S^3\s S^2$ on which the map $\alpha\mapsto d_{\alpha,0}\restriction\nu$ is constant.
As $\{ \beta_{\alpha,0}\mid \alpha\in S^3\}$ is cofinal in $\kappa$,
the choice of the coloring $c$ provides an ordinal $\eta<\lambda$ and a map $t:\eta\rightarrow\lambda$ such that,
for every $\tau<\lambda$, for some $\alpha\in S^3$, $d_{\alpha,0}\restriction\eta=t$ and $d_{\alpha,0}(\eta)>\tau$.
The same analysis is true for any final segment of $S^3$ and hence, by Clause~(1) and the pigeonhole principle,
we may fix some $t:\eta\rightarrow\lambda$ such that,
for every $\tau<\lambda$, for cofinally many $\alpha\in S^3$, $d_{\alpha,0}\restriction\eta=t$ and $d_{\alpha,0}(\eta)>\tau$.
Clearly, $\eta\ge\nu$.

As $\{\vec d,S,\Sigma\}\in M_\delta$ and $S^3\cap M_\delta=\emptyset$, by elementarity, the set of $\xi\in S\cap M_\delta$ such that $d_{\xi,0}\restriction\eta=t$
is cofinal in $\gamma$.
As $\gamma\in\partial(e)$, $G:=\{\bar\gamma<\gamma\mid e(\bar\gamma,\delta)\le{{\eta}}\}$ is bounded below $\gamma$.
So, we may find $\xi\in S\cap \gamma$ above $\max\{\epsilon,\varepsilon,\sup(G)\}$ such that $d_{\xi,0}\restriction\eta=t$.
Set $\tau:=\max\{e(\xi,\delta),d_{\xi,0}(\eta)\}$,
and then pick $\alpha\in S^3$ such that $d_{\alpha,0}\restriction\eta=t$ and $d_{\alpha,0}(\eta)>\tau$.
From this point on, a verification identical to that of Claim~\ref{claim312}
shows that $\xi$ is an element of $A^i_{\alpha,0}\cap S$ above $\epsilon$, contradicting the fact that $\alpha\in S^3\s S^1$.
\end{proof}

In summary, we have shown that there exists a club $C\s\kappa$ such that:
\begin{enumerate}
\item[(a)] For every $\alpha\in C$, for every $i<\lambda$, $\sup(A^i_\alpha)=\alpha$;
\item[(b)] For every $S\in\mathcal S$, for every sequence $\langle X_ j\mid  j<\zeta\rangle$ of cofinal subsets of $\kappa$ with $0<\zeta<\theta$,
the set $\{ \alpha\in S\mid \forall i<\lambda\forall j<\zeta\,\sup(A^i_\alpha\cap X_ j)=\alpha\}$ is stationary;
\item[(c)] For all $(\alpha,\beta)\in[\mathbb S]^2$ and $i,i'<\lambda$,
$\sup(A^i_\alpha\cap A^{i'}_\beta)<\alpha$.
\end{enumerate}

Suppose now that either $\kappa=\lambda^+$ or $\lambda^{<\lambda}=\lambda$,
and let us prove that $\clubsuit_{\ad}(\mathcal S,\lambda,{<}\theta)$ holds. For this, it suffices to define for every $\alpha\in \mathbb S\setminus C$,
a sequence $\langle a_\alpha^i\mid i<\lambda\rangle$ of pairwise disjoint cofinal subsets of $\alpha$ such that
the amalgam of $\langle \langle a_\alpha^i\mid i<\lambda\rangle\mid \alpha\in \mathbb S\setminus C\rangle $
and $\langle \langle A_\alpha^i\mid i<\lambda\rangle\mid \alpha\in \mathbb S\cap C\rangle $ will form an almost-disjoint system.
To this end, let $\alpha\in \mathbb S\setminus C$.
\begin{claim} For every $\eta<\lambda$, $\sup\{\xi<\alpha\mid \Delta(d_{\xi,0},d_{\alpha,0})\ge\eta\}=\alpha$.
\end{claim}
\begin{proof} Let $\eta<\lambda$, $t:=d_{\alpha,0}\restriction\eta$ and $\epsilon<\alpha$;
we need to find $\xi$ with $\epsilon<\xi<\alpha$ such that $t\s d_{\xi,0}$.
Now, recall that by the construction of $\vec d$, $d_{\alpha,0}=c_\beta$ for some ordinal $\beta\in\kappa\setminus\varrho$.
Consequently, $t=c_\beta\restriction\eta\in T^*$.
So since either $\kappa=\lambda^+$ or $\lambda^{<\lambda}=\lambda$,
Claim~\ref{claim3331} implies that $|T^*|\le\lambda=\cf(\alpha)$. Then, since the surjection $f$ was chosen to satisfy
$\{f(\xi+1)\mid \epsilon<\xi<\epsilon+|T^*|\}=T^*$,
we may find some $\xi$ with $\epsilon<\xi<\xi+1<\alpha$ such that $f(\xi+1)=t$.
As $\bigcup\mathcal S\s \mathbb S$,
we get from Equation~\eqref{defofbetaalpha2} that $d_{\xi+1,0}=c_{\beta_{\xi+1,0}}\supseteq t$,
and hence $\Delta(d_{\xi+1,0},d_{\alpha,0})\ge\eta$.
\end{proof}

Using the preceding claim, fix a strictly increasing sequence $\langle \xi_\eta^\alpha\mid \eta<\lambda\rangle$ of ordinals, converging to $\alpha$,
such that, for every $\eta<\lambda$, $\Delta(d_{\xi_\alpha^\alpha,0},d_{\alpha,0})\ge\eta$.
Then, let $\langle a_\alpha^i\mid i<\lambda\rangle$ be some partition of $\{\xi_\eta^\alpha\mid \eta<\lambda\}$ into $\lambda$ many sets of size $\lambda$.

As each $a_\alpha^i$ has order-type $\lambda$, the verification of almost-disjointness of the merged systems boils down to verifying the following case.
\begin{claim} Let $\alpha\in \mathbb S\setminus C$ and $\beta\in \mathbb S\cap C$ above $\alpha$.
Let $i,{i'}<\lambda$. Then $\sup(a_\alpha^i\cap A_\beta^{i'})<\alpha$.
\end{claim}
\begin{proof} Suppose not. Fix $ j'<\zeta(\beta)$ such that
$a_{\alpha}^i\cap A^{i'}_{\beta, j'}$ is cofinal in $\alpha$.
Set $\eta:=\Delta(d_{\alpha,0},d_{\beta, j'})$. By the choice of $a_\alpha^i$, $\{ \xi\in a^i_\alpha\mid \Delta(d_{\xi,0},d_{\alpha,0})\le\eta\}$ is bounded in $\alpha$,
and hence the following set is cofinal in $\alpha$:
$$Y:=\{ \xi\in a_{\alpha}^i\cap A^{i'}_{\beta, j'}\mid \Delta(d_{\xi,0},d_{\alpha,0})>\eta\}.$$

For every $\xi\in Y$,  $\Delta( d_{\xi,0},d_{\alpha,0})>\eta=\Delta(d_{\alpha,0},d_{\beta, j'})$,
and hence $\Delta(d_{\xi,0},d_{\beta, j'})=\eta$.
Set $\tau:=\hat d_{\beta, j'}(\eta)$. As $Y\s\alpha\cap A_{\beta, j'}$, altogether $Y\s \{\xi<\alpha\mid e(\xi,\beta)\le \tau\}$.
As $\alpha\in \mathbb{S}\s \partial(e)$, $Y$ is bounded in $\alpha$. This is a contradiction.
\end{proof}
This completes the proof.
\end{proof}

\subsection{Variations}
A second read of the proof of Theorem~\ref{maintheorem}
makes it clear that the conclusion remains valid even after relaxing Clause~(3) in the hypothesis
to $\cov(\cf(\mathcal T_c,{\supseteq}),\allowbreak\lambda,\theta,2)\le\kappa$.
In the other direction,
by waiving Clause~(3) completely, the above proof yields the following:
\begin{theorem}\label{thm34} Suppose $\lambda<\kappa$ is a pair of infinite regular cardinals,
and $\mathbb S$ is a nonreflecting stationary subset of $E^\kappa_\lambda$.

If there exists a strongly unbounded coloring $c:\lambda\times\kappa\rightarrow\lambda$,
then there exists a club $C\s\kappa$ and a matrix
$\langle A_\alpha^i\mid \alpha\in\mathbb S\cap C, i<\alpha\rangle$ such that:
\begin{enumerate}
\item For every $\alpha\in\mathbb S\cap C$,
$\langle A^i_\alpha\mid i<\alpha\rangle$ is a sequence
of pairwise disjoint cofinal subsets of $\alpha$;
\item For every stationary $S\s\mathbb S$,
there are stationarily many $\alpha\in S\cap C$ such that $\sup(A^i_\alpha\cap S)=\alpha$ for all $i<\alpha$;
\item For all $(\alpha,\alpha')\in[\mathbb S\cap C]^2$, $i<\alpha$ and $i'<\alpha'$,
$\sup(A^i_\alpha\cap A^{i'}_{\alpha'})<\alpha$.
\end{enumerate}

In the special case that $\kappa=\lambda^+$ or $\lambda^{<\lambda}=\lambda$, one can take $C$ to be whole of $\kappa$.\qed
\end{theorem}

Let $\clubsuit_{\ad^*}(\mathcal S,\mu,{<}\theta)$
denote the strengthening of $\clubsuit_{\ad}(\mathcal S,\mu,{<}\theta)$
obtained by replacing Clause~(3) of Definition~\ref{clubad} by:
\begin{enumerate}
\item[($3^*$)] For all $A\neq A'$ from $\bigcup_{S\in\mathcal S}\bigcup_{\alpha\in S}\mathcal A_\alpha$, $|A\cap A'|<\cf(\sup(A))$.
		\end{enumerate}
In the special case that $\kappa=\lambda^+$, one can use in the proof of Theorem~\ref{maintheorem} a \emph{locally small} coloring $e:[\kappa]^2\rightarrow\lambda$ (such as the map $\rho_1$ from \cite[\S6.2]{TodWalks}), and then get:

\begin{theorem} Suppose:
\begin{enumerate}
\item $\lambda$ is an infinite regular cardinal,
\item there exists a tight strongly unbounded coloriung $c:\lambda\times\lambda^+\rightarrow\lambda$, and
\item $\mathcal S$ is a partition of $E^{\lambda^+}_{\lambda}$ into stationary sets.
\end{enumerate}

Then $\clubsuit_{\ad^*}(\mathcal S,\lambda,{<}\lambda)$ holds.\qed
\end{theorem}

\section{Theorem~B}\label{Sec4}
In this section, we give two sufficient conditions for a strong form of $\clubsuit_{\ad}$ to hold.
The strong form under discussion is a double strengthening of $\clubsuit_{\ad}(\mathcal S,\mu,{<}\kappa)$,
and it reads as follows.
\begin{defn}  Let $\mathcal S$ be a collection of stationary subsets of a regular uncountable cardinal $\kappa$,
and $\mu$ be a nonzero cardinal $<\kappa$.
The principle
$\clubsuit_{\ad^*}(\mathcal S,\mu,\kappa)$ asserts the existence of a sequence $\langle \mathcal A_\alpha\mid \alpha\in\bigcup\mathcal S\rangle$ such that:
\begin{enumerate}
\item For every $\alpha\in\acc(\kappa)\cap\bigcup\mathcal S$, $\mathcal A_\alpha$ is a pairwise disjoint family of $\mu$ many cofinal subsets of $\alpha$;
\item For every sequence $\langle B_i\mid i<\kappa\rangle$ of cofinal subsets of $\kappa$, for every $S\in\mathcal S$, there are stationarily many $\alpha\in S$ such that,
for all $A\in\mathcal A_\alpha$ and $i<\alpha$,  $\sup(A\cap B_i)=\alpha$;
\item For all $A\neq A'$ from $\bigcup_{S\in\mathcal S}\bigcup_{\alpha\in S}\mathcal A_\alpha$, $|A\cap A'|<\cf(\sup(A))$.
\end{enumerate}
\end{defn}

An inspection of the proof \cite[\S3]{MR4068775} yields the following useful fact:

\begin{fact}[{\cite{MR4068775}}]\label{claim512}
For an infinite cardinal $\lambda$, the following are equivalent:
\begin{enumerate}
\item ${\ensuremath \mspace{2mu}\mid\mspace{-12mu} {\raise0.6em\hbox{$\bullet$}}}(\lambda^+)$ holds, i.e.,
there exists a sequence $\langle x_\beta\mid \beta<\lambda^+\rangle$
of elements of $[\lambda^+]^\lambda$
such that, for every cofinal $X\s\lambda^+$, there exists $\beta<\lambda^+$ such that $x_\beta\s X$;
\item There exists a sequence $\langle x_\beta\mid \beta<\lambda^+\rangle$
of elements of $[\lambda^+]^\lambda$ satisfying the following.
For every sequence $\langle A_\alpha\mid \alpha<\lambda^+\rangle$ of elements of $[\lambda^+]^{\le\lambda}$ such that $|A_\alpha\cap A_\beta|<\lambda$ for all $\alpha<\beta<\lambda^+$,
for every cofinal $X\s\lambda^+$, there exists $\beta<\lambda^+$ such that $x_\beta\s X$ and,
for every $a\in[\lambda^+]^{<\cf(\lambda)}$, $|x_\beta\setminus\bigcup_{\alpha\in a}A_\alpha|=\lambda$.
\end{enumerate}
\end{fact}

\begin{theorem} Suppose that ${\ensuremath \mspace{2mu}\mid\mspace{-12mu} {\raise0.6em\hbox{$\bullet$}}}(\lambda^+)$ holds for an infinite regular cardinal $\lambda$.

For every partition $\mathcal S$ of $E^{\lambda^+}_\lambda$ into stationary sets,
$\clubsuit_{\ad^*}(\mathcal S,\lambda,\lambda^+)$ holds.
\end{theorem}
\begin{proof} Let $\vec x=\langle x_\beta\mid\beta<\lambda^+\rangle$ be given by Fact~\ref{claim512}(2).
Fix a bijection $\pi:\lambda\leftrightarrow\lambda\times\lambda$ and then let $\pi_0,\pi_1$ be the unique maps from $\lambda$ to $\lambda$
to satisfy $\pi(j)=(\pi_0(j),\pi_1(j))$ for all $j<\lambda$.
For all nonzero $\alpha<\lambda^+$, fix a surjection $e_\alpha:\lambda\rightarrow\alpha$.

Define a sequence $\langle A_\alpha\mid\alpha<\lambda^+\rangle$ by recursion on $\alpha<\lambda^+$, as follows.
Set $A_\alpha:=\emptyset$.
Next, given a nonzero $\alpha<\lambda^+$ such that
$\langle A_{\bar\alpha}\mid\bar\alpha<\alpha\rangle$ has already been defined,
put $$J_\alpha:=\{ j<\lambda\mid |x_{e_\alpha(\pi_0(j))}\cap\alpha\setminus\bigcup\{ A_{e_\alpha(j')}\mid {j'\le j}\}|=\lambda\}.$$
Then, pick an injective sequence $\langle \xi_{\alpha,j}\mid j\in J_\alpha\rangle$
such that, for each $j\in J_\alpha$, $$\xi_{\alpha,j}\in x_{e_\alpha(\pi_0(j))}\cap\alpha\setminus\bigcup\{ A_{e_\alpha(j')}\mid {j'\le j}\}.$$
If $\{ \xi_{\alpha,j}\mid j\in J_\alpha\ \&\ \pi_1(j)=i\}$ happens to be cofinal in $\alpha$ for every $i<\lambda$, then we say that $\alpha$ is \emph{good}, and let
\begin{itemize}
\item $A_\alpha^i:=\{ \xi_{\alpha,j}\mid j\in J_\alpha\ \&\ \pi_1(j)=i\}$ for every $i<\lambda$,
and
\item $A_\alpha:=\{\xi_{\alpha,j}\mid j\in J_\alpha\}$.
\end{itemize}
Otherwise, we just let $A_\alpha$ be any cofinal subset of $\alpha$ of order-type $\cf(\alpha)$,
and let $\langle A_\alpha^i\mid i<\cf(\alpha)\rangle$ be any partition of $A_\alpha$ into cofinal subsets of $\alpha$.

\begin{claim}\label{claim511} For all $\bar\alpha<\alpha<\lambda^+$, $|A_{\bar\alpha}\cap A_\alpha|<\lambda$.
\end{claim}
\begin{proof} If $\alpha$ is not good, then $\otp(A_\alpha)=\cf(\alpha)\le\lambda$, and the conclusion follows.
Next, suppose that $\alpha$ is good. Find $j'<\lambda$ such that $e_\alpha(j')=\bar\alpha$.
Then $A_{\bar\alpha}\cap A_\alpha\s\{ \xi_{\alpha,j}\mid j\in J_\alpha\cap j'\}$
\end{proof}

Next, given a cofinal $X\s\lambda^+$,
for every $\epsilon<\lambda^+$, by Claim~\ref{claim511} and the choice of $\vec x$,
we may let $\beta_\epsilon$ denote the least $\beta<\lambda^+$
to satisfy both $x_{\beta}\s X\setminus\epsilon$ and
$|x_{\beta}\setminus\bigcup_{\alpha\in a}A_\alpha|=\lambda$ for every $a\in[\lambda^+]^{<\lambda}$.

Fix a set $E\in[\lambda^+]^{\lambda^+}$ on which the map $\epsilon\mapsto\beta_\epsilon$ is strictly increasing.
Consider the club $$D:=\{ \delta\in\acc^+(E)\mid \forall\epsilon\in E\cap\delta\, (\beta_\epsilon\cup x_{\beta_\epsilon}\s \delta)\}.$$

\begin{claim} Let $\delta\in D$. For every $i<\lambda$, $\sup(A_\delta^i\cap X)=\delta$.
\end{claim}
\begin{proof} Let $i<\lambda$ and let $\epsilon<\delta$.
We shall show that there exists $j\in J_\delta$ such that
$\xi_{\delta,j}$ is an element of $A_\delta^i\cap X\setminus\epsilon$.

Here we go.
By possibly increasing $\epsilon$, we may assume that $\epsilon\in E\cap\delta$.
Set $k:=e_\delta^{-1}(\beta_\epsilon)$,
and pick the unique $j<\lambda$ such that $\pi(j)=(k,i)$.
Then
$$x_{e_\delta(\pi_0(j))}\cap\delta\setminus\bigcup\{ A_{e_\delta(j')}\mid {j'\le j}\}=x_{\beta_\epsilon}\setminus \bigcup_{\alpha\in a}A_\alpha$$
for the set $a:=e_\delta[j+1]$ which is an element of $[\lambda^+]^{<\lambda}$. Consequently, $j\in J_\delta$,
and since $\pi_1(j)=i$,
$\xi_{\delta,j}$ is an element of $x_{\beta_\epsilon}\s X\setminus\epsilon$ that lies in $A_\delta^i$.
\end{proof}

It follows that for every partition $\mathcal S$ of $E^{\lambda^+}_\lambda$
into stationary sets, $\langle \{ A_\delta^i\mid i<\lambda\}\mid \delta\in E^{\lambda^+}_\lambda\rangle$
witnesses $\clubsuit_{\ad^*}(\mathcal S,\lambda,\lambda^+)$.
\end{proof}

\begin{lemma}\label{l51} Suppose that $\clubsuit_{\ad^*}(\{S\},\mu,\lambda^+)$ holds for some stationary subset $S$ of a successor cardinal $\lambda^+$.
Then $\clubsuit_{\ad^*}(\mathcal S,\mu,\lambda^+)$ holds for some partition $\mathcal S$ of $S$ into $\lambda^+$ many stationary sets.
\end{lemma}
\begin{proof} Let $\langle \{ A_\alpha^i\mid i<\mu\}\mid \alpha\in S\rangle$ be an array witnessing that $\clubsuit_{\ad^*}(\{S\},\mu,\lambda^+)$ holds.
Let $\mathcal I$ denote the collection of all $T\s S$ such that $\langle \{ A_\alpha^i\mid i<\mu\}\mid \alpha\in T\rangle$ fails to witness that $\clubsuit_{\ad^*}(\{T\},\mu,\lambda^+)$ holds.
It is not hard to see that $\mathcal I$ is a $\lambda^+$-complete proper ideal on $S$.
By Ulam's theorem, then, $\mathcal I$ is not weakly $\lambda^+$-saturated,
meaning that we may fix a partition $\mathcal S$ of $S$ into $\lambda^+$-many $\mathcal I^+$-sets.
Then $\langle \{A_\alpha^i\mid i<\mu\}\mid \alpha\in S\rangle$ witnesses that $\clubsuit_{\ad^*}(\mathcal S,\mu,\lambda^+)$ holds.
\end{proof}
\begin{defn}[\cite{MR2048518}] $\diamondsuit(\mathfrak b)$ asserts that for every function $F:{}^{<\omega_1}2\rightarrow{}^\omega\omega$,
there exists a function $g:\omega_1\rightarrow{}^{\omega}\omega$ with the property that for every function $f:\omega_1\rightarrow2$,
the set $\{ \alpha<\omega_1\mid F(f\restriction\alpha)\le^* g(\alpha)\}$ is stationary.
\end{defn}

\begin{cor} Suppose that $\diamondsuit(\mathfrak b)$ holds.
Then:
\begin{enumerate}
\item $\clubsuit_{\ad^*}(\mathcal S,1,\omega_1)$ holds for some partition $\mathcal S$ of $\omega_1$ into uncountably many stationary sets;
\item There exist $2^{\aleph_1}$ many pairwise nonhomeomorphic Dowker spaces of size $\aleph_1$.
\end{enumerate}
\end{cor}
\begin{proof}  (1) By \cite[Theorem~5.5]{MR2048518}, $\diamondsuit(\mathfrak b)$ implies that 	$\clubsuit_{\ad^*}(\{\omega_1\},1,\omega_1)$ holds.
Now, the conclusion follows from Lemma~\ref{l51}.

(2) By Clause~(1) and Theorem~\ref{manyd} below.
\end{proof}

\renewcommand{\thesection}{A}
\section{Appendix: Many Dowker spaces}\label{many}

In this section, $\kappa$ denotes a regular uncountable cardinal.
By \cite[\S3]{paper48}, if $\clubsuit_{\ad}(\mathcal S,1,2)$ holds
for a partition $\mathcal S$ of some nonreflecting stationary subset of $\kappa$
into infinitely many stationary sets, then there exists a Dowker space of size $\kappa$.
Here, we demonstrate the advantage of $\mathcal S$ being large.
\begin{theorem}\label{manyd} Suppose that $\clubsuit_{\ad}(\mathcal S,1,2)$ holds,
where  $\mathcal S$ is a partition of a nonreflecting stationary subset of $\kappa$
into infinitely many stationary sets. Denote $\mu:=|\mathcal S|$.
Then there are $2^\mu$ many pairwise nonhomeomorphic Dowker spaces of size $\kappa$.
\end{theorem}
\begin{proof}
Fix an injective enumeration $\langle S^\zeta_{n}\mid \zeta<\mu, n<\omega\rangle$ of the elements of $\mathcal S$.
As $\clubsuit_{\ad}(\mathcal S,1,2)$ holds,
we may fix a sequence
$\langle A_{\alpha}\mid\alpha\in\bigcup\mathcal S\rangle$ such that:
\begin{itemize}
\item[(i)] For every $\alpha\in\bigcup\mathcal S$, $A_\alpha$ is a subset of $\alpha$, and for every $\alpha'\in\alpha\cap\bigcup\mathcal S$, $\sup(A_{\alpha'}\cap A_\alpha)<\alpha'$;
\item[(ii)] For all $B_0,B_1\in[\kappa]^{\kappa}$ and $(\zeta,n)\in\mu\times\omega$,
the following set is stationary:
$$G(S^\zeta_n,B_0,B_1):=\{\alpha\in S^\zeta_n\mid  \sup(A_\alpha\cap B_0)=\sup(A_\alpha\cap B_1)=\alpha\}.$$
\end{itemize}

For every nonempty $Z\s\mu$, we shall want to define a topological space $\mathbb X^Z$.
To this end, fix a nonempty $Z\s\mu$.
For every $n<\omega$, let $S^Z_{n+1}:=\biguplus_{\zeta\in Z}S^\zeta_{n+1}$,
and then let $S^Z_0:=\kappa\setminus\biguplus_{n<\omega} S^Z_{n+1}$.
For every $\alpha<\kappa$, let $n^Z(\alpha)$ denote the unique $n<\omega$ such that $\alpha\in S^Z_n$.
For each $n<\omega$, let $ W^Z_n:=\bigcup_{i\leq n}S^Z_i $.
Then, define a sequence $\vec {L^Z}=\langle L^Z_\alpha\mid\alpha<\kappa\rangle$ via:
$$L^Z_\alpha:=\begin{cases}
W^Z_{n^Z(\alpha)-1}\cap A_\alpha,&\text{if }n^Z(\alpha)>0\ \&\ \sup(W^Z_{n^Z(\alpha)-1}\cap A_\alpha)=\alpha;\\
\emptyset,&\text{otherwise.}
\end{cases}$$

Denote $S^Z:=\{ \alpha\in\acc(\kappa)\mid \sup(L^Z_\alpha)=\alpha\}$.
Finally, let $\mathbb X^Z=(\kappa,\tau^Z)$
be the ladder-system space determined by $\vec{L^Z}$, that is, a subset $U\s\kappa$ is $\tau^Z$-open iff, for every $\alpha\in U\cap S^Z$, $\sup(L^Z_\alpha\setminus U)<\alpha$.

\begin{claim}\label{lemma61} Let $Z$ and $Z'$ be nonempty subsets of $\mu$. Then:
\begin{enumerate}
\item For all $n<\omega$ and $\alpha\in S^Z_{n+1}$, $L^Z_\alpha\s W^Z_n$;
\item If $Z\setminus Z'$ is nonempty, then $S^Z\setminus S^{Z'}$ is stationary;
\item For all $\alpha\neq\alpha'$ from $S^Z$, $\sup(L^Z_\alpha\cap L^Z_{\alpha'})<\alpha$;
\item For all $B_0,B_1\in[\kappa]^\kappa$, there exists  $m<\omega$ such that, for every $n\in\omega\setminus m$, the following set is stationary:
\[ \{ \alpha\in S^Z_n \mid \sup(L^Z_\alpha\cap B_0)=\sup(L^Z_\alpha\cap B_1)=\alpha  \};\]
\item $S^Z$ is a nonreflecting stationary set.
\end{enumerate}
\end{claim}
\begin{proof} (1) Clear.

(2) Suppose that $Z\setminus Z'\neq\emptyset$, and pick $\zeta\in Z\setminus Z'$.
As $W^Z_0=S^Z_0\supseteq S^\zeta_0$, the former is cofinal.
So, $S^Z\setminus S^{Z'}$ covers the stationary set $G(S^\zeta_1,W^Z_0,\kappa)$.

(3) For all $\alpha\neq\alpha'$ from $S^Z$, $\sup(L^Z_\alpha\cap L^Z_{\alpha'})\le\sup(A_\alpha\cap A_{\alpha'})<\alpha$.

(4) Pick $\zeta\in Z$. Given two cofinal subsets $B_0,B_1$ of $\kappa$,
find $m_0,m_1<\omega$ be such that $|B_0\cap S^Z_{m_0}|=|B_1\cap S^Z_{m_1}|=\kappa$. Set $m:=\max\{m_0,m_1\}+1$. Then, for every $n\in\omega\setminus m$,
$$G(S_n^\zeta,B_0\cap S^Z_{m_0},B_1\cap S^Z_{m_1})\s \{ \alpha\in S^Z_n \mid \sup(L^Z_\alpha\cap B_0)=\sup(L^Z_\alpha\cap B_1)=\alpha\}$$ and hence the latter is stationary.

(5) By Clause~(4), $S^Z$ is stationary. As $S^Z\s\bigcup_{n<\omega}S^Z_{n+1}\s\bigcup\mathcal S$,
and since $\bigcup\mathcal S$ is a nonreflecting stationary set, so is $S^Z$.
\end{proof}

By the preceding claim,
and the results of \cite[\S3]{paper48}, for every nonempty $Z\s\mu$, $\mathbb X^Z$ is a Dowker space. Thus we are left with proving the following:

\begin{claim} Suppose that $Z$ and $Z'$ are two distinct nonempty subsets of $\mu$.
Then $\mathbb X^Z$ and $\mathbb X^{Z'}$ are not homeomorphic.
\end{claim}
\begin{proof} Without loss of generality, we may pick $\zeta\in Z\setminus Z'$. Towards a contradiction,
suppose that $f:\kappa\leftrightarrow\kappa$ forms an homeomorphism from $\mathbb X^Z$ to $\mathbb X^{Z'}$.
As $f$ is a bijection, there are club many $\alpha<\kappa$ such that $f^{-1}[\alpha]=\alpha$.
By Claim~\ref{lemma61}(2), then, we may pick some $\alpha\in S^Z\setminus S^{Z'}$ such that $f^{-1}[\alpha]=\alpha$.
Set $\beta:=f(\alpha)$.

$\blacktriangleright$ If $\beta\notin S^{Z'}$, then $U:=\{\beta\}$ is a $\tau^{Z'}$-open neighborhood of $\beta$.

$\blacktriangleright$ If $\beta\in S^{Z'}$, then $\beta>\alpha+1$ and the ordinal interval $U:=[\alpha+1,\beta+1]$ is a $\tau^{Z'}$-open neighborhood of $\beta$.

In both cases, $U\s\kappa\setminus\alpha$, so that $f^{-1}[U]\s f^{-1}[\kappa\setminus\alpha]=\kappa\setminus\alpha$.
As $f$ is continuous and $U$ is a $\tau^{Z'}$-open neighborhood of $f(\alpha)$, $f^{-1}[U]$ must be a $\tau^Z$-open neighborhood of $\alpha$, contradicting the fact that $f^{-1}[U]$ is disjoint from $L^Z_\alpha$.
\end{proof}
This completes the proof.
\end{proof}
\section*{Acknowledgments}
The first author is partially supported by the European Research Council (grant agreement ERC-2018-StG 802756) and by the Israel Science Foundation (grant agreement 2066/18).
The second author is supported by the European Research Council (grant agreement ERC-2018-StG 802756).

redpajama_set_name: RedPajamaArXiv


text:
Live Music / Music / Reviews / May 18, 2017
Living Colour @ The Gov: The Hardest Working Men In Show Business – Live Music Review
Governor Hindmarsh Hotel, Wed 17 May.
While many once fantastic bands of their generation have either been diminished by line-up changes, destroyed by substance abuse, or simply descended into rote nostalgia, Living Colour have been hard at work, touring extensively for the past four years, playing every venue they can fill, and consequently they sound better than ever.
A Living Colour show is always spontaneous because the band is restless with too much skill to settle for just playing their songs as recorded. Vernon Reid's signature lightning fast solos are as fleeting as they are unrepeatable, and that same improvisatory streak runs through Will Calhoun's drum fills, Corey Glover's vocal melodies and screams, and Doug Wimbish's bass slaps. These guys could play the same setlist every night and produce substantially different results.
Tonight they encountered a packed and energised Governor Hindmarsh Hotel that egged them on and the result was incendiary. Opening with Robert Johnson's Preachin' Blues, they quickly got to work stretching the blues to suit their aggressive style. Doug Wimbish introduced Wall, the closer from 1993's Stain, and a not-so-subtle dig at what's his face in the White House, with its litany of societal divisions and its plea that, "The wall between us all must fall."
Next came a one-two punch from their still remarkable debut album, 1988's Vivid, in the form of Middle Man and Desperate People, both of which were delivered with a ferocity that provoked an ecstatic reaction from the crowd.
From there they bulldozed their way through Stain's Mind Your Own Business and Ignorance Is Bliss before locking into a muscular groove for their cover of The Notorious B.I.G.'s Who Shot Ya? The crowd was game, and turned Love Rears Its Ugly Head into a massive sing-along.
The band stretched out, granting Doug Wimbish and Will Calhoun extended solos in the second act, giving them a chance to showcase their respective talents while allowing the other members a likely needed breather, especially Corey Glover, who still wrings every drop out of his improbably undiminished range. He's also a consummate entertainer, dancing, cracking jokes, and generally more than holding his own against his band's trio of world-class instrumentalists.
Down the stretch, Living Colour turned Elvis Is Dead into a playful argument with Corey entreating the audience to convince Vernon of the King's passing by chanting along while Vernon chided back "I don't believe them…they're getting quieter," and, "Whatever," before they slipped a verse of Hound Dog in the middle, complete with Elvis dance moves. From there, they broke straight into their up-tempo version of Type, the lead single from 1991's Time's Up.
Living Colour ended the show as they opened, once again referencing the elephant in every room with a cover of David Bowie's I'm Afraid Of Americans before closing out with their still painfully relevant signature single, Cult Of Personality, prompting a sigh of recognition with lyrics like, "I exploit you, still you love me…I tell you one and one makes three."
After a quick break, they returned to play the title track from their sophomore album, Time's Up, segueing into a jam on James Brown's Get Up (I Feel Like Being A) Sex Machine, and then ending the night with Cream's Sunshine Of Your Love.
For a band so heavy and politically-minded, they sure know how to show people a good time; there were smiling faces everywhere as the performance ended. Living Colour thanked the audience profusely from the stage, and proceeded to humbly work the merchandise table, signing autographs and chatting to all comers. They promised to come back, as their new album Shade is being released in September, so hopefully we'll be seeing more of them soon.
All images courtesy of Peter Marsella
Governor Hindmarsh Hotel, Wed 17 May. While many once fantastic bands of their generation have either been diminished by line-up changes, destroyed by substance abuse, or simply descended into rote nostalgia, Living Colour have been hard at work, touring extensively for the past four years, playing every venue they can fill, and consequently they sound better than ever. A Living Colour show is always spontaneous because the band is restless with too much skill to settle for just playing their songs as recorded. Vernon Reid's signature lightning fast solos are as fleeting as they are unrepeatable, and that same improvisatory streak runs through Will Calhoun's drum fills, Corey Glover's vocal melodies and screams, and Doug Wimbish's bass slaps. These guys could play the same setlist every night and produce substantially different results. Tonight they encountered a packed and energised Governor Hindmarsh Hotel that egged them on and the result was incendiary. Opening with Robert Johnson's Preachin' Blues, they quickly got to work stretching the blues to suit their aggressive style. Doug Wimbish introduced Wall, the closer from 1993's Stain, and a not-so-subtle dig at what's his face in the White House, with its litany of societal divisions and its plea that, "The wall between us all must fall." Next came a one-two punch from their still remarkable debut album, 1988's Vivid, in the form of Middle Man and Desperate People, both of which were delivered with a ferocity that provoked an ecstatic reaction from the crowd. From there they bulldozed their way through Stain's Mind Your Own Business and Ignorance Is Bliss before locking into a muscular groove for their cover of The Notorious B.I.G.'s Who Shot Ya? The crowd was game, and turned Love Rears Its Ugly Head into a massive sing-along. The band stretched out, granting Doug Wimbish and Will Calhoun extended solos in the second act, giving them a chance to showcase their respective talents while allowing the other members a likely needed breather, especially Corey Glover, who still wrings every drop out of his improbably undiminished range. He's also a consummate entertainer, dancing, cracking jokes, and generally more than holding his own against his band's trio of world-class instrumentalists. Down the stretch, Living Colour turned Elvis Is Dead into a playful argument with Corey entreating the audience to convince Vernon of the King's passing by chanting along while Vernon chided back "I don't believe them…they're getting quieter," and, "Whatever," before they slipped a verse of Hound Dog in the middle, complete with Elvis dance moves. From there, they broke straight into their up-tempo version of Type, the lead single from 1991's Time's Up. Living Colour ended the show as they opened, once again referencing the elephant in every room with a cover of David Bowie's I'm Afraid Of Americans before closing out with their still painfully relevant signature single, Cult Of Personality, prompting a sigh of recognition with lyrics like, "I exploit you, still you love me…I tell you one and one makes three." After a…
Great performance from a band sounding better than ever.
Tags: Corey Glover Cream Cult Of Personality David Bowie Desperate People Doug Wimbish Elvis Is Dead Get Up (I Feel Like Being A) Sex Machine Hound Dog I'm Afraid Of Americans Ignorance Is Bliss James Brown Living Colour Love Rears Its Ugly Head Matt Saunders Middle Man Mind Your Own Business Peter Marsella Preachin' Blues Robert Johnson Shade Stain Sunshine Of Your Love The Clothesline The Gov The Notorious B.I.G. Time's Up Type Vernon Reid Vivid Wall Who Shot Ya? Will Calhoun
When he's not playing his drums, moonlighting as a singer/songwriter or working his day job as an IT analyst, Matt Saunders writes reviews for MusicSA, BSide Magazine and The Clothesline. He minored in journalism at Uni, so reviewing music and events is a way to indulge his love of the arts, journalism and Adelaide while turning people on to some of the cool things going on around town.
HEATSTROKE by Hazel Barkworth: Some Like It Hot ~ Hachette Australia Book Review June 8, 2020

redpajama_set_name: RedPajamaCommonCrawl


text:
Book Review: 'Resthaven' by Erik Therme
Iowa author Erik Therme, from what he shares in his bio data, seems like a fascinating and very handsome father of two who has 'thrashed in garage bands, inadvertently harbored runaways, and met Darth Vader.' That bit of info makes us like him even before we read his young adult adventure books (this is his second available on Kindle). His books are short, well crafted, full of great ideas and are tales that will offer enough of the 'don't turn off the lights after reading it' element to satisfy not only the teenage audience to whom they appear to be directed, but also to adults who yearn for a brief escape form reality to become involved in a fine mystery.
As far as his choice of genre is concerned Erik has said, 'RESTHAVEN, at heart, is a young adult novel, so I tried not to make it too terrifying. I have two teenage daughters (one that's obsessed with horror movies) so I wanted to write something they would enjoy. The story follows a pack of girls who venture into an abandoned retirement home for a scavenger hunt, and everything quickly goes wrong. I've always wanted to tell a story that unfolds in a creepy, old building, and RESTHAVEN fit the bill nicely.'
Erik's ability to capture the teenage approach to language and life in general is like coached by his daughters, but to share a bit of how well he has honed in on this style, read the opening: 'Listen. That's my father's favorite word. I'm pretty sure that's why he left my mother and me: we never listened. What does he expect? My mother hides behind a computer sixty hours a week, and I'm only fifteen. It's not in our nature to listen. Case in point: I think my mother just asked a question I didn't catch— which is bad— because she hates repeating herself almost as much as I do. My best defense is to keep staring out the car window, pretending I didn't hear. "Well?" she asked. "Yeah," I said, "I get it. You hate me." "Kaylee, I don't hate you. And we both know you brought this on yourself." "Me? How is this my fault? I haven't done anything—" "Exactly. It's been three weeks since we moved here, and in that time you've made zero effort to make friends or leave the apartment. If it wasn't for Anna, you wouldn't do anything but come home from school and sulk in your room." I slunk down further in the passenger seat. "Yeah, and if it wasn't for Anna I wouldn't even be here right now." "Well, I think it was very generous of these girls to invite you over tonight, so at least try and make an effort. It's only for a few hours." I scowled out my window, knowing it was pointless to argue. Once my mother made up her mind it took an act of God, or an argument with my father, to change it. "I'll make you a deal," I said, scrambling upright. "We'll go back to the apartment, make popcorn, and watch old, sappy movies the rest of the night. Deal? OK, deal. Let's go." There is the flavor that pervades this smart little novel.
The plot is well delineated in Erik's synopsis – 'The last thing Kaylee wants to do is participate in a childish scavenger hunt--especially inside the abandoned retirement home on the edge of town. When she finds a bruised, deaf boy hiding inside one of the rooms, she vows to lead him to safety . . . only to discover the front doors are now padlocked, and her friends are nowhere to be found. Kaylee is about to learn that not everything that goes 'bump in the night' is imaginary, and sometimes there are worse things to fear than ghosts.'
This book, as his first book MORTOM, is destined to be a success – and probably garner even more awards. Refreshing and smart, but most of all, exceptionally entertaining. Highly Recommended. Grady Harp, May 16
Editor's note: This review has been published with the permission of Grady Harp. Like what you read? Subscribe to the SFRB's free daily email notice so you can be up-to-date on our latest articles. Scroll up this page to the sign-up field on your right.
Book Review: 'Watchdog: How Protecting Consumers Can Save Our Families, Our Economy, and Our Democracy' by Richard Cordray (Foreword by) Elizabeth Warren
Book Review: 'The Me Too Girl' by Lance and James Morcan
Book Review: 'The Batter's Box: A Novel of Baseball, War, and Love' by Andy Kutler
Book Review: 'Chasing Life (CHASE WEN THRILLER)' by Brandt Legg
'Cotto/Gottfried': William F. Buckely: Who was he, really? Peter Brimelow explains.
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← Boost for the Burrows Centre.
Daily events. (Regular weekly events in part 2, below).
10 .15am Bideford Probus meet at Royal Hotel, tel 01271 473337.
2pm-3.30 pm Bideford Friends meet at Burton Art Gallery, 01805 622666 Ageing Well.
10am – midday coffee, craft and chat Kingsley Hall, Westward Ho!
10am -1pm Lundy Art Group at Bluelights Hall Appledore.
2-4pm Sew Together Baptist Church Westward HO! Ageing Well 01805 622666.
9.30am – 10.30am Bideford Library Childminder's Storytimes (Term Time only.) 01237 476075.
10-12pm Appledore Community Art Group at Appledore Community Hall.
11am -1pm Bideford Library . Nat West Community Banker, 0771163255.
6.45pm – 8.30pm. Breakaway Social Club for adults with learning/physical disabilities at the Baptist Church Hall, Westward Ho! (not bank holidays.) 01237 475051.
7pm Tai Chi at Bideford Methodist Church. 01237 472532.
7pm – 8pm Walking Football at Kingsley School. Men and Women. Tel 07895 035091.
7pm – 9pm Table Tennis Club at Bideford Youth Centre.
7pm Appledore Band Practice; training band, followed by senior band at 7. 45 pm St Mary's Church Hall. Contact Hugh, Secretary, 473723.
7.15pm Appledore Singers rehearse at Appledore Baptist Church. 420652.
7.45pm Bideford Band Senior Practice at Band Room Victoria Park. Tel Becky, 01237 475653.
10 -1pm Lavington Church coffee and lunches.
10.30am Tai Chi for health. Littleham Village Hall. Lauren Hutchings, 07557 676278. Free, with donations welcome.
11.45-12.45 pm Tai chi. Northam Community Centre. Ageing Well 01409 622666.
2-3.30pm Salvation Army 'Fun and Fellowship' Club meets at Baptist Hall, Bideford.
2.30pm 'Lift off for ladies.' Westward Ho! Baptist Church, 01237 425471.
6.30pm Bideford Band Beginners' and Youth Band Practice, Victoria Park Band Room. 01237 475653.
7- 8pm Walking Football at Kingsley School – Women only. tel Rose Young French, 07895 035091.
7.30pm Bideford Camera Club meet at Chubb/Churchill Hall. 01237 421391.
7.30pm Northam Choral Society rehearsals at Northam Methodist Hall. Tel 01237 478406. northamchoral.org.uk New members welcome.
10-12pm Bideford Healing Group at Sea Cadets' Bldg, Victoria Park.
10am – 1pm Work Hub at Bideford Library, 01237 476075.
11am -12pm Walking tennis for over 40's at Atlantic Racquets, Caddsdown.
10.30-12pm 'Feel Better with a Book' at Bideford Library, 01237 476075.
10.30am-11.30am and 11.30am-12.30pm Tai Chi SW. Free for over 50s. G4 Gym, Dadden Court Ind Est. 01237 478802.
10.30am Walking for Health in Victoria Park. Meet at Cafe du Parc. 421528.
2pm Tea with friends at St Mary's.
7.30pm Two Rivers Wind Ensemble rehearsal at Bideford Band Room.
7.30pm Phoenix Morris practise at Baptist Church Hall Bideford. 01237 473798.
8pm Appledore Band Senior rehearsals at St Mary's Church Hall Appledore. Sec. 01237 473723.
9.30am -10am Under fives story time at Bideford Library, 01237 476075.
10-11am Tai Chi Beginners. Marlborough Court, Bideford. £2.50 donation. Ageing Well. 01805 622666.
10.15am – 12noon Northam and District Men's Forum, Methodist Church Hall, Cross St. Mike, 01237 459324.
10.30am Walking for Health along Tarka Trail. Clarence Wharf car park. 421528.
2-3pm Seated exercise for over 60s at Westward Ho! Baptist Hall 01805 622666.
2- 3.30pm 'Tea and IT' at Bideford Library. Bring your tablet/phone etc. Devon Age UK. 01598 740060 to book a place.
6.45pm Westward Ho! Bridge Club at Trinity Church Hall.01237 470990.
8pm Bideford Folk Club at Joiners' Arms.
9.30am – 10.30am Bideford Library Childminder's Storytimes (Term Time only). 01237 476075.
2-4pm 'Sew Together' at Westward Ho! Baptist Church. 01805 622666 Ageing Well.
2.30- 4.30pm Torridge Table Tennis Club and afternoon table tennis club. 01237 477932 at Bideford Youth Centre.
4-6pm Needlecrafts at Bideford Library, 01237 476075.
7.45pm Modern Sequence Dancing, Kingsley Hall, W Ho! 01769 540309.
8pm Ceilidh Club at Northam Hall. 01237 476632.
10am -12noon Children's lego Club at Bideford Library. 01237 476075.
10am -12pm Table tennis club at Bideford Youth Centre. 01237 477932.
10.30am -12.30pm Scrabble Club at Bideford Library. 01237 476075.
7.45pm ND Cancer Care Trust Quiz at Robins' Nest Football Club. Tel Neil 01237 473798.
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text:
Del Mar 2018: What to watch at the races this summer
Accelerate with jockey Victor Espinoza heads toward the finish line to win the 2017 San Diego Handicap.
(Hayne Palmour IV / San Diego Union-Tribune)
By Tod Leonard
A year ago, the folks who run the racetrack at Del Mar were equal parts giddy and edgy.
The start of the summer meeting was the bell that opened the gate on their marathon run toward hosting their first Breeders' Cup in November, with all the glamour and headaches that come with it.
On the track, there was great anticipation for the performances of the world's top-ranked thoroughbred — the Bob Baffert-trained Arrogate.
And the track itself was a story, with Del Mar nervously opening the meet with the hope that an extensive dirt surface overhaul would produce safer trips for horses and jockeys.
Two out of those three worked out better than most could have imagined.
The Breeders' Cup was an enormous success, and six horse deaths on the grounds was an encouraging reduction to the 17 fatalities the previous summer — though some trainers and handicappers grumbled that a softer track seemed to affect some horses more than others.
Maybe one of those was Arrogate, who couldn't rise to the headlines, and figuratively stumbled badly in losses in the San Diego Handicap, Pacific Classic and, ultimately, the Breeders' Cup Classic.
It will be tough to match that kind of tension and drama this summer, but maybe that's a good thing. Del Mar is supposed to be like a day at the beach, not rush hour in L.A.
"I feel like my vacation is starting," Baffert joked in front of his barn Monday morning.
So break out the beach chair, dig your toes into the sand and consider what the next seven weeks of frolic and competition might look like.
Trainer and jockey battles
The only Baffert horse here this year that carries Arrogate's stature actually surpasses him — reigning Triple Crown winner Justify, and it's already been determined he won't run in the meet over concerns about his ankle. (A ceremonial parade on the track is in the works.)
Baffert still has his usual impressive pool of talented 2-year-olds and a couple of contenders for the Pacific Classic that will keep him busy, but probably not up all night.
Despite Arrogate's troubles, the Hall of Famer had a strong 2017 meet, leading not in wins (he had 12), but in money earned ($1.54 million). Twenty-three percent of his starters won and 53 percent were in the top three.
The stables that produce the most starters belong to Richard Baltas, Phil D'Amato, Doug O'Neill and Peter Miller — and not surprisingly, the four of them finished in the top six in earnings, with D'Amato and Baltas tying for the training title with 18 wins apiece. (Miller won last year's fall meet here.)
O'Neill and D'Amato were the hottest trainers in the Santa Anita spring meet, notching 31 and 29 wins, respectively. Miller had 26.
Keith Desormeaux was a good trainer for place and show finishers last summer. Of his 56 starters, 48 percent placed in the money. (Nine percent won.)
Among the jockeys, the faces are mostly the same, and Frenchman Flavien Prat has established himself as Southern California's dominant rider. He won both Del Mar titles last year (with 35 and 18 victories), topped the Santa Anita spring meet and was second (by one win) to Evin Roman in that track's winter meeting.
Prat was challenged in the spring at Santa Anita by Mexican Geovanni Franco, who edged Joe Talamo for second place. The veteran who got only 23 mounts at Del Mar last summer figures to get a bunch more after notching 28 spring victories at Santa Anita — second only to Prat's 39. Franco has three stakes victories this year — all aboard D'Amato's Fault.
As an apprentice last summer with a 5-pound weight allowance, Roman tied Rafael Bejarano for No. 2 in wins with 29. However, after winning Santa Anita's winter meet he was just 3-for-44 in the spring. Apprentices vying to be his successor include Asa Espinoza — the 18-year-old nephew of Hall of Famer Victor Espinoza — who tied for fifth this spring at Santa Anita with 21 wins.
Stalking the stakes
There are 41 stakes races scheduled, including six Grade I challenges.
The first biggie is Saturday's Grade II San Diego Handicap, the prep race for the meet's marquee attraction, the Pacific Classic (Aug. 18).
Del Mar racing secretary David Jerkens expects the field to include Accelerate, the John Sadler-trained 5-year-old who shocked Arrogate in last year's San Diego and this year has captured the Santa Anita Handicap and Santa Anita Gold Cup, and Baffert's Mubtaahij, who faded to eighth in the Breeders' Cup Classic but has since notched five straight top-three finishes.
Two other potential Pacific Classic starters who likely will prep in other races are Michael McCarthy's City of Light, a 4-year-old who has two Grade I victories, and Pavel, a 4-year-old trained by O'Neill who captured the Grade I Stephen Foster last month at Churchill Downs.
Baffert's possible Pacific Classic entries include Mubtaahij. West Coast, a six-time winner who was third in last year's Breeders' Cup Classic and second in the Dubai World Cup, hasn't had an official workout since that race 3½ months ago.
"Let's be honest, the last few years we've had star power," Jerkens said of the Classic. "We had Arrogate, and Beholder vs. California Chrome (in 2016). I don't know if we're going to have that superstar, but we may have a more competitive race and a bigger field."
On the distaff side, there isn't a better filly in the country right now than Jerry Hollendorfer's Unique Bella, who, after capturing the Beholder Mile last month, is expected to run in the Clement L. Hirsch Stakes on July 29.
The next great 2-year-old?
Del Mar has been a proving ground for 2-year-olds, and it doesn't get better than the memories of future Triple Crown winner American Pharoah breaking his maiden in the 2014 Del Mar Futurity.
Baffert knew he had something special in Pharoah early, and he's got at least one 2-year-old in his barn, Roadster, who has turned heads with his workouts at Santa Anita.
Among other promising colts are Hollendorfer's Instagrand, who won in his first start at Los Alamitos; the O'Neill-trained Lebron J, who was purchased for $1.2 million; and Simon Callaghan's Owning, a debut winner at Los Alamitos.
Handicappers will take keen interest in Wednesday's 5-furlong seventh race for fillies, which features Hollendorfer's Brill (a $1-million yearling purchase), Miller's San Luis Rey Downs-based Angel Alessandra and O'Neill's Bizwhacks.
Baffert has a handful of 2-year-old fillies, including Chasing Yesterday, who is a half-sister to American Pharoah.
$1 million betting bonus in play
There's a good way to keep fans from heading for the parking lot early — give them a potentially huge late-in-the-day payoff.
Among the new betting options is a 50-cent late Pick Five for the final five races. On Saturday and Sunday, a single ticket with the five winners would earn a $1-million payout (the weekday bonus is still a healthy $500,000).
The track explains that if a late Pick Five paid $650,000 to the winner, Del Mar would chip in the additional $350,000 to make it $1 million.
Labor Day — the meet's closing day Sept. 3 — will be treated as a weekend.
Beyond the late Pick Five, also offered this year is an early, 50-cent Pick Four starting with the second race, and a $2 win, place, show parlay for a minimum of two and maximum of six races.
Del Mar's 79th summer season
When: Wednesday through Sept. 3, with racing Wednesdays through Sundays, plus Monday, Sept. 3.
Where: Del Mar Racetrack, at the Del Mar Fairgrounds west of I-5 at the intersection of Via de la Valle and Jimmy Durante Boulevard.
Post time: 2 p.m. daily, except for 4 p.m. on Fridays (3:30 p.m. on Aug. 24 and 31).
Gates open: Two hours before first post except 2½ hours on opening day and Pacific Classic day (Aug. 18).
Admission: $6 for Stretch Run (grandstand) and infield ($15 on opening day). Active military and dependents are free every day with ID. Clubhouse admission is $10 ($30 on Opening Day). Diamond Club members (free to join) are admitted free every Wednesday (not counting opening day) and half-price every other day. Diamond Club Seniors (62 and over) receive free Stretch Run admission every Thursday. Children 17 and under are free but must be accompanied by parent or guardian. Free program with every paid admission.
Early Bird Wagering: 9-10 a.m. each racing day except Fridays when it continues until 1 p.m.
Seating: Reserved seats are $5 (Wednesday-Thursday), $8 (Friday), $10 (Saturday, Sunday and Labor Day), $15 (Pacific Classic Day, Aug. 18) and $30 (opening day).
Parking: $10 general and early bird, $15 preferred, $25 valet ($40 opening day).
Wagering menu: $2 Win, Place and Show, $1 Exacta and $2 Quinella on all races; $2 WPS Parlay starting with all races except the last (at least two races; no more than six); $2 Rolling Double on all races except last; Fifty-cent Pick-5 on first five and last five races; Fifty-cent Pick-4 on races 2-5 and last four races; $2 Pick Six on last six races; Fifty-cent Trifecta on all races with at least four betting interests; Ten-cent Superfectas on all races with at least six betting interests; Fifty-cent Rolling Pick Three starting with the first race and ending with the last three; $1 Place Pick All starting with first race (unless there are 11 races when it starts with second race), and $1 Super High Five on final race (if at least eight betting interests).
For the kids: Camp Del Mar is a day-camp program for children 5-12 available every racing day. The $24 fee includes activities, a Camp Del Mar T-shirt, lunch and snacks. No reservations are required.
Main telephone: (858) 755-1141
Ticket office: (858) 792-4242 or 1-877-LETSBET
Information line: (858) 793-5533.
On the Web: www.dmtc.com
tod.leonard@sduniontribune.com; Twitter: @sdutleonard
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Tod Leonard
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vTuner is your guide to Internet streamed media. The vTuner platform connects your Internet enabled product to the wide world of streamed music, talk, and video. vTuner has a proven and reliable platform already used by over 60 partners. The vTuner service is used for a wide variety of products – including Internet radios, A/V receivers, TVs, mobile handsets, portable media players, PC programs, IPTV service, photo frames, hotel in-room service and more. The possibilities are endless. The amount of content we offer keeps growing all the time. At the moment, vTuner gives you access to: radio, TV stations, audio and video podcasts as well as other content like premium services, weather, horoscopes, etc. vTuner is a privately held company with offices in the USA and Europe.


While decarbonisation of the transport sector and the role of mobility remain under Brussels' spotlight as main key enablers of the energy transition, people seem to forget the following significant data: heating and cooling of buildings are responsible for 36% of the total CO2 emissions in the EU. Similarly, the energy-intensive industry needs to cut CO2 emissions by 40% as defined by the 2030 EU targets.
But what if the decarbonisation of heating and cooling in production processes was already possible? What if currently available large heat pumps (exceeding capacities of 100kW) can provide heat up to 100°C? What if a number of visionary researchers and businesses had already picked up the challenge and had successfully integrated industrial and commercial heat pumps in their facilities and production processes?
No, this is not a dream - participants of this morning's breakfast debate were very well awake when discussing this reality. Decarbonisation of the industry can happen now thanks to heat pumps! If fully used, their potential is estimated at approx. 10% of the total industrial energy demand.
Policymakers, MEP Dr. Paul Rübig (AT/EPP) and MEP Seán Kelly (IE/EPP), discussed with representatives of industry, civil society and researchers about the next policy actions needed to seize the opportunities that high capacity and high-temperature heat pumps can provide to production processes in many applications. Businesses should play a leading role to enable the large deployment of heat pumps, thus fostering the energy transition of industrial and commercial sector.
Eric Delforge, Chair of the EHPA's Industrial and Commercial Heat Pump Working Group, inspired the participants by presenting the potential of large heat pump applications in a sustainable and smart energy system. He then led the audience through the demonstration that waste-less heating and cooling in the production process is possible. The installation of such closed and renewable system is possible thanks to large heat pump appliances. Now, "we need bold policymakers", he said, to push energy efficiency in the industries, in the same way that the EU is pushing for energy efficiency of fridges and lightbulbs back in time.
Dr. Veronika Wilk, research engineer at the Austrian Institute of Technology and coordinator of DryFiciency, spoke on behalf of the research and development sector, presenting some concrete applications of heat pumps. The EU financed project DryFiciency has been developing pilot high-temperature heat pumps for industrial drying processes (capable of reaching 160 C). According to Dr Wilk, investments "will be paid back in 3 to 4 years", while also "significantly reducing CO2 emissions in several industrial sectors". The project is now at the stage of building heat pumps in the demo sites of Agrana and Wienerberger, "finally, we are going from paper to reality", said Veronika.
Collin Bootsveld, from Colruyt Group, talked about the costs of integrating heat pump appliances in the commercial sector. Overall, investments costs of heat pumps are comparable to the investments costs for installing boilers. Collin said that the future food stores of Colruyt Group will be only using heat pumps for heating and refrigeration. "No more gas-fired boilers" - by choosing heat pumps, Colruyt Group has chosen "the planet as their client".
Alexander Anton, Secretary General of the European Dairy Association (EDA), presented the issue of energy efficiency in the dairy sector. According to him, big companies and SMEs are turning to renewable when restructuring their facilities, not only for economic reasons but also to improve the quality of their products by preserving the environment. While planning the transition, heat pumps are appreciated by businesses for their versatility: "In a dairy site, you need cooling and heating – this is the 'biosphere' in which heat pumps can show their full potential".
Thomas Nowak, Secretary General of European Heat Pump Association (EHPA), who moderated the event, reaffirmed that decarbonising EU's industrial sector is one of the major challenges we are facing. Despite standard technology being able to provide process energy and heat for district energy systems up to 100°C, this potential contribution of heat pumps has long been overlooked. Fortunately, recent research and development have lifted this upper-temperature boundary. Latest high-temperature heat pump units can provide around 160°C, and this is not the end of this development.
Naturally, participants agreed on the need for a proper recognition of heat pump technology to deploy it where possible in the industrial and commercial sector, thus using the full potential of waste heat to the maximum extent and include renewable technologies into industrial heating and cooling processes. The economic advantage of using heat pumps in businesses is a reality, as in the described cases of Colruyt and the Dairy industry.
In this view, to better disseminate the potentialities of the industrial and commercial heat pumps, EHPA is preparing a second edition of the ''Large scale heat pump in Europe'' (download the previous version). This brochure presents examples of realized and successful projects involving large heat pumps. The publication will be presented the 15-16 May 2019, during the Heat Pump Forum in Brussels.
The video-record of the event is available on the SME Europe Facebook Page.
EHPA promotes awareness and deployment of heat pump technology in Europe. All activities aim at creating a market environment that facilitates a faster deployment of heat pump technology to unleash its benefits on a European level: efficient heating and cooling using renewable energy. EHPA also coordinates the Heat Pump Keymark – a European certification scheme for all heat pumps, combination heat pumps and hot water heater.
For more information, please visit: www.ehpa.org.


CANCELLED - De Dikke Vette Grote Oud en Nieuw Vitamine ACU Disco!
New Years Eve edition of Vitamine ACU disco! Go loose in the new year with our selection of alternative Vitamine ACU disco DJ's that will play this night!


Where to wash clothes, to shoot from the Kalashnikov rifle, to listen to national songs - we gathered most popular questions asked by foreigners to employees of tourist centers in Kiev.
According to Kiev City Tourist Information Center, about 30 tourist centers are in Kiev now where visitors can get information about the city. In summer it was activity peak of this centers. Author of this article worked the whole season in one of them and made a list of the most popular questions and answers asked by foreigners who came to the Ukrainian capital first time.
Which currency exchange office can I trust? Advise us pease, where is the best rate.
The map is full of tourist information centers. Where are they?
Where is the big shopping mole?
Where can I eat Ukrainian food?
Where can I buy auto parts?
Where can I look at ethno show - Ukrainian dances and songs?
Where to go with kids?
Where can I book Kiev guided tours?
I am bored here. Could you go with me for dinner? And introduce me to your friends, especially girls? I will pay!
Have you visited this massage (showing popular massage advertising in Kiev with half-naked girls on it)?
Can I visit the opera? Where can I listen to classical music?
Is today a day-off in Kiev historical museum?
Can I take part in Chernobyl tour tomorrow?
Is it possible to get to the top of the Mother Motherland statue?
When the nearest football game is? How to get to the Olympic Stadium?
When can I go to a Kiev shooting range?
Ukrainian flag have caused a storm of positive emotions, but did you know if I went out with a flag of my country, the people would have booed me?!