text:
\section{Introduction}

The main theorem of this paper is Theorem \ref{thm_strict_set_bound}, which concerns the existence of strictly monotonic sequences in $\mathbb{R}^n$ (for some definition of strictly monotonic). The proof of Theorem \ref{thm_strict_set_bound} also requires Theorem~\ref{thm_crossintersecting}, a theorem of independent interest. Section \ref{secn_pillage_games} describes an application of our results to stable sets in pillage games (this was the original motivation for Theorem \ref{thm_strict_set_bound}). We begin by giving some background.

\subsection{Non-strict monotonicity}
\label{secn_nonstrict}

A theorem of Erd\H{o}s\ and Szekeres \cite{szekeres} tells us that within a sequence of $ab + 1$ real numbers, we can always find a monotonically increasing subsequence of length $a+1$ or a monotonically decreasing subsequence of length $b+1$. The bound $ab + 1$ is best possible, as can be seen by considering the sequence
\begin{equation}
 \label{real_sequence_es}
 (b,b-1,\ldots,1,2b,2b-1,\ldots,b+1,\ldots,ab,ab-1,\ldots,(a-1)b+1).
\end{equation}
The original proof of Erd\H{o}s\ and Szekeres used geometrical reasoning. One can also deduce it from Dilworth's theorem (or an immediate corollary of it; see Lemma \ref{minantichainsize}) by considering a partial order where $x \leq y$ in the partial order if $x \leq y$ and $x$ occurs before $y$ in the sequence. Then a chain in this partial order corresponds to an increasing subsequence and an antichain corresponds to a decreasing subsequence. We also give a distinct proof due to Seidenberg \cite{seidenberg} below.

\begin{defn_monotonic}
 A sequence of points $(x^j)_{j=1}^{\ell}$ with $x^j \in \mathbb{R}^n$ is \emph{monotonic in direction} $c \in \{-1,1\}^n$ if for each $i \in [n]$, the $i$th coordinate sequence $(x_i^j)_{j=1}^\ell$ is (not necessarily strictly) decreasing or increasing according to whether $c_i = -1$ or $c_i = 1$ respectively.
\end{defn_monotonic}

We will sometimes omit the direction, so a monotonic sequence in $\mathbb{R}^n$ is one that is monotonic in some direction.

\begin{defn_mono_in_set}
 A set $S \subset \mathbb{R}^n$ contains a monotonic sequence of length $\ell$ (in direction $c$) if there are distinct points $\{x^1,\ldots,x^\ell\} \subset S$ such that the sequence $(x^j)_{j=1}^\ell$ is monotonic (in direction $c$).
\end{defn_mono_in_set}

There is a rough equivalence between sequences in $\mathbb{R}$ and sets in $\mathbb{R}^2$. A set in $\mathbb{R}^2$ can be ordered by the first coordinate (making an arbitrary choice of ordering when two points share a first coordinate) and projected in the second coordinate to get a sequence in $\mathbb{R}$. Conversely, a sequence of real numbers $(x^j)_{j=1}^\ell$ can be mapped to a set in $\mathbb{R}^2$ via $x^j \mapsto (j,x^j)$. These generalize to a rough equivalence between sequences in $\mathbb{R}^{n-1}$ and sets in $\mathbb{R}^n$. The Erd\H{o}s-Szekeres theorem thus gives conditions guaranteeing a monotonic sequence in a set in $\mathbb{R}^2$.

If $(x^1,\ldots,x^\ell)$ is a monotonic sequence in direction $c$, then $(x^\ell,\ldots,x^1)$ is a monotonic sequence in direction $-c$, so for sequences in sets, we only need to consider one of $c$ or $-c$. With this in mind, define
$$ C_n = \{ c \in \{-1,1\}^n : c_1 = 1 \}, $$
and enumerate this set as $C_n = \{c^1,\ldots,c^{2^{n-1}}\}$.

The generalization of the Erd\H{o}s-Szekeres theorem to $\mathbb{R}^n$ is as follows.

\begin{propn_nonstrict_set_bound}
 \label{propn_nonstrict_set_bound} (Non-strict monotonicity.)
Let $\ell_i \geq 1$ for $1 \leq i \leq 2^{n-1}$, and let $S \subset \mathbb{R}^n$ have size
\[ |S| > \prod_{i = 1}^{2^{n-1}} \ell_i. \]
Then $S$ contains a monotonic sequence of length $\ell_i+1$ in some direction $c^i$.
\end{propn_nonstrict_set_bound}

The bound in Proposition \ref{propn_nonstrict_set_bound} is best possible; we give a construction of size $\prod_i \ell_i$ containing no such sequence in Section \ref{secn_nonstrict_constr}. The proposition was proved by De Bruijin \cite{multidim} in the case $\ell_i = \ell$ for all $i$ by using $n-1$ applications of the Erd\H{o}s-Szekeres theorem. For non-constant $\ell_i$, Proposition \ref{propn_nonstrict_set_bound} can be proved by a counting argument of Seidenberg \cite{seidenberg}, which we give below.

\begin{proof}
Let $S \subset \mathbb{R}^n$, $|S| = t$. Order the points by the first coordinate (so we consider $S$ as a sequence of points $(x^j)_{j=1}^t$ in $\mathbb{R}^n$). Assign each position $j, 1 \leq j \leq t$, a $2^{n-1}$-tuple of numbers $(r_1^j,\ldots,r_{2^{n-1}}^j)$, where $r_i^j$ is the maximum length of a subsequence in $S$ in direction $c^i$ ending at $x^j$. Then no tuple of numbers is repeated: given sequence positions $1 \leq j < k \leq t$, the point $x^k$ must lie in some direction $c^i$ from the point $x^j$. Then the sequence in direction $c^i$ of length $r_i^j$ ending at $x^j$ can be extended to a sequence containing $x^k$, and thus $r_i^j < r_i^k$. If no sequence has length $\ell_i + 1$ then $r_i^j \leq \ell_i$ for all $j$, and so by distinctness of the tuples, we have $|S| = t \leq \prod_i \ell_i$.
\end{proof}

\subsection{Strict monotonicity}
\label{secn_strict_sets}

Suppose we wish to find a strictly increasing, strictly decreasing or constant subsequence in a sequence in $\mathbb{R}$ (we must allow constant subsequences). The Seidenberg counting argument shows that a sequence in $\mathbb{R}$ with no such subsequence of length $\ell+1$ has maximum length $\ell^3$. This is best possible; consider the example (\ref{real_sequence_es}) with each $x$ replaced by $\ell$ consecutive copies of $x$.

\begin{defn_strict_monotonic}
 A sequence of points $(x^j)_{j=1}^{\ell}$ with $x^j \in \mathbb{R}^n$ is \emph{strictly monotonic in direction} $d \in \{-1,0,1\}^n$ if for each $i \in [n]$, the $i$th coordinate sequence $(x_i^j)_{j=1}^\ell$ is strictly decreasing, constant, or strictly increasing according to whether $d_i = -1$, $0$ or $1$ respectively.
\end{defn_strict_monotonic}

As before, we will sometimes omit the direction when talking about strictly monotonic sequences.

\begin{defn_strict_mono_in_set}
 A set $S \subset \mathbb{R}^n$ contains a strictly monotonic sequence of length $\ell$ (in direction $d$) if there are distinct points $\{x^1,\ldots,x^\ell\} \subset S$ such that the sequence $(x^j)_{j=1}^\ell$ is strictly monotonic (in direction $d$).
\end{defn_strict_mono_in_set}

We need consider only one of each $d$ or $-d$ for each $d \in \{-1,0,1\}^n$, so define
$$ D_n = \{ d \in \{-1,0,1\}^n : d \neq (0,\ldots,0), d_{i_0} = 1 \mbox{ where } i_0 = \min \{ i : d_i \neq 0 \} \}. $$

Consider a set $S \subset \mathbb{R}^n$. If we order the points by a coordinate to get a sequence $S'$ in $\mathbb{R}^{n-1}$, then a \emph{strictly} monotonic subsequence of $S'$ does not necessarily correspond to a strictly monotonic sequence in $S$ (it now matters what happens to points sharing the coordinate that we order by). Thus one cannot apply the counting argument of Seidenberg to bound the size of a set $S \subset \mathbb{R}^n$ with no strictly monotonic sequence of length $\ell+1$. Further, even if we start with a sequence in $\mathbb{R}^n$, the counting argument only gives a bound of $\ell^{3^n}$, which is far from best possible. Thus we need new techniques to work with strict monotonicity.

\subsection{Strict monotonicity in sets}

Suppose we wish to construct a large set in $\mathbb{R}^n$ with no strictly monotonic sequence of length $\ell+1$. Call such a set a \emph{good} set. Here we describe a natural construction which is in fact largest possible (see Section \ref{secn_strict_constr} for the exact construction).
As mentioned for Proposition \ref{propn_nonstrict_set_bound}, there is a set $E_n \subset \mathbb{R}^n$ of size $|E_n| = \ell^{2^{n-1}}$ with no monotonic sequence of length $\ell+1$, and so it is certainly good. In fact the construction in Section \ref{secn_nonstrict_constr} for $E_n$ contains no pair of points that share a coordinate for any coordinate position. Suppose $F_n$ is another good set such that every pair of points in $F_n$ share a coordinate in some coordinate position. If we replace each point in $E_n$ with a very small copy of $F_n$ to get a new set $G_n$ ($G_n$ is the ``product'' of $E_n$ and $F_n$; this is made more precise in Section \ref{secn_strict_constr}), then any strictly monotonic sequence in $G_n$ must either have all the coordinate sequences non-constant (thus taking at most one point from each copy of $F_n$), or it must lie strictly inside some fixed copy of $F_n$. In the first case, it corresponds to some monotonic sequence in $E_n$, and thus has length at most $\ell$. In the second case it has length at most $\ell$ since $F_n$ is good. Thus $G_n$ is also good. One candidate for $F_n$ is given by the recursive definition $F_1 = \{0\}$ and $F_n = f_n(G_{n-1})$, where $f_n : (x_1,\ldots,x_{n-1}) \mapsto (x_1,\ldots,x_{n-1},0)$. This recursive construction then gives $|G_n| = |E_n| |F_n| = \ell^{2^{n-1}} |G_{n-1}| = \ell^{2^n-1}$. Our main theorem shows that this is in fact best possible.

Now let $(\ell_d)_{d \in D_n}$ be a collection of maximal lengths with $\ell_d \geq 2$ for all $d$. We will show that the maximum size of a set with no strictly monotonic sequence in direction $d$ of length $\ell_d + 1$ for all $d$ is essentially the same recursive construction, with suitable choices at each stage to maximize the size of the set produced.

Let the function
$$ N : \{-1,0,1\}^n \rightarrow \mathcal{P}[n] $$
give the set of positions of the non-zero coordinates. Note that $|\{d \in D_n : N(d) = I\}| = 2^{|I|-1}$. Define $(m_I)_{I \subset [n]}$ and $(\lambda_I)_{I \subset [n]}$ via $\lambda_\emptyset = 1$ and
\begin{equation}
m_I = \prod_{d \in D_n : N(d) = I} \ell_d, \label{defn_mI}
\end{equation}
\begin{equation}
 \lambda_I = m_I \cdot \max_{i \in I} \lambda_{I \setminus \{i\}}. \label{defnlambda}
\end{equation}
For $I \subset [n]$, $\lambda_I$ should be thought of (this will be shown) as the maximum size of a good set $S$ when the coordinates of the points can only vary in $I$ (for all $x,y \in S$, $x_i = y_i$ for $i \notin I$). Similarly, $m_I$ should be thought of as the maximum size of a good set $S$ when the coordinates can only vary in $I$, and no two points share a common coordinate from $I$ (for all $x,y \in S$, $x_i = y_i$ if and only if $i \notin I$).

If $\ell_d = \ell$ for all $d \in D_n$, then $m_I = \ell^{2^{|I|-1}}$, $\lambda_{I} = \ell^{2^{|I|}-1}$ and $\lambda_{[n]} = \ell^{2^n-1}$.

\begin{thm_strict_set_bound}
 \label{thm_strict_set_bound} (Strict monotonicity in sets.)
Let $(\ell_d)_{d \in D_n}$ satisfy $\ell_d \geq 2$ for all $d$, and let $\lambda_{[n]}$ be as above. Let $S \subset \mathbb{R}^n$ have size
\[
|S| > \lambda_{[n]}.
\]
Then $S$ contains a strictly monotonic sequence of length $\ell_d+1$ in some direction $d \in D_n$.

In particular, if $|S| > \ell^{2^n-1}$ then $S$ contains a strictly monotonic sequence of length $\ell+1$.
\end{thm_strict_set_bound}

Theorem \ref{thm_strict_set_bound} is best possible; we give a construction of size $\lambda_{[n]}$ with no such sequence in Section \ref{secn_strict_constr}.
If we do not impose $\ell_d \geq 2$ for all $d$, then $\lambda_{[n]}$ may not be a correct bound. For example, with $n=3$, take the collection
$\ell_{(0,1,-1)} = \ell_{(1,0,-1)} = \ell_{(1,-1,0)} = 2$, and
$\ell_d = 1$ otherwise.
Then $\lambda_{[n]} = 2$, but
$ \{(1,0,0),(0,1,0),(0,0,1)\} $
is a good set of size 3.

\subsection{Strict monotonicity in sequences}
\label{secn_strict_monotonicity_in_sequences}

Theorem \ref{thm_strict_set_bound} bounds how large a \emph{set} can be without containing a strictly monotonic sequence. We would like an analogue of this theorem for sequences. Unlike the non-strict case, such an analogue is not a triviality.

Let $S = (x^j)_{j=1}^{|S|}$ be a sequence of points in $\mathbb{R}^n$. Each of the $3^n$ directions in $\{-1,0,1\}^n$ are now non-equivalent for the purposes of the existence of a subsequence in this direction. Suppose for each $d \in \{-1,0,1\}^n$, we forbid a subsequence of length $\ell_d + 1$ in direction $d$. Map the sequence $S$ to a set $T \subset \mathbb{R}^{n+1}$ as described in Section \ref{secn_nonstrict}, i.e., $x^j \mapsto (j,x^j)$. The set of maximum lengths for $T$ is now $(\ell_d^*)_{d \in D_{n+1}}$, where
\begin{equation*}
 \ell_{(d_0,d_1,\ldots,d_n)}^* =
\begin{cases}
 \ell_{(d_1,\ldots,d_n)} & \text{ if } d_0 = 1 \\
 1 & \text{otherwise (i.e., }d_0 = 0).
\end{cases}
\end{equation*}
Define $(\lambda_{I}^*)_{I \subset [n+1]}$ as in Section \ref{secn_strict_sets} for the lengths $(\ell_d^*)_{d \in D_{n+1}}$. We would like to apply Theorem \ref{thm_strict_set_bound}, but we cannot do this as stated, since we do not have $\ell_d^* \geq 2$ for all $d$. However, we will show that the proof of Theorem \ref{thm_strict_set_bound} still applies for this special case.

\begin{thm_strict_seq_bound}
 \label{thm_strict_seq_bound} (Strict monotonicity in sequences.)
Let $(\ell_d)_{d \in \{-1,0,1\}^n}$ satisfy $\ell_d \geq 2$ for all $d$, and let $\lambda_{[n+1]}^*$ be as above. Let $S$ be a sequence in $\mathbb{R}^n$ of length
\[ |S| > \lambda_{[n+1]}^*. \]
Then $S$ contains a strictly monotonic subsequence of length $\ell_d+1$ in some direction $d \in \{-1,0,1\}^n$.

In particular, if $|S| > \ell^{2^{n+1}-1}$ then $S$ contains a strictly monotonic subsequence of length $\ell+1$.
\end{thm_strict_seq_bound}

Theorem \ref{thm_strict_seq_bound} is best possible. The construction $G$ given in Section \ref{secn_strict_constr} for the set of lengths $(\ell_d^*)_{d \in D_{n+1}}$ has the property that no two points share the same first coordinate, so $G$ can be ordered by the first coordinate and projected in the remaining $n$ coordinates to get a sequence of points in $\mathbb{R}^n$.

\section{Lower bound constructions}
\label{secn_lower_bound_constrn}

\subsection{Construction for non-strict monotonicity}
\label{secn_nonstrict_constr}

Let $(\ell_i)_{i=1}^{2^{n-1}}$ be a collection of maximum lengths for the set of directions $C_n$. We will construct a set of size $\prod_i \ell_i$ with no sequence of length $\ell_i + 1$ in direction $c^i$, for all $1 \leq i \leq 2^{n-1}$. This shows the value appearing in Proposition \ref{propn_nonstrict_set_bound} is best possible.

For convenience, write
$$ L_k = \prod_{i=1}^k \ell_i $$
(and $L_0 = 1$). Define, for a set $A \subset \mathbb{R}^n$ and a vector $s \in \mathbb{R}^n$, the set translation
$$ A + s = \{ x + s : x \in A \}. $$
Define recursively, for $1 \leq k \leq 2^{n-1}$, the following collection of sets in $\mathbb{R}^n$.
\begin{align*}
 A_0 &= \{0\} \\
 A_{k,m} &= A_{k-1} + m L_{k-1} c^k\text{, \ for }0 \leq m \leq \ell_k - 1. \\
 A_k &= A_{k,0} \cup \cdots \cup A_{k,\ell_k-1}
\end{align*}
For $A \subset \mathbb{R}^n$, define
$$ w(A) = \max_{x,y \in A} ||x - y||_\infty, $$
where $||z||_\infty = \max_{1 \leq i \leq n} |z_i|$.

\begin{esconstruction}
 \label{esconstruction} (Properties of $A_k$)
\begin{enumerate}[label=(i)]
\item $A_k \subset \mathbb{Z}^n$,
\item $w(A_k) \leq L_k - 1$,
\item $|A_k| = L_k$,
\item for each $i$, $1 \leq i \leq k$, $A_k$ does not have a monotonic sequence of length $\ell_i+1$ in the direction $c^i$. Further $A_k$ does not have any non-trivial monotonic sequences in any of the directions $c^{k+1}, \ldots, c^{2^{n-1}}$.
\end{enumerate}
\end{esconstruction}

\begin{proof}
 (i) This is immediate from the construction.

 (ii) We proceed by induction on $k$. The statement is true for $A_0$. Let $x,y \in A_k$, say $x = x_0 + m_1 L_{k-1} c^k$, $y = y_0 + m_2 L_{k-1} c^k$, where $x_0, y_0 \in A_{k-1}$. Then
\begin{align*}
 ||x - y||_\infty & \leq ||x_0 - y_0|| + |m_1 L_{k-1} - m_2 L_{k-1}| \cdot || c^k ||_\infty \\
 & \leq L_{k-1} - 1 + (\ell_k-1) L_{k-1} \\
 & = L_k - 1.
\end{align*}

(iii) We proceed by induction on $k$. The statement is true for $A_0$. It is sufficient to show that $A_{k,m_1} \cap A_{k,m_2} = \emptyset$ for $m_1 \neq m_2$. Then, $|A_k| = \ell_k |A_{k-1}|$ and we are done. Indeed, suppose $x \in A_{k,m_1} \cap A_{k,m_2}$ for $m_1 \neq m_2$. Then $x - m_1 L_{k-1} c^k \in A_{k-1}$ and $x - m_2 L_{k-1} c^k \in A_{k-1}$. Hence
\begin{align*}
 w(A_{k-1}) & \geq || (x - m_1 L_{k-1} c^k) - (x - m_2 L_{k-1} c^k) ||_\infty \\
 & = |m_1-m_2| L_{k-1} \\
 & > L_{k-1} - 1,
\end{align*}
contradicting (ii).

(iv) We proceed by induction on $k$. The statement is true for $A_0$. Suppose $x \in A_{k,m_1}$, $y \in A_{k,m_2}$ with $m_1 < m_2$. Then again by (ii) we have that $(x,y)$ is a sequence in direction $c^k$. Therefore, if we have a sequence of points inside $A_k$, then either it must lie entirely inside an $A_{k,m_0}$ for some $m_0$ in direction $c^i$ for some $i < k$ (and thus have length at most $\ell_i$ by the inductive hypothesis); otherwise it lies in direction $c^k$ and can take at most one point from each $A_{k,m}, 0 \leq m \leq \ell_k-1$.
\end{proof}

Properties (iii) and (iv) of Lemma \ref{esconstruction} show that we can take the set $A_{2^{n-1}}$ as our construction.

\subsection{Construction for strict monotonicity}
\label{secn_strict_constr}

We construct sets showing that the bound is best possible for Theorem \ref{thm_strict_set_bound} (and Theorem \ref{thm_strict_seq_bound}) by induction on $n$. The case $n=1$ is simply $\ell_d$ distinct points in $\mathbb{R}$ for the unique $d \in D_1$.

Let $(\ell_d)_{d \in D_n}$, $(\lambda_I)_{I \subset [n]}$ be as in Theorem~\ref{thm_strict_set_bound}. In this section, a sequence will mean a strictly monotonic sequence. The set we construct will have size $\lambda_{[n]}$ with no sequence of length $\ell_d + 1$ in direction $d$ for all $d \in D_n$.

In equation (\ref{defn_mI}), $\{ d \in D_n : N(d) = [n] \} = C_n$, and so
$$ m_{[n]} = \prod_{c \in C_n} \ell_c. $$
Let $i_0 \in [n]$ be such that
$$ \lambda_{[n]} = m_{[n]} \lambda_{[n] \setminus \{i_0\} }. $$
Construct a new set of lengths $(\ell_d')_{d \in D_{n-1}}$ via
$$ \ell_d' = \ell_{(d_1, \ldots, d_{i_0-1}, 0, d_{i_0}, \ldots, d_{n-1})}. $$
By the inductive hypothesis with the collection $(\ell_d')_{d \in D_{n-1}}$ there is a set $F \subset \mathbb{R}^{n-1}$ of size $|F| = \lambda_{[n] \setminus \{i_0\}}$, which contains no sequence of length $\ell_d' + 1$ in direction $d$ for all $d \in D_{n-1}$. Let $F'$ be the embedding and scaling of $F$ into $\mathbb{R}^n$,
\begin{align*}
F' &= \{ (1/(w(F)+1))(x_1, \ldots, x_{i_0-1}, 0, x_{i_0}, \ldots, x_{n-1} ) : (x_1,\ldots,x_{n-1}) \in F \}.
\end{align*}
The set $F'$ has no sequences of length $\ell_d + 1$ in direction $d$ for all $d \in D_n \setminus C_n$, and no non-trivial sequences in any direction $d \in C_n$. Further $w(F') < 1$.

Construct another set of lengths $(\ell_i'')_{i=1}^{2^{n-1}}$ via
$$\ell_i'' = \ell_{c^i}.$$
Then as in Section~\ref{secn_nonstrict_constr}, there is a set $E \subset \mathbb{R}^n$ of size $|E| = \prod_i \ell_i'' = \prod_{c \in C_n} \ell_c = m_{[n]}$ with no sequence of length $\ell_c + 1$ in direction $c$ for all $c \in C_n$, and no non-trivial sequence in any direction $d \in D_n \setminus C_n$.

Lemma \ref{sesconstruction} shows that the following construction works.
\begin{equation}
 \label{defnG}
 G = \bigcup_{x \in E} (F' + x)
\end{equation}

\begin{sesconstruction}
 \label{sesconstruction} (Properties of $G$)
\begin{enumerate}[label=(i)]
 \item $|G| = \lambda_{[n]}$.
\item $G$ has no strictly monotonic sequence of length $\ell_d+1$ in direction $d$ for all $d \in D_n$.
\end{enumerate}
\end{sesconstruction}

\begin{proof}
(i) This argument is similar to that for property (iii) in Lemma \ref{esconstruction}. We have that $w(F') < 1$ and $||x - y||_\infty \geq 1$ for all $x,y \in E$, $x \neq y$, since $E \subset \mathbb{Z}^n$. Thus in (\ref{defnG}), $(F' + x) \cap (F' + y) = \emptyset$ for $x \neq y$, and so
$$ |G| = |F| |E| = \lambda_{[n] \setminus \{i_0\} } \cdot m_{[n]} = \lambda_{[n]}. $$

(ii) This argument is similar to that for property (iv) in Lemma \ref{esconstruction}. Let $(y^1, \ldots, y^t)$ be a strictly monotonic sequence in $G$ in direction $d \in D_n$, where $y^j \in F' + x^j$. There are two possibilities for this sequence. If it has a constant coordinate sequence, then the points must lie in some copy of $F'$, i.e., $x^1 = \cdots = x^t = x$ for some $x$, and $d_{i_0} = 0$. Then the construction of $F$ guarantees that $t \leq \ell_d$. Otherwise it has no constant coordinate, and so $(x^1,\ldots,x^t)$ is a sequence in $E$ in direction $d$, and so by construction of $E$, $t \leq \ell_d$.
\end{proof}

\section{Proof of the upper bound}
\label{secn_proof_of_the_upper_bound}

We begin by giving a theorem of independent interest, required for the proof of the main theorem.

\begin{defnintersectingpoints}
 Two points $x,y \in \mathbb{R}^n$ are \emph{intersecting} if they agree in some coordinate, i.e., $x_i = y_i$ for some $i \in [n]$.
\end{defnintersectingpoints}

\begin{thm_crossintersecting}
 \label{thm_crossintersecting}
 Let $\{a^1,b^1\},\ldots,\{a^t,b^t\}$ be a collection of $t$ pairs of points in $\mathbb{R}^d$ such that each pair $a^j,b^j$ is non-intersecting, but all $2t$ points are otherwise pairwise intersecting. Then $t \leq 2^{d-1}$.
\end{thm_crossintersecting}

This bound can be achieved by taking as pairs $\{a^j,b^j\}$ the opposing corners in the d-dimensional cube $\{0,1\}^d$.

Our proof uses exterior algebras, and is reminiscent of a proof of a theorem on intersecting sets given by Alon \cite{alon}. We describe them here briefly; for a comprehensive introduction the reader can consult for example Marcus \cite{marcus}. Given a real $n$-dimensional vector space $V$ with basis $\{e_1,\ldots,e_n\}$, the exterior algebra $\Lambda V$ is a $2^n$ dimensional vector space with basis $\{ e_A : A \subset [n] \}$ and an associative bilinear operation $\wedge$. For $A = \{i_1,\ldots,i_r \}, i_1 < \cdots < i_r$, we identify
$$ e_A = e_{i_1} \wedge \cdots \wedge e_{i_r}. $$
The operation $\wedge$ is defined to satisfy $e_i \wedge e_j = -e_j \wedge e_i$, and we extend by linearity. In particular, for a set of vectors $U = \{u_1, \ldots, u_m\} \subset V$, the wedge product $u_1 \wedge \cdots \wedge u_m$ is non-zero if and only if $U$ is an independent set of vectors.

\begin{proof}[Proof of Theorem \ref{thm_crossintersecting}]
Without loss of generality, the coordinates of the points take values in $[m]$, i.e., $\{a^1, b^1, \ldots, a^t, b^t\} \subset [m]^d$.

We consider the exterior algebra over the real vector space $\mathbb{R}^{2d}$. Label a set of basis elements for $\mathbb{R}^{2d}$ as
$$ \{e_1,\ldots,e_d,f_1,\ldots,f_d \}. $$
For each $i$, let $\text{lin}(e_i,f_i)$ be the subspace of $\mathbb{R}^{2d}$ spanned by the vectors $\{e_i,f_i\}$, and let
$$ \{ v_i^j : 1 \leq j \leq m \} \subset \text{lin}(e_i,f_i) $$
be a set of $m$ vectors in general position, i.e., any 2 of them are linearly independent. For $x \in [m]^d$, let $v_x$ be the vector
$$ v_x = \bigwedge_{i=1}^d v_i^{x_i}. $$
Then for $x,y \in [m]^d$, $v_x \wedge v_y = 0$ if and only if $x$ and $y$ intersect. Hence,
\begin{align*}
v_{a_i} \wedge v_{a_j} &= 0 \text{ \ for all } i,j, \\
v_{b_i} \wedge v_{b_j} &= 0 \text{ \ for all } i,j, \\
v_{a_i} \wedge v_{b_j} &= 0 \text{ \ if and only if } i \neq j.
\end{align*}
We now show that the vectors $\{ v_{a_j}, v_{b_j} : 1 \leq j \leq t \}$ are linearly independent. Suppose for some constants $\alpha_j, \beta_j$ we have that
$$ \sum_{j=1}^t \alpha_j v_{a_j} + \sum_{j=1}^t \beta_j v_{b_j} = 0. $$
For given $k$, the wedge product of the left hand side of this expression with $v_{a_k}$ is $\beta_k v_{b_k} \wedge v_{a_k}$. Since $v_{b_k} \wedge v_{a_k} \neq 0$, we must have $\beta_k = 0$. This is true for all $\alpha_k$ and $\beta_k$. This shows linear independence.

The vectors $v_{a_j}, v_{b_j}$ lie in the vector space spanned by the $2^d$ vectors of the form $x_1 \wedge \cdots \wedge x_d$, where $x_i \in \{e_i,f_i\}$. Thus by linear independence, we have $2t \leq 2^d$, i.e., $t \leq 2^{d-1}$.
\end{proof}

Here are two technical lemmas that we will need in the proof of Theorem \ref{thm_strict_set_bound}. Lemma \ref{minantichainsize} is also an immediate corollary of Dilworth's theorem.

\begin{intersectinggridsubset}
 \label{intersectinggridsubset}
 Let $m,d$ be integers, and let $A \subset [m]^d$ be intersecting (i.e., for all $x,y \in A$ there exists $i \in [d]$ such that $x_i = y_i$). Then $|A| \leq m^{d-1}$.
\end{intersectinggridsubset}

\begin{proof}
 Consider $[m]^d$ as the finite vector space $\mathbb{F}_m^d$. Partition $\mathbb{F}_m^d$ into the sets
$$\{(0,\ldots,0),(1,\ldots,1),\ldots,(m-1,\ldots,m-1)\} \subset \mathbb{F}_m^d $$
and all translates. There are $m^{d-1}$ such sets, and any intersecting subset of $\mathbb{F}_m^d$ takes at most one point from each set.
\end{proof}

\begin{minantichainsize}
 \label{minantichainsize}
Let $S$ be a partially ordered finite set, with maximum chain length $\ell$. Then $S$ contains an antichain $A$ of size at least $|A| \geq |S| / \ell$.
\end{minantichainsize}

\begin{proof}
The set $T$ of maximal elements in $S$ is an antichain, and the set $S \setminus T$ has maximal chain length $\ell-1$. Proceeding by induction on $\ell$ with the set $S \setminus T$ gives a partition of $S$ into $\ell$ antichains, one of which has size at least $|S| / \ell$.
\end{proof}

\subsection{Intersecting flats}

We consider \emph{axis-aligned affine subspaces} in $\mathbb{R}^n$, which we will refer to as \emph{flats}. Label such a flat $u$ via a string of $n$ numbers and $\star$s, say, $u \in (\mathbb{R} \cup \{\star\})^n$, where $\star$ is a wildcard. Thus a point $x \in \mathbb{R}^n$ lies in $u$ if for all $i \in [n]$, either $x_i = u_i$ or $u_i = \star$. For example, $(2,1,\star)$ represents a line in $\mathbb{R}^3$ parallel to the $z$-axis through the point $(2,1,0)$. Observe that the dimension of the flat is the number of $\star$ coordinates.

\begin{defnintersectingflats}
 A set of flats $W \subset (\mathbb{R} \cup \{\star\})^n$ is \emph{intersecting} if for all $u,v \in W$ we have $u_i = v_i \neq \star$ for some $i \in [n]$.
\end{defnintersectingflats}

Equivalently, a set of flats $W$ is intersecting if any pair of points taken from a single flat or a pair of flats in $W$ intersect.

\begin{defnminimalflats}
 \label{defnminimalflats}
An intersecting set of flats $W \subset (\mathbb{R} \cup \{\star\})^n$ is \emph{minimal} if no flats in $W$ can be enlarged (by replacing a coordinate with a $\star$) while $W$ remains intersecting.
\end{defnminimalflats}

Equivalently, $W$ is intersecting and minimal if $W$ is intersecting and further for all $u \in W$ and all $i \in [n]$ with $u_i \ne \star$, there is some $v \in W$ such that $u_i = v_i$ and for $j \ne i$, either $u_j \ne v_j$ or $u_j = v_j = \star$.

Given a set of pairwise intersecting points $V \subset \mathbb{R}^n$, we can construct an intersecting and minimal set of flats containing all the points of $V$ as follows. Initially, let $W = V$, considering $W$ as an intersecting set of flats (each of dimension $0$). Either $W$ is minimal, or we can enlarge one of the flats by replacing a coordinate with a $\star$, taking only one copy if this produces a duplicate flat. Continue in this manner until no further enlargements can be made.

The set of 3 flats listed below is an example of such an intersecting and minimal system in $\mathbb{R}^4$.
\begin{center}
\begin{tabular}{llll}
1 & $\star$ & 1 & $\star$\\
$\star$ & 0 & 1 & $\star$\\
1 & 0 & $\star$ & $\star$
\end{tabular}
\end{center}

We can use Theorem \ref{thm_crossintersecting} to bound the number of non-$\star$ values appearing in each coordinate position for a minimal set of intersecting flats.

\begin{intersectingflats}
 \label{intersectingflats}
 Let $W$ be an intersecting and minimal set of flats in $\mathbb{R}^n$. Then for all $i \in [n]$,
$$ | \{ w_i : w_i \neq \star, w \in W \} | \leq 2^{n-2}. $$
\end{intersectingflats}

\begin{proof}
Without loss of generality, we will bound the number of values occuring in the first coordinate. We may assume that the values occuring here are $[t]$. For each $j \in [t]$, there are $u^j, v^j \in W$ such that $u^j_1 = v^j_1 = j$, but either $u^j_i \neq v^j_i$ or $u^j_i = v^j_i = \star$ for $2 \leq i \leq n$. Project the last $n-1$ coordinates of $u,v$ to form $a^j, b^j \in \mathbb{R}^{n-1}$ respectively, where we replace $\star$s in $u$ with 1, and $\star$s in $v$ with 2. (There is nothing special about these two values other than that they are distinct.) Then $a^j$ and $b^j$ do not intersect. However, for $j \neq k$, $a^j,b^j$ must intersect $a^k,b^k$ since $W$ is intersecting and $u^j_1, v^j_1 \ne u^k_1, v^k_1$. The collection of pairs $\{a^1,b^1\},\ldots,\{a^t,b^t\}$ satisfies the conditions in Theorem \ref{thm_crossintersecting} and the result follows.
\end{proof}

\subsection{Proofs of the main theorems}

\begin{proof}[Proof of Theorem \ref{thm_strict_set_bound}]
We proceed by induction on $n$. The result is clearly true for $n=1$, so we may assume $n \geq 2$. Let $(\ell_d)_{d \in D_n}$, $(\lambda_I)_{I \subset [n]}$ be as in Theorem~\ref{thm_strict_set_bound}. Let $U \subset \mathbb{R}^n$ be a set of points containing no strictly monotonic sequence of length $\ell_d + 1$ in direction $d$, for all $d \in D_n$. We aim to show that $|U| \leq \lambda_{[n]}$.

Each direction $c \in C_n$ corresponds to a partial order on $U$, where we say $x < y$ if $(x,y)$ is a strictly monotonic sequence in direction $c$. The maximum chain length in the order corresponding to $c$ is $\ell_c$. Recursively construct a sequence of sets $U_0, \ldots, U_{2^{n-1}}$ as follows. Let $U_0 = U$. For $1 \leq i \leq 2^{n-1}$, partially order $U_{i-1}$ with $c^i$. Lemma \ref{minantichainsize} guarantees an antichain $U_i \subset U_{i-1}$ of size $|U_i| \geq |U_{i-1}| / \ell_{c^i}$. Set $V = U_{2^{n-1}}$. No two points in $V$ form a sequence in any of the directions $c \in C_n$, and so every pair of points agree in some coordinate ($V$ is pairwise intersecting). We have
$$ |V| \geq |U| / \prod_{c \in C_n} \ell_c = |U| / m_{[n]}. $$
It remains to show that
$$ |V| \leq \max_{i \in [n]} \lambda_{[n] \setminus \{i\}} = \lambda_{I_0} $$
(for some $I_0 \subset [n]$).

Define a function
$$ F : ( \mathbb{R} \cup \{\star\} )^n \rightarrow \mathcal{P}[n] $$
to give the coordinate positions of the $\star$-coordinates of a flat, i.e., $i \in F(w)$ if and only if $w_i = \star$.

Our inductive hypothesis tells us the following. Let $w \in (\mathbb{R} \cup \{\star\})^n$ be a flat and let $J = F(w) \subset [n]$ be the set of $\star$-coordinates of $w$. Let $X \subset V$ be the points of $V$ that lie in the flat $w$. Then $|X| \leq \lambda_{J}$. Indeed, let $t = |J|$ and let $\pi_J : \mathbb{R}^n \rightarrow \mathbb{R}^t$ project in the coordinate positions $J$. For $d \in D_t$, let $g(d) \in D_n$ be the direction such that $\pi_J(g(d)) = d$ and $g(d)_i = 0$ for $i \not\in J$. Define the set of lengths $(\ell_d')_{d \in D_t}$ via
$$\ell_d' = \ell_{g(d)}.$$
This gives a collection $(\lambda_I')_{I \subset [t]}$ with $\lambda_{[t]}' = \lambda_J$. Then by the inductive hypothesis with this collection of lengths, $|X| = |\pi_J(X)| \leq \lambda_{[t]}' = \lambda_J$.

In particular, let $W$ be an intersecting and minimal set of flats that contain all the points of $V$. Then
\begin{equation}
 \label{eqn_sum_of_lambda}
 |V| \leq \sum_{w \in W} \lambda_{F(w)}.
\end{equation}
It is sufficient to show that the right hand side of (\ref{eqn_sum_of_lambda}) is at most $\lambda_{I_0}$.
If all the points of $V$ lie in a flat of dimension $n-1$, i.e., all points share a single fixed coordinate position $i_0$, then $|W| = 1$ and we are done by the maximality of $\lambda_{I_0}$. So we will assume from now on that there is no single fixed coordinate, and all the flats have dimension at most $n-2$.

Let $J \subset [n]$ be a set of $|J| = s$ coordinate positions. Take a chain of subsets $J = J_s \subset J_{s+1} \subset \cdots \subset J_{n-1} \subset [n]$ with $|J_t| = t$. Since $\ell_d \geq 2$ for all $d \in D_n$, we have that $m_{J_t} \geq 2^{2^{t-1}}$ and hence
\begin{equation}
\label{lambdaforellequals2}
 \lambda_{I_0} \geq
\lambda_{J_{n-1}} \geq
\lambda_{J_{n-2}} 2^{2^{n-2}} \geq
\cdots \geq
\lambda_{J} 2^{2^{n-2} + 2^{n-3} + \cdots + 2^s} =
\lambda_{J} 2^{2^{n-1}-2^s}.
\end{equation}
Thus $\lambda_{J} / \lambda_{I_0} \leq 2^{2^s-1} / 2^{2^{n-1}-1}$ ($2^{2^s-1}$ is the value of $\lambda_{J}$ when $\ell_d = 2$ for all $d$). Thus from (\ref{eqn_sum_of_lambda}), it is sufficient to prove the inequality
\begin{equation}
 \label{eqn_sum_of_lambda_2}
f(W) := \sum_{w \in W} 2^{2^{|F(w)|}-1} \leq 2^{2^{n-1}-1}.
\end{equation}

We consider the specific cases $2 \leq n \leq 4$, before proving (\ref{eqn_sum_of_lambda_2}) for general $n \geq 5$. The cases of small $n$ are necessary to consider as we are inducting on $n$.

\paragraph {$n=2$.} By assumption $W$ consists only of flats of dimension $n-2$, i.e., points. By Lemma \ref{intersectingflats}, at most one value appears in each coordinate. However $W$ is minimal (Definition \ref{defnminimalflats}), so there are at least two values appearing in each coordinate. This contradiction implies that $W$ cannot exist (the only valid $W$ is one containing a single 1-dimensional flat).

\paragraph {$n=3$.} We aim to show that $f(W) \leq 2^{2^2-1} = 8$. By Lemma \ref{intersectingflats}, there are at most 2 non-$\star$ values used in each coordinate in $W$, say, $\{1,2\}$. If all flats have dimension 0, then $f(W) = |W| \leq 2^3$ and we are done.

Otherwise, say, $W$ contains the flat $(1,1,\star)$. By minimality, $W$ must contain a flat of the form $(1,a,c)$ and a flat of the form $(b,1,d)$, with $a,b \ne 1$, and no other forms are possible. Since $W$ is intersecting, $c=d \in \{1,2\}$. Further by minimality, $a = b = \star$. Thus $f(W) = 2|W| = 6$.

\paragraph {$n=4$.} We aim to show that $f(W) \leq 2^{2^3-1} = 128$. By Lemma \ref{intersectingflats}, there are at most 4 non-$\star$ values used in each coordinate in $W$, say, $\{1,2,3,4\}$.

Suppose first that there are only flats of dimension 0 or 1. Construct a bipartite graph $G$ with vertex classes $(X,Y)$ as follows. Let $X \subset \{1,2,3,4,\star\}^4$ be the set of all flats with exactly one $\star$ coordinate. Let $Y = \{1,2,3,4\}^4$. Add an edge between $x \in X$ and $y \in Y$ if $x$ and $y$ differ only in the unique $\star$-coordinate of $x$. $G$ is thus a 4-regular bipartite graph. The set of flats $W$ corresponds to an independent set of vertices in $G$. By Hall's Theorem, $G$ has a matching of size $|W|$, whose set of end vertices in $Y$ is an intersecting subset of $\{1,2,3,4\}^4$. This has maximum size $4^3$ by Lemma \ref{intersectinggridsubset}, hence $|W| \leq 4^3$ and $f(W) \leq 2|W| \leq 128$ as required.

So suppose there is a flat of dimension 2, say, $(1,1,\star,\star)$.
Similarly to the case $n=3$, by minimality $W$ must contain flats of the form $(1,a,c,d)$ and $(b,1,e,f)$ with $a,b \ne 1$, and no other flats are possible. Hence further by minimality, $a=b=\star$.
Denote the set of flats of the first form by $W_1 \subset W$ and those of the second form by $W_2 \subset W$, i.e.,
\begin{align*}
 W_1 &= \{ (1,\star,a,b) : a,b \in \{1,2,3,4,\star\} \} \cap W, \\
 W_2 &= \{ (\star,1,a,b) : a,b \in \{1,2,3,4,\star\} \} \cap W.
\end{align*}
So $W = \{(1,1,\star,\star)\} \cup W_1 \cup W_2$.
The number of 1-dimensional flats in each of $W_1$ and $W_2$ is at most $4^2$.
Suppose $(1,\star,a,\star)$ is a 2-dimensional flat in $W_1$. Then $a$ can take at most 1 value, since otherwise the flats of this form cannot all intersect $W_2$ ($W_2$ is non-empty). The same is true for flats of the form $(1,\star,\star,a)$, so there are at most two 2-dimensional flats in $W_1$, and similarly at most two 2-dimensional flats in $W_2$.
Thus
$$ f(W) \leq 2^{2^2-1} + 2 \cdot 4^2 \cdot 2 + 4 \cdot 2^{2^2-1} = 104 \leq 128. $$

\paragraph {$n\geq5$.}
Let $I \subset [n]$ be a set of size $|I| = s$. By Lemmas \ref{intersectinggridsubset} and \ref{intersectingflats} the number of flats $w \in W$ with $F(w) = I$ is at most $2^{(n-2)(n-s-1)}$. Thus
$$ f(W) \leq \sum_{s=0}^{n-2} \binom{n}{s} \cdot 2^{(n-2)(n-s-1)} \cdot 2^{2^s-1}. $$
It is easy to check that this quantity is at most $2^{2^{n-1}-1}$ for $n \geq 5$. Indeed one can calculate explicitly the values for $n=5,6,7$. For $n \geq 8$, bound the expression by $2^n \cdot 2^{(n-2)(n-1)} \cdot 2^{2^{n-2}-1}$. Looking at exponents, it is sufficient to show that $n^2 - 2n + 2 \leq 2^{n-2}$. This is true for $n=8$, and the right hand side grows faster than the left for all $n \geq 8$.
\end{proof}

\begin{proof}[Proof of Theorem \ref{thm_strict_seq_bound}]
Let $S$ be a sequence with no strictly monotonic subsequence of length $\ell_d + 1$ ($\ell_d \geq 2$) in direction $d$, for all $d \in \{-1,0,1\}^n$. Map it to a set $T \subset \mathbb{R}^{n+1}$ as described in Section \ref{secn_strict_monotonicity_in_sequences}. The proof of Theorem \ref{thm_strict_set_bound} is still valid for the set $T$, with the possible exception of inequality (\ref{lambdaforellequals2}). However, the only $J$ that we need to check (\ref{lambdaforellequals2}) for are those that occur as $J = F(w)$ for some $w \in W$, where $W$ is the intersecting and minimal set of flats that cover $V$. Indeed, let $w \in W$ and $J = F(w)$. Then we must have $w_1 = \star$ since $T$ contains no two points sharing the same first coordinate (the first coordinate is the index of the point in $S$), and hence $1 \in J$. Let $J \subset J_{s+1} \subset \cdots \subset J_{n-1} \subset [n]$ be a chain with $|J_t| = t$, as before. Then since $1 \in J_t$ for all $t$, we again have $m_{J_t} \geq 2^{2^{t-1}}$, and the inequality holds.
\end{proof}

\section{Pillage games}
\label{secn_pillage_games}

Theorem \ref{thm_strict_set_bound} was first motivated by an application to economic theory. We describe it here because it has some combinatorial interest.
Jordan \cite{initialjordan} introduced the concept of a pillage game. The set of \emph{players} in the pillage game is the set $[n] := \{ 1, \ldots, n \}$. A \emph{coalition} is a subset $I \subset [n]$ of the players. The set of \emph{allocations} is an $(n-1)$-dimensional simplex
$$ A = \{x \in \mathbb{R}^n : x_i \geq 0 \mbox{ for all } i \in [n] \mbox{ and } \sum_{i=1}^n x_i = 1 \}. $$
A \emph{distribution of wealth} of the players is a point $x \in A$.
The \emph{power function}, defined below, gives the strength of a coalition of players who have a certain distribution of wealth.

\begin{defnpowerfunction}
 The \emph{power function} is a map $\pi : \mathcal{P}[n] \times A \rightarrow \mathbb{R}$ satisfying
\begin{enumerate}[label=(p.1)]
\item if $C \subset C'$ then $\pi(C',x) \geq \pi(C,x)$ for all $x \in A$;
\item if $x_i' \geq x_i$ for all $i \in C$ then $\pi(C,x') \geq \pi(C,x)$; and
\item if $C \ne \emptyset$ and $x_i' > x_i$ for all $i \in C$ then $\pi(C,x') > \pi(C,x)$.
\end{enumerate}
\end{defnpowerfunction}

The above axioms thus specify monotonicity conditions: (p.1) says that the power of a coalition does not decrease if new members are added; (p.2) says that the power of a coalition does not decrease if the wealth of some of the members is increased without decreasing the wealth of other members; (p.3) says that if the wealth of each member of a coalition is strictly increased then the power of the coalition must also strictly increase.

\subsection{Domination and stable sets}
\label{secn_domination_and_stable_sets}

\begin{defndomination}
 Let $x, x' \in A$. Define the sets $W = \{i : x_i' > x_i\}$ (winners) and $L = \{i : x_i' < x_i \}$ (losers). Then $x'$ \emph{dominates} $x$ if
$$ \pi(W,x) > \pi(L,x). $$
\end{defndomination}

The interpretation of this is that $x'$ dominates $x$ if the coalition of players whose wealth strictly increases in going from $x$ to $x'$ is more powerful than the coalition of players whose wealth strictly decreases in going from $x$ to $x'$, when the wealth distribution is $x$.

\begin{defnstable}
 A set $S \subset A$ is a \emph{stable set} if it satifies
\begin{itemize}
\item (internal stability) no element of $S$ is dominated by any other element of $S$; and
\item (external stability) each element of $A \setminus S$ is dominated by some element of $S$.
\end{itemize}
\end{defnstable}

Note that a stable set need not exist.

Jordan \cite{initialjordan} makes the following observation. Let $S \subset A$ be a stable set. Let $\{x^1, x^2, x^3, x^4\} \subset S$ be four points in the stable set, and for $k=1,2,3$ define
\begin{align*}
 W_k & = \{ i : x_i^{k+1} > x_i^k \}, \\
 L_k & = \{ i : x_i^{k+1} < x_i^k \}.
\end{align*}
We have by internal stability of $S$ that
\begin{itemize}
 \item $\pi(W_{k-1},x^k) \geq \pi(L_{k-1},x^k)$ for $k=2,3,4$ since $x^k$ does not dominate $x^{k-1}$; and
 \item $\pi(W_k,x^k) \leq \pi(L_k,x^k)$ for $k=1,2,3$ since $x^k$ does not dominate $x^{k+1}$.
\end{itemize}
In particular we have that
\begin{align*}
\pi(W_1,x^2) &\geq \pi(L_1,x^2), &
\pi(W_2,x^2) &\leq \pi(L_2,x^2), \\
\pi(W_2,x^3) &\geq \pi(L_2,x^3), &
\pi(W_3,x^3) &\leq \pi(L_3,x^3).
\end{align*}
If $W_1 = W_2 = W_3 = W$ and $L_1 = L_2 = L_3 = L$ then
\begin{align*}
\pi(W,x^2) &= \pi(L,x^2), &
\pi(W,x^3) &= \pi(L,x^3).
\end{align*}
However this violates axiom (p.3) of the power function, since we must have $\pi(W,x^2) < \pi(W,x^3)$ and $\pi(L,x^2) > \pi(L,x^3)$.
Therefore, given four points in a stable set, we cannot have $W_1 = W_2 = W_3$ and $L_1 = L_2 = L_3$.
More generally, it can be seen that we cannot have $W_1 \subset W_2 \subset W_3$ and $L_1 \supset L_2 \supset L_3$ either \cite{kerberrowatpersonal}.
In his original paper \cite{initialjordan} Jordan observed that this implies that any stable set must be finite.

In fact, the situation with $W_1 = W_2 = W_3 = W$ and $L_1 = L_2 = L_3 = L$ is precisely the case that $(x^1,x^2,x^3,x^4)$ is a strictly monotonic sequence in direction $d$, where
$$ d_i =
\begin{cases}
 1 & \text{if } i \in W, \\
 -1 & \text{if } i \in L, \\
 0 & \text{otherwise.}
\end{cases}
$$
(Pairs of disjoint sets $(W,L)$ and directions $d \in \{-1,0,1\}^n$ can both be used to specify an equivalent direction.)
So a stable set cannot contain a strictly monotonic sequence of length 4.

In $A$ any direction $(W,L)$ that appears between a pair of points must have both $W \ne \emptyset$ and $L \ne \emptyset$. There are $c = (3^n-2^{n+1}+1)/2$ pairs $\{W,L\}$ with $W,L \subset [n]$, $W \cap L = \emptyset$, and $W \neq \emptyset \neq L$. Kerber and Rowat \cite{ramseybound} used Ramsey theory to bound the maximum size of a stable set $S$, as follows. Consider the complete graph $K_{|S|}$ on $|S|$ vertices. Associate with each of the $c$ possible directions in $S$ a colour $i$, $1 \leq i \leq c$. Colour the edge $xy$, $x,y \in S$ with the colour associated with the direction of $(x,y)$ or $(y,x)$. Then $|S| < R_c(4)$, the maximum size of a complete graph coloured with $c$ colours with no monochromatic clique of size 4. Indeed, such a monochomatic clique would give a strictly monotonic sequence of length 4, which cannot exist by the observation above.

We have the bounds $3^c < R_c(4) \leq c^{2c+1}$ (the lower bound is a product construction that can be found in \cite{ramsey_lower_bound}. The upper bound follows from a pigeonhole argument; see for example \cite{ramsey_upper_bound}). The upper bound on $S$ can be improved if we make use of the fact that the colours in the graph are induced by the directions of a set of points in $\mathbb{R}^n$. Indeed, by Theorem \ref{thm_strict_set_bound} with $\ell=3$, we immediately have the following.

\begin{stableupperbound}
 \label{stableupperbound}
Let $S \subset A$ be a stable set for $n$ players. Then
$$ |S| \leq 3^{2^n-1}. $$
\end{stableupperbound}

In fact with a little more work one can tighten this bound to $3^{2^n-n-2} \cdot 2^n + n$. We demonstrate some of these ideas for $n=3$ in the next theorem.

\begin{stableupperbound3}
 \label{stableupperbound3}
Let $S \subset A$ be a stable set for $3$ players. Then
$$ |S| \leq 27. $$
\end{stableupperbound3}

Kerber and Rowat \cite{threeagentpillage} have a tighter (in fact, best possible) bound of 15 under certain niceness conditions on the power function. Here we do not make any assumptions on the power function beyond the three axioms given.

\subsection{Towards constructing large stable sets}

Theorem \ref{stableupperbound} gives a super-exponential upper bound for the size of a stable set. In fact, the proof only makes use of internal stability. The next theorem gives a subset $S \subset A$ of a similar super-exponential size together with a power function defined on $S$ which makes $S$ into an internally stable set. We have not been able to determine whether or not the power function could be extended to one defined on all of $A$, while making $S$ into an externally stable set as well. However, this theorem at least shows that we cannot hope to improve substantially on the bound in Theorem \ref{stableupperbound} by considering internal stability alone.

\begin{internallystableconstr}
 \label{internallystableconstr}
Let $n$ be given, and let $d = \lfloor (n-1) / 2 \rfloor$. There exists a set $S \subset A$ of size
$$ |S| = 3^{\frac{1}{2} \binom{2d}{d}} $$
together with a function $\pi : \mathcal{P}[n] \times S \rightarrow \mathbb{R}$ satisfying axioms (p.1), (p.2), (p.3) that makes $S$ an internally stable set.
\end{internallystableconstr}

\subsection{Proof of Theorem \ref{stableupperbound3}}

\begin{proof}
Let $S \subset A$ be a stable set. Let $S' \subset S$ be $S$ minus any corner points that it may contain, i.e.,
$$ S' = S \setminus \{ (1,0,0), (0,1,0), (0,0,1) \}. $$

Suppose, say, that $c = (0,1,0) \not \in S$. Then by external stability, $c$ is dominated by some point $x \in S$, where $x_1,x_3 \geq 0$ and $x_2 < 1$. Let $F \subset \{1,3\}$, $F \ne \emptyset$ be such that $x_i > 0$ for $i \in F$. Then by the definition of domination,
\begin{equation}
 \label{pi_for_corner}
 \pi(F, c) > \pi(\{2\}, c).
\end{equation}
This condition then forbids any pair of points forming a sequence in direction $(1,-1,1)$ in $S'$. Indeed, suppose $y,z \in S'$ with $(y,z)$ such a sequence.
By axioms (p.2) and (p.1) we have that
\begin{align*}
 \pi( \{1,3\}, y ) & \geq \pi( \{1,3\}, c ) \geq \pi( F, c ), \\
 \pi( \{2\}, y )   & \leq \pi( \{2\}, c ).
\end{align*}
But then by (\ref{pi_for_corner}) we must have that $\pi( \{1,3\}, y ) > \pi( \{2\}, y )$, i.e., $z$ dominates $y$, contradicting the internal stability of $S$.

If we do have the point $c = (0,1,0)$ in $S$ then there can be no 3 points in a sequence in $S'$ in the direction $(1,-1,1)$. Indeed, suppose $x,y,z$ formed a sequence in the direction $(1,-1,1)$, or equivalently, the direction $(\{1,3\},\{2\})$. Then $(c,x)$ is a sequence in the direction $(W,\{2\})$, where $W \subset \{1,3\}$. By the observation in Section \ref{secn_domination_and_stable_sets}, $S$ cannot then be a stable set.

So in each of the three directions $(1,1,-1),(1,-1,1),(-1,1,1),$ the maximum length of a sequence of points in $S'$ is either 1 or 2. As in the proof of Theorem \ref{thm_strict_set_bound} we can find a subset $|S''| \geq |S'| / 2^3$ with no pair of points lying in any of these directions. This means that for all $x,y \in S''$ we have $x_i = y_i$ for some $i \in [3]$.

We now show that $|S''| \leq 3$. If all the points in $S''$ share some coordinate, say $x_1 = a$ for all $x \in S''$, then the points in $S''$ form a sequence in direction $(0,1,-1)$, and so $S'' \leq 3$. Otherwise, for some $a,b,c \in [0,1]$, $S''$ contains the points
$$
 ( a     , b , 1-a-b ),
 ( a     , c , 1-a-c ),
 ( a+c-b , b , 1-a-c ),
$$
with $b \ne c$. No further distinct points can be added that share a coordinate with each of the three points above. Thus again $|S''| \leq 3$.

Putting these results together, we have that
$$ |S| \leq 3 + |S'| \leq 3 + 2^3 |S''| \leq 27. $$
\end{proof}

\subsection{Proof of Theorem \ref{internallystableconstr}}

\begin{proof}
 We construct such a subset of $A \subset \mathbb{R}^n$.
For $c \in C_{2d}$, let $p(c)$ count the number of $1$-coordinates of $c$. Define a collection of lengths $(\ell_i)_{i=1}^{2^{2d-1}}$ via
$$ \ell_i =
\begin{cases}
 3 & \text{if } p(c^i) = d, \\
 1 & \text{otherwise.}
\end{cases} $$
By the construction in Section \ref{secn_nonstrict_constr} there is a set $T \subset \mathbb{R}^{2d}$ of size $\prod_i \ell_i = 3^{\frac{1}{2} \binom{2d}{d}}$ with no sequence of length $\ell_i+1$ in direction $c^i$. Moreover, for every two distinct points $x,y \in T$, $x_i \ne y_i$ for all $i \in [2d]$. In particular, every pair of points in $T$ must lie in some direction $c \in C_{2d}$ with $p(c)=d$.

Scale and translate the set so that it lies in $[0,1/(2d)]^{2d} \subset \mathbb{R}^{2d}$. Then map it to a set $S$, $S \subset A \subset \mathbb{R}^n$, by adding either one or two last coordinates (depending on whether $n = 2d+1$ or $n = 2d + 2$) such that the sum of the coordinates of any point is $1$.

We now define a power function on $S$. Let $B \subset [2d]$ with $|B| = d$ and let $x \in S$, with $x$ the image of the point $x' \in T$. Partially order the points of $T$ such that $u < v$ if $u_i < v_i$ for all $i \in B$, so by the properties of $T$, $u_i > v_i$ for all $i \in [2d] \setminus B$. Note that $T$ contains no chain of length 4 by construction. Let $q(B,x)$ denote the furthest distance up a chain that $x'$ lies, i.e.,
$$ q(B,x) =
\begin{cases}
 3 & \text{if there are $u,v \in T$ with $u < v < x'$,} \\
 2 & \text{if otherwise there is some $u \in T$ with $u < x'$,} \\
 1 & \text{otherwise.}
\end{cases} $$
Now define $\pi : \mathcal{P}[n] \times S \rightarrow \mathbb{R}$ via
$$ \pi(C, x) =
\begin{cases}
 \max_{i \in C} x_i     & \text{if $|C \cap [2d]| < d$,} \\
 q(C \cap [2d],x)       & \text{if $|C \cap [2d]| = d$,} \\
 \max_{i \in C} x_i + 3 & \text{if $|C \cap [2d]| > d$}
\end{cases} $$
where we take $\max_{i \in \emptyset} x_i = 0$.

This satisfies axiom (p.1), since $0 \leq \max_{i \in C} x_i \leq 1$ and $1 \leq q(B,x) \leq 3$.

Axioms (p.2) and (p.3) are satisfied when $|C \cap [2d]| \ne d$, as can easily be checked. So let $C$ be such that $|C \cap [2d]| = d$. Let $x,y \in S$, $x \ne y$, with $x_i \leq y_i$ for all $i \in C$. Then in fact $x_i < y_i$ for all $i \in C \cap [2d]$, and so $x < y$ in the partial order given by $C$. Thus $q(C \cap [2d], x) < q(C \cap [2d], y)$. Thus axioms (p.2) and (p.3) are satisfied in this case as well.

It remains to check internal stability, i.e., that no point in $S$ dominates any other point of $S$. Let $x, y \in S$ be two distinct points, with $W = \{ i : y_i > x_i \}$ and $L = \{ i : y_i < x_i \}$, and let $x', y'$ be their original points in $T$. By construction of $T$, $|W \cap [2d]| = |L \cap [2d]| = d$. In the partial order on $T$ induced by $W \cap [2d]$, as described above, we have that $x' < y'$, and so $q(W \cap [2d], x) \leq 2$. Similarly $q(L \cap [2d], x) \geq 2$. Thus $y$ does not dominate $x$, and we are done.
\end{proof}

\paragraph{Acknowledgements} The author would like to thank Manfred Kerber and Colin Rowat for drawing attention to this question and Andrew Thomason for his helpful suggestions and comments.

\bibliographystyle{elsarticle-num}

redpajama_set_name: RedPajamaArXiv


text:
Tag: Europe. Showing results 1 - 10 of 25
Scholarship helps promising opera singer's dreams in Europe
6:01am, Tuesday 30 November 2021
Natasha Te Rupe Wilson's interest in classical music came to fruition as a teenager.
Ngāpuhi
Te Arawa
University of Auckland
Toi moko returned to Aotearoa, described as reconciliation for Māori
3:39pm, Tuesday 3 November 2020
The tūpuna (ancestors) began their journey home in ceremonies at the Berlin Ethnological Museum in Germany in October.
Te Papa Museum
Alien Weaponry work on their 2020 arsenal
9:41pm, Tuesday 31 December 2019
Te Reo Māori metal band, Alien Weaponry have recently returned home from their latest tour in Europe. The group incorporates te reo Māori, whakataukī and Māori history in their music, a concept that has made European metal fans bang their heads and slap their chests.
Māori interests 'not entirely straightforward' in NZ-EU free trade negotiations
3:58pm, Friday 9 August 2019
New Zealand is currently in negotiations with the European Union (EU) over a potential Free Trade Agreement (FTA).
Ministry of Foreign Affairs & Trade
Ngāti Rānana long-time member remembered
1:33pm, Friday 5 July 2019
A longstanding member of Ngāti Rānana (London's Māori Club), Joylene Winitana Fenikowski, 51, passed away earlier this week in England and will be brought back to Aotearoa by her whānau to her beloved rohe of Te Mahia.
Ngāti Rānana
Rongomaiwahine
Brothers suicide drives Oceania Mountain Biking champ
7:16pm, Sunday 23 June 2019
Tuhoto Ariki Pene's pre-race ritual involves meditation while remembering his late brother who he lost to suicide.
Overseas 'haka' businesses remove footage
9:19pm, Tuesday 4 June 2019
The growth of the internet is exposing Māori culture to the world but incidents of cultural appropriation are also becoming more common in the digital era. Founder of Māori Movement, Ngarino Te Waati, speaks to Te Ao Māori News about having his own image taken without consent.
NZ and UK sign interim trade deal
6:09pm, Tuesday 22 January 2019
New Zealand businesses will be no worse off, whatever the outcome of the Brexit process according to Prime Minister Jacinda Ardern after her meeting with UK Prime Minister Theresa May.
How Brexit could affect NZers
6:28pm, Wednesday 16 January 2019
Former NZ Labour Party general secretary Tim Barnett says it could be much harder for New Zealanders seeking employment in Europe if Britain leaves the EU. Today UK MPs rejected Prime Minister Theresa May's Brexit deal, leaving the future of the country up in the air.
Hato Hohepa return home from Europe
7:26pm, Sunday 23 December 2018
Hato Hohepa students have returned home from a once-in-a-lifetime, three-week trip to Europe. The group have become "facebook-famous" with their popular performance videos going viral. An auspicious time as the Māori boarding school celebrates their 150th anniversary.
Hato Hōhepa
Kura Tuarua - Secondary School

redpajama_set_name: RedPajamaCommonCrawl


text:
Alito 5 Must Stay the Course
June 27, 2022 May 6, 2022 by Patrick J. Buchanan
Pro-abortion Democrats say a woman's "right to choose" must remain paramount and sacrosanct. But the "right to choose" what? As President Joe Biden just described it bluntly this week, it is the right to choose to "abort a child." In February, five Supreme Court Justices voted in camera to overturn …
America: a Land of Ceaseless Conflict
October 24, 2020 October 23, 2020 by Patrick J. Buchanan
What these episodes suggest is that the idea of bipartisan comity, or some new era of national unity should Biden win, is self-delusion. When Amy Coney Barrett was nominated to a judgeship on the U.S. Court of Appeals in 2017, Sen. Dianne Feinstein was taken aback by the Notre Dame …
Ten Days That Shook the Presidency
October 6, 2020 by Patrick J. Buchanan
Trump has four weeks to turn it around. And his task, while easy to describe, is not so easy to accomplish. He needs to persuade undecided and soft Biden voters that Joe is simply not up to the job of president. What a difference a week can make. Saturday, Sept. …
Will Justice Amy Star in 'The Five'?
May 6, 2022 September 28, 2020 by Patrick J. Buchanan
America's court wars, in which the coming battle over Barrett's nomination may prove decisive, go back half a century. By nominating Federal Judge Amy Coney Barrett to the Supreme Court, Donald Trump kept his word, and more than that. Should she be confirmed, he will have made history. Even his …
All the Chips Are on the Table Now
September 25, 2020 by Patrick J. Buchanan
On the court issue, Democrats are exhibiting something akin to panic. They are warning that if a conservative jurist like Barrett is confirmed, Democrats may retaliate by "packing" the Supreme Court — increasing the number of justices from nine to 11 and installing two new liberals — if they win …
Last Best Chance to Capture Supreme Court
Are there Republicans who would really walk away from this last, best chance to secure the court, simply because the process offends their sense of proper procedure? This may be a hard vote for Murkowski, Collins and a few other Republicans. But to vote down a qualified conservative nominee, on …


The Marquee – July 2018 The Marquee – Volume 16 No. 5 featuring interviews with YOB, moe., and Slightly Stoopid with Colorado-only album reviews of In The Whale, Paul Kimbiris, Darren Garvey, The Martin Project and Paul DeHaven. Plus, Colorado's most complete concert calendar. Period!


The oddest thing about the innocuous 1977 comedy Melody of Love is that the director is Mau Dui-Fai, famed Category III pioneer. Not only that, but the guy may have directed stuff harder than Cat. III, according to this Bullets Over Chinatown post.
Melody of Love is harmless fun with Danny Lee in a starring role. The film was presumably made for the Taiwan market. I say that as the DVD only had a Mandarin language track and most other Shaw titles made after 1972 were released on DVD with both Cantonese and Mandarin language tracks. Additionally, this HKMDB link lists the country of production as Taiwan; it just doesn't feel like a local Hong Kong Shaws production for some reason.
Danny Lee is Dr. Xiao Erke and Liu Shang-chien and Liu Lu-Hua are two other doctors in this story of three young men romancing three young women.
I'm not going to try to make this sound like more than it is as it's basically "Three's Company"-level comedy for the 90-or-so-minutes of this thing.
I mean, I'm not spoiling anything by warning you that the same gag of a guy trapped on a car door as it pulls away is played out twice in this film. In fact, what this film calls to mind is not so much a sitcom but one of those live action Disney films from the 1970s. I half expected Tim Conway to show up at some point.
As for the three ladies in the film, former Shaw Brothers child star Niu Niu is appealing but I was more interested in Siu Yiu, star of 1977's Mighty Peking Man. The third lady is played by Chen Szu-Chia, who was in The Magic Blade (1976).
Melody of Love is hardly a masterpiece but it's a refreshing non-Wong Jing-style comedy that is a pleasant time-killer.
Add to that, it's further proof of the versatility of leading man Danny Lee; he really could do any genre of film. He's no Olivier but he is appealing in a very basic way.
You can order Melody of Love with Danny Lee on DVD here.


Best of three 13-point games. Level play. No time limit.
Record result on green handicap card.
Contact your next opponent as soon as his/her name is known and arrange to play your next match as soon as possible.


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Shop our selection of Bar and Buffet tools,supplied by some of the most trusted vendors in the foodservice industry. Whatever the size of your venue from a large restaurant to a neighborhood bar, you'll find all the Cooling-Containers you need.
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In response to the developing medical catastrophe in Haiti, InsideSurgery is sharing information with the American College of Surgeons, American Association of Orthopedic Surgeons, many NGOs and the wider medical blogging community though a 60 member listserv.
Please check the links below or contact this site through insidesurgery at gmail then the dot then com if you wish help in disseminating information or wish information on how to volunteer your services.
Non-ACS members can register at https://acspbls.resiliencesoftware.com/Haiti-registration.
The system will automatically add Haiti as a location, and surgeons can start adding cases immediately.
ACS members who have not registered to use the case log system can register at http://acscaselogregister.org/. Once logged in, members can add "Haiti" as a location for cases associated with relief activities.
The ACS has gone live today with a comments and questions forum where deployed physicians can give updates or those deploying can leave questions for ACS surgeons.
The ACS has now gone live with a database for those wishing to register as volunteers. All medical professionals are encouraged to use this database.
The ACS is leading medical volunteer efforts for Haiti and has now gone live with a social media page listing relevant twitter feeds and social media contacts. Information and links can be found at Operation Giving Back.
I've served as a surgeon on mission trips to an established clinic in Ouanminthe, Haiti. The clinic has an operating room, large (8-10 bed under "normal" circumstances) pre-op/post-op ward, and is supported by a school in the immediate vicinity that could house and feed a large staff and patients. The clinic is secured by a fence, and the school site is partially fenced. the school has a large "bunkhouse", kitchen facility, gymnasium, and large, very flat, school yard that could be used as a staging area, camp site, recreational facility, etc.
Regrettably, Ouanminthe is 12-13 hours away from the immediate quake area by land, but would only be 40-45 minutes by air. The airport would support light fixed-wing planes and there are innumerable sites appropriate for helicopters. It lies on the border with the DR, adjacent to the town of Dajabon, making it fairly accessible from the DR without having to negotiate the Port-a-Prince airport. The exodus has begun from the stricken area, and the victims are soon to arrive in cities like Ouanaminthe regardless whether transported in a coordinated effort or by desperation for care.
Institution Univers (clinic and school) are anxious and committed to helping their families and countrymen, and this would serve as an excellent resource for an established aid organization.
We will know more in a few days when access routes from Port au Prince open up and more patients are triaged out to us. Milot is 70 miles north of PAP , 8 mile from Cap Haitien airport. Commercial flights are fully booked. Teams that did not already have reserved seats have to travel by private aircraft so we are looking for donated craft. If you can help us there is would be greatly appreciated! Also need private craft to fly in supplies to Cap Haitien. We are well stocked with supplies at the moment but that will change very quickly. Trauma and ortho supplies especially needed right now. Amputations are what we are seeing most. Crutches very important.


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An asteroid, with a diameter of about 500 m, hits Lappajärvi at about 50 km per second. The impact is equivalent to a million Hiroshima atomic bombs, but without the radioactive radiation. The explosion kills everything within a few hundred kilometres.


Friendly staff and great working cars, I've gotten a few cars with this company and never had to repair any of them.
Owner is a crazy maniac that cursed me out for asking questions. Told me if I want d to test drive a car I had have cash in hand. I told him forget it, and he berated me. Eventually he told me he would bury me on his propert if I showed up to look at the car. Pretty insane. Steer clear of this loony SOB!
Thiện Long Lawn Care Services, LLC.
Imidiate response time Excellent work ethics Brought the beauty of our Lawn Back to life Thank you Thien Long!!
Very professional work and awesome customer service!! Highly recommended.


I need to get some for a buddy who is wanting to shoot long range. Who sells them?
Thanks a ton but I am going to be out of town when you are here. I think I saw some at Wholesale sports a few months ago so I may need to go check it out.
x2 on P+D. If you can't secure some, let me know as I have more than I need right now.
If you want to get rid of some Bobby then send me a PM. I need 7mm bullets around in the 160gr weight. I think they are 168 if I remember. Hunting bullets only. If you have too many and want to dump a box or 2 let me know.


The elite high school and university admissions process in the US and Canada is a complex, often difficult to navigate, and vastly different from admissions processes in other countries. CS Educacional serves its clients with tailor-made programs to provide students at all levels of study (middle school through Ph.D.) with customized experiences that can meet their educational, extra-curricular (sports, music, art, dance, acting, etc.) and personal needs. CS Educacional eliminates the confusion that can hinder, or even derail application procedures and athletic recruiting processes and assists our clients in navigating through the complexities that come with having a myriad of options, ensuring nothing stops you from your goal.
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Marc Donaldson launches new site in-line with re-branding and colour scheme.
If you happen to know me you will know that I have been working really hard over the last couple of days to re-brand & re-launch my own personal website.
This website was built with new development methologies / design guidelines, new technologies, and of course has new functionality to help engage my visitors, as my previous site was not as flexible and looked very dated.
Well, I am extremely happy to say that I am glad that I bit the bullet, and told myself to schedule in some time to work on it as i am very pleased with the outcome so far.
I did find it was a difficult process at first coming up with a design that could work for both mobile / desktop and portable platforms and still keep the content readable,fluid and easy to use for my website visitors.
Not being a expert designer I had to look at other creative / corporate and personal websites for inspiration which took a bit of time as I had to look at these designs and then add in my own flair to make it have originality.
I hope you like my new look and find the site helpful. This is only the start of my new site and I will be making ongoing changes to the site.
Stay tuned! Oh and watch out for new posts to the blog too..


World Mosquito Day, observed annually on 20 August, is a commemoration of British doctor Sir Ronald Ross's discovery in 1897 that female mosquitoes transmit malaria between humans. Ross is responsible for the annual observance, having declared shortly after his discovery that the day should be known as World Mosquito Day in the future.
The London School of Hygiene & Tropical Medicine holds Mosquito Day celebrations every year, including events such as parties and exhibitions, a tradition dating back to as early as the 1930s.
^ a b "World Mosquito Day 2010". Department for International Development. 20 August 2010. Archived from the original on 21 November 2012. Retrieved 21 November 2012.
^ "Health and Science - Abuzz over malaria on World Mosquito Day". AlertNet. Retrieved 21 November 2012.
This page was last edited on 2 June 2018, at 07:05 (UTC).


text:
49ers-Cowboys history: 40 years ago today was…
49ers-Cowboys history: 40 years ago today was 'The Catch' that launched a dynasty
Monday marks the 40th anniversary of 'The Catch' Dwight Clark made to lead the 49ers past the Cowboys, a matchup that will be renewed this Sunday
San Francisco 49ers receiver Dwight Clark hauls in a six-yard touchdown pass from Joe Montana with 51 seconds left in the 1982 NFC Championship game against the Dallas Cowboys. Immortalized in sports lore as "The Catch," Clark's reception capped an 87-yard drive that propelled the 49ers into the Super Bowl. The Cowboys play the 49ers in Sunday's NFC Championship game, the first such matchup since "The Catch." (Phill Huber/Dallas Morning News/KRT)
By Cam Inman |
PUBLISHED: January 10, 2022 at 9:01 a.m. | UPDATED: January 10, 2022 at 9:01 a.m.
Sunday's playoff-clinching comeback surely elicited memories of the 49ers' cherished history.
Their next opponent sure should.
Today marks the 49ers' 40th anniversary of "The Catch," Dwight Clark's iconic touchdown off a Joe Montana throw beat the Dallas Cowboys – this Sunday's wild-card opponent.
Want to relive that?
The 1981 team's NFC Championship Game win, that is, all due respect to Sunday's 27-24 overtime win at the Los Angeles Rams.
Streaming on Peacock TV is the "Joe Montana: Cool Under Pressure" documentary. Episode 2 offers tremendous anecdotes from not only Montana and Clark but an up-and-coming coach named Bill Walsh.
Walsh not only rehashes the "Sprint Right Option" game-winning call but is shown lambasting the Cowboys' trash talk leading up to their playoff rematch.
"This goddamn Dallas team, they can't keep their mouth shut, you know?" Walsh says in a grainy video in front of a whiteboard during a team meeting. "They always like to make press releases out of Dallas how they're going to kick somebody's ass, and they're at it again.
"I'm sick of them. I hope you guys are feeling the same way I am."
Montana sure did, and he recalled during the documentary how, after juking Ed "Too Tall" Jones and completing a midfield pass to Clark, Montana turned to Jones and said: "Respect that (expletive)."
Joe Montana poses with a statue of himself unveiled at Levi's Stadium in Santa Clara, Calif., Sunday, Oct.21, 2018. The statue pairs depicting Dwight Clark "The Catch", the duos iconic touchdown play from 1982. (Karl Mondon/Bay Area News Group)
Of course, Montana eluded Jones and two other Cowboys in the final minute to launch a third-down, 6-yard touchdown pass to Clark with 51 seconds left in their 28-27 win at Candlestick Park – with a young Tom Brady with his mom in the upper deck.
"You guys earned it and deserve it, because you're the best team in football," owner Eddie DeBartolo said in the joyous locker room, per the Montana documentary. "I'm so proud of you. I don't think anything top this. Anything in my life. And I love you all, and I thank you."
Added Walsh: "Looking back on it, that was Camelot. That was Camelot. That was the greatest experience for everyone involved that they're ever have in their life, to go through a season like that. That was the turning point. That was the origin of a dynasty."
Clark, via footage from the locker-room interview: "That was a play we worked on all year. It really hasn't worked like it did today. I was the secondary receiver. Freddie (Solomon) was supposed to get the ball. It became wide open. Joe made a great throw."
The documentary then cuts away to Walsh in the meeting room, describing the "Sprint Right Option" to players, and Montana notes how they not only scored with the play earlier that game against the Cowboys, but how Walsh made them run the play repeatedly in practice at camp.
Clark, before passing away in 2018, said in the documentary of Montana's throw: "My amazement with Joe is always that, in the frickin' situation that sends you to the Super Bowl, he's rolling right, Freddie slips and he's covered, there's three Cowboys in front of him, so he can't run, has the calm to say, 'I've got to get these guys up in the air.' Pumps them, they jump, and he throws it in the exact spot it had to be. It was just a perfect pass, a magical moment from Joe Montana."
Said Montana: "There was a big sigh of relief, because it was such an important game for us to get past. Because of the preparation that went into it, knowing what the results would be if we won, made it that much more special.
"It was great to beat the Cowboys but now we were going to the Super Bowl."
They went on to win the franchise's first of five Super Bowls, the fifth of which came after the 1994 49ers beat the Cowboys in their last playoff meeting until this Sunday's.
Cam Inman


'Photographer perks' are a great thing. Aka having an excuse to whip my camera out in the best moments on road trips. Last weekend, my gals and I traveled to Marfa & Big Bend, TX for four adventure days. We slept in a teepee, didn't reach Prada (but found a Target), were deemed superheroes by a Marfa native, swam in a natural hot springs off the Rio Grande, and sought out some more big rocks to scale (see last year's RT blog post, The Great Expedition). Honestly, it may seem borderline obnoxious, but I love the ability to capture these moments and to have them documented forever. I won't get these days back with these friends at this age, and if I can shoot my friends and Creation in all their beauty and giggles and shades of colors, without getting caught up in vanity and the plagues of social media, I will. So, this is an assurance for all the skeptics: we did in fact enjoy nature and each other on this trip, even with the occasional Instagram post.
p.s. I am ITCHING to do a real shoot in Marfa. The way the sun sets on the town and all its eclectic structures....WHAT A DREAM. So, if I have any adventurous clients out there, holla at me.


A very rare Glenmorangie distilled in 1972 then aged for 22 years and bottled at cask strength in 1994 from cask 1831 and a limited edition of only 250 bottles. A very small amount of Glenmorangie, distilled in 1972, has now been selected for this vintage. After a long and slow maturation in carefully selected casks of American oak, this vintage has been bottled directly from the individual cask. Each cask matures the whisky in a slightly different manner and will only yield around 250 bottles. This gives subtle nuances of taste, making the Single Barrel Vintage bottling a truly rare and individual expression of Glenmorangie. A very rare and great vintage single malt whisky from the Glenmorangie distillery! A must for collectors of Glenmorangie and rare Single Malts.


I've spent my whole life being told that Christ paid for my sins and I believed that. I believed if I did my very best, then Christ would make the difference. When I was a teenager I remember being told he didn't just suffer for my sins, but for all the pain and sorrow I would ever feel. He had felt it all and could take it from me. I always found comfort in knowing the Christ truly understood how I felt when I was sad, angry, hurting, etc. I didn't feel as alone, but it's always been hard for me to understand how he could just take it away, how he could take my pain at this very moment. I felt it at times, but still didn't really understand the atonement.
I think it became real to me about a year ago. I'm not sure if someone recommended this or I just thought of it, but I tried visualizing Christ standing there in front of me with his arms wide open, his face filled with love, a welcoming light glowed from within him. I approached him with a box that held all my anger, frustration, and hurt I was feeling due to some recent events out of my control. I walked up to him, handed him the box and he turned and carried it away and at that moment I felt lighter, like a heavy load had been lifted.
Recently I found myself once again struggling. This time due to something that happened long ago. Something that I thought I had forgotten or moved on from, but here it was again – All the heartache from long ago. I felt exhausted and longed for relief. I went for a walk trying to clear my head and as I was praying I realized I didn't need to carry this anymore. I felt inspired to visualize Christ once again and give him my heartache and burdens. I visualized myself approaching the Savior and this time I had a backpack filled with the heavy burdens of disappointments and pain upon my back. I was weary from carrying these burdens and as I approached the Savior I took the bag off my back and I placed it by his feet and he wrapped me in his loving arms and just held me. I at last felt comfort and peace.
I'm sharing this in hopes that it may help someone else remember that Christ loves them and he truly understands. Take your burdens to him. Try visualizing him right there in front of you. We aren't meant to spend this life carrying all the heartache and pain. Give it to Christ and you will become free to become all you are meant to be, not held down by all the heavy burdens you have been carrying.


Once you have your Apple Device connected to iCloud Unlock Buddy the Software will enter sync mode similar to the one used by iTunes and gather vital data about your device to use that information in preparation for the unlocking process to ensure all preparations are taken to Backup and Wipe your specific Device Model.
Nov 28, 2016 · But is iCloud Unlocking possible by using Unlock iCloud Software? Read this article and find out the whole truth about Unlock iCloud Software and finally which is the only way to turn OFF the iCloud Activation Lock! What Unlock iCloud Software you can find on Web Search Engines. Try google "Unlock iCloud Software"!


text:
Q: aws-sdk getSignedUrl for videos in Safari or IOS does not work My issue is that Safari (and it seems like all browsers on IOS) does not show a video when i pipe the response to the frontend. The videos are however shown on Android and win/linux Desktops on all browsers.
Note, i believe the videos were shown previously in IOS and Safari, this could have been a recent bug or new change to IOS/Safari policy?
My backend looks something like:
...
const aws = require('aws-sdk');

aws.config.update({
  accessKeyId: process.env.ACCESS,
  secretAccessKey: process.env.SECRET,
  region: process.env.S3_REGION
});

const s3 = new aws.S3();

function getS3Url(filename) {
  var params = { Bucket: process.env.BUCKET, Key: filename };
  return new Promise((resolve, reject) => {
    if (filename) {
      s3.getSignedUrl('getObject', params, function(err, url) {
        if (err) {
          reject(err.message ? err.message : err);
        }
        resolve(url);
      });
    } else {
      resolve('/none.png');
    }
  });
}

// Return file
router.get('/file/:filename', (req, res) => {
  const { filename = '' } = req.params;

  getS3Url(filename).then((url) => {
    request(url).pipe(res);
  }).catch((err) => {
    return res.status(codeToInt(err.code, 404)).send(err.message ? err.message : err);
  });
});

I also save the mediaType "video/mp4" which is used in the source tag.
The resulting file-url in my frontend will be something like:
src="www.myhost.com/api/file/123" type="video/mp4"
The media before beeing uploaded to S3 can be viewed in Safari, so it must be the way I pipe the response with getSignedUrl?

A: Good solution
The solution was to add Cloudfront to AWS.
I did not need to implement my own streaming service, instead i returned the signed urls to the frontend, which also solves another issue i had with using my server as a middleman when sending images/videos

"Almost" solution
My issue seems to be that IOS/Safari only supports streamed videos
I implemented a streaming response if the client is IOS/Safari:
function getS3StreamUrl(filename) {
  var params = { Bucket: process.env.AWS_BUCKET, Key: filename };
  return s3.getObject(params).createReadStream();
}

const isAnnoyingClient = isIOS || isSafari;

if (isAnnoyingClient) {
  getS3StreamUrl(filename).pipe(res);
} else {
  getS3Url(filename).then((url) => {
    request(url).pipe(res);
  }).catch((err) => {
    return res.status(codeToInt(err.code, 404)).send(err.message ? err.message : err);
  });
}

But now, my new issue is that videos in IOS/Safari will not loop, they will only play once..

redpajama_set_name: RedPajamaStackExchange


The regular Scottoiler kit works well and is easy to mount on a DR650. I didn't want to have to fill the tank every 400-800 miles since I ride that much weekly sometimes. So the touring kit seemed like the best idea.
The DR has plenty of room on the license plate bracket. You just have to cut off the bumps molded into it that the plate mounts to stock. I just used a hacksaw layed flat against it and cut them off. From there Just drill the center lower hole and bolt the tank on. The top holes go through the opening left from the license plate mount but with a washer on the back they hold on just fine.
The hoses route easily along side the stock wiring and had no issues. I just used the stock wiring hold down locations and fed the hoses through them. The vacuum source is easy to find at the fuel petcock. I just cut the line and put the tee on it.
The oiler tube took a little bit of messing around. I cut the brackets up that came with the kit and mounted them to hold the tube in a good location. I had to cut away a little bit of the chain guide but not enough to weaken it.
Now the oiler is set up and working good. The tube isn't really noticeable on the swing arm. Right now it is clean compared to a dirty bike but soon it will be just as dirty and be pretty much invisible.


While we aren't overly experienced cruisers like some of our friends who have been on multiple cruise lines to multiple locations, but we have done a fair amount of cruising this year. This has prompted some questions from folks who haven't cruised but think they might like to. So we thought we'd do a post with some of those questions. Feel free to ask us more in the comments below.
On your first cruise, did you go with friends, or just you and Stu? Do you think going with friends makes it easier, or does going by yourself just make it that much more part of the adventure?
We usually cruise alone, but did one cruise with a couple of firefighter buddies (and their wives). Stu and his late wife did cruise to Hawaii with friends. The secret, for us, when cruising with friends or family is to set boundaries. You don't have to do everything together. We usually just meet for dinner unless an excursion happens to suit us all. The more flexible you are, the more fun it will be. Don't lock yourself into being together 24/7. Relax!
How do you normally book your cruises... Thru a travel agent or do you do it yourself by watching for cruise deals?
We don't use a travel agent, at least right now. If we had a good friend in the business or if our travel plans were complicated, we might. Most cruise lines will assign you a personal planner to help you, if needed. When we have had to cancel/reschedule a cruise, we have used them. But typically we book online or we book while on a cruise (where we can get on-board credit for that cruise).
Some folks only book discount or last minute cruises through a variety of outlets. We subscribe to several newsletters but have never booked that way.
Did seeing nothing but sea bother you?
If that is a concern, I would get an inside/interior cabin or one with just an ocean view. When you go to sites like Cruise Critic, you can see deck plans and often even what the room would look like. Spend your days at sea away from the edges of the boat. There is plenty to do without walking on the outer decks. That might help any fears that could crop up.
Have you ever gotten sea sick?
I got close once. We were in the Bering Sea, coming back from our Alaska inside passage trip. I had eaten a lot of fruit that date (think citric acid), the sea was rough and after I ate an appetizer of more fruit...oops! I went to my cabin, drank ginger ale and rested that night and the next day.
Stu seems to never be bothered by the boat movement. Last January the seas were a bit rough. I just stayed in the cabin, resting as needed and ordering room service. Love their grilled cheese sandwiches with chips!
We see many folks wearing the patches to fight seasickness. We don't but we do carry a small zipper bag with medical things we might need. Seasickness pills, anti-diarrhea pills, allergy/sinus pills, pain pills, aloe with lidocaine (great for both sunburn and bug bites), hydrocortisone cream, triple-antibiotic cream and both daytime and nightime cold meds. We put this together after we spent almost $40 for Stu to get some cold meds in the ship store, a necessity so he could scuba dive the next port. They are all small containers and now that we have it, we may never need it. LOL!
What about vaccinations? Have you had to get any?
The only vaccination we have had to get for our cruises was because we had been in South America recently. We had to get the Yellow Fever vaccination before going to Barbados. The best thing to do is talk to your primary care physician about the countries you plan to visit.
Which is best - a passport card or a full passport? It confuses me to figure out which I need when.
We have both. We always carry our passport card, even in the US. It's a great means of providing valid identification.
Passport Book. $110 Valid for all international travel.
Passport Card. $30 Valid only for return to the U.S. by land or sea from Canada, Mexico, Bermuda, and the Caribbean.
Passport Book & Card. $140 Apply for both documents at the same time.
Since we are cruising by sea with the US at both ends, we tend to carry both. Why? We can go ashore with just the card for ID. If something were to happen to us while on shore that would necessitate our flying home, the cruise line can get our passport(s) out of our stateroom safe (if our spouse wasn't available to do so).
And on your first cruise ever, did you feel a bit intimidated by 1) simply being on a humongous ship going out to sea, and 2) not knowing where anything was and the schedules?
The good news is every floor has a map of the ship with a YOU ARE HERE marker. It shows what is on each deck, where the elevators are (you will typically be in one of the elevator areas when looking at these), pretty much everything except the restrooms (we just ask a staff member). You also will often be handed a deck plan when you get your stateroom card. Carry that with you!
Speaking of stateroom cards, your cabin number will NOT be on it. That is for security reasons, so either mark it on your deck plan that you are carrying with you or better yet, memorize it. Yes, we have had to go to Guest Services and ask what our room number was when we forgot. LOL!
On both Princess and Carnival cruises, there is a daily flyer put in your room at bedtime turndown that outlines all the activities for the next day. This page shows a good example of what is in the Carnival Fun Times. You can also pick them up all over the ship during the day, just in case.
Did you have any expenses shipboard in addition to the cruise cost? Did you do tips on board?
First, tips. Most cruise lines have gone to pre-paid gratuities. Those will be added to your cruise cost right at the beginning of the cruise. This method allows them to pay a tip to the folks behind the scenes. Here is Carnival's schedule.
Often you may find that you would like to tip a little extra to staff that you feel have gone out of their way to help you during your vacation. For us it might be the head room steward, the head waiter and possibly a helper, the coffee bar waiter you saw the most or the server at your favorite bar.
But the food in the main dining room, all the buffets, the lido/pool area and most of room service items are all included (although a small tip is recommended for room service). Coffee, tea, iced tea, lemonade and in the morning, juices, are free all day. Same for the self-serve ice cream that is open 24 hours.
All the extra items are added to your room charge (all done with your stateroom card - no cash needed until last day if you cash tip as mentioned above). You can view your statement at any time via the TV in your room, kiosks located in several areas or by going to guest services. Typically you put your charges on a credit card, some folks will use a debit card and other put down cash. Obviously, if you exceed your limit - bank account - cash deposit, you will need to find a way to pay your balance on the last day.
So on a normal 6 to 7 day cruise, how much do you estimate expenses, aside from the original cruise cost?
We aren't heavy drinkers, we rarely gamble (I never do), we book our excursions in advance (and thus pay in advance for them when we book them). We allow for tips for staff, tips for Stu's diving guides, a little for drinks (valid prices as of Nov 2015), some shopping both on the ship and at ports (varies by what you want to do - for us, I used to allow $100-150 for shopping, now it's more like $50 and I don't usually spend that), tips for porters on either end of the trip (both land and air) plus one steakhouse visit ($35 each). The spa is pricey and I rarely use it but I have started paying the $60 for internet for the week.
Of all your cruises, which do you think you most enjoyed?
That's hard to say since they are all different. Because we winter in Florida, it's very easy for us to cruise the Caribbean. It works out well because we've been to all the ports, some multiple times. Stu scuba dives and I write!
We enjoy our partial transit Panama Canal out of Ft Lauderdale a few years back, we'd still like to do the full transit at some point. Alaska was amazing and we'd like to do it again. We'd also like to do a Scandinavian cruise, possibly on Princess or Royal Caribbean. We are doing a seven day Royal Caribbean next February, we'll see how we like them.
What do you like about cruising?
That's easy. You don't have to do ANYTHING - no cooking, no cleaning, no bed making, no cleanup, no laundry, NOTHING has to be done unless you want to!
I hope this has answered a few questions, feel free to post more in the comments below and we will do our best to answer them.


text:
// Copyright (C) 2017 History in Paderborn App - Universität Paderborn
//
// Licensed under the Apache License, Version 2.0 (the "License");
// you may not use this file except in compliance with the License.
// You may obtain a copy of the License at
//
//      http://www.apache.org/licenses/LICENSE-2.0
//
// Unless required by applicable law or agreed to in writing, software
// distributed under the License is distributed on an "AS IS" BASIS,
// WITHOUT WARRANTIES OR CONDITIONS OF ANY KIND, either express or implied.
// See the License for the specific language governing permissions and
// limitations under the License.

using Xamarin.Forms;

namespace PaderbornUniversity.SILab.Hip.Mobile.UI.Controls
{
    public class ZeroPaddingButton : Button
    {
    }
}

redpajama_set_name: RedPajamaGithub


Desc: Sterling Silver hunter jumper necklace. The Sterling Silver chain is heavy duty cable link and has a spring ring clasp. The pendant is fashioned to resemble an English stirrup iron with a rider and horse jumping through the center of the stirrup and over a jump. The bale is fashioned to resemble stirrup leathers. The necklace has a bright polish finish. This necklace is made in the United States of America.


text:
Базилика Максенция—Константина () — самое большое здание, когда-либо построенное на Римском форуме и самая последняя постройка императорского Рима. Заложена в 308 году императором Максенцием, через два года после захвата власти в Риме. Закончена его преемником — Константином Великим после победы над Максенцием в битве у Мульвиева моста 28 октября 312 года. В VI в. здание стали называть «Римским храмом» (), а ещё позднее —  «Новой базиликой»  ().

Базилика воздвигнута на месте складов для специй () на северной стороне форума и располагалась параллельно «Священной дороге» (), главный (короткий) фасад был обращён на восток, к амфитеатру Флавиев (Колизею). Здание строили из кирпича с последующей облицовкой мрамором на бетонном фундаменте (100 Х 65 метров) .

Базилика состояла из трёх больших нефов (центральный: 80 метров в длину и 25 в ширину), протянувшихся с востока на запад и перекрытых цилиндрическими кессонированными сводами (верхняя точка сводов располагалась на высоте 29 метра). Их пересекали с севера на юг три трансепта (23 метра в длину и 17 в ширину каждый). Восемь мраморных колонн высотой 14,5 м поддерживали перекрытие, которые в пересечении нефов и трансептов образовывали крестовые своды. Это была совершенно новая композиция, не встречавшаяся ранее в римской архитектуре, как в отношении размеров, так и в отношении конструкции.

Центральный неф площадью 2000 кв.м в западной оконечности заканчивался большой апсидой, в которой находился «Колосс» огромная мраморная фигура сидящего на троне императора Константина. Её фрагменты ныне экспонируются во дворе Палаццо деи Консерватори (Palazzo dei Conservatori) Капитолийского музея.

По всей длине восточного фасада здания проходила аркада. Южный фасад, обращённый к Священной дороге, по распоряжению императора Константина также был оформлен двухэтажной аркадой, в центре которой (по оси срединного трансепта) был устроен главный вход, оформленный портиком на четырёх порфировых колоннах (тетрастиль). Напротив главного входа в базилику, с северной стороны построили ещё одну полукруглую апсиду (она сохранилась). В ней при Константине заседал трибунал (ранее заседания проводили в западной апсиде) . Внутри стены базилики были облицованы мраморными плитами, а пол выложен цветным мрамором. Стояли мраморные и вызолоченные бронзовые скульптуры и жертвенники. Считается, что в качестве образцов строителям были указаны колоссальные Термы Каракаллы и Диоклетиана. Оригинальной является внешнее «скатное» покрытие базилики. Если ранние римские здания имели простую двускатную кровлю, то Базилика Максенция-Константина имела многоскатное покрытие, облицованное сверкающими на солнце позолоченными бронзовыми пластинами (в VII в. сняты для кровли собора Св. Петра в Ватикане) .

В 1349 году свод центрального нефа обрушился в результате землетрясения. Остался лишь северный неф с тремя огромными арками. Единственная из восьми  колонн, которая пережила землетрясение, в 1614 году была перенесена по указанию Папы  Павла V на площадь Санта-Мария-Маджоре и установлена в её центре с бронзовой статуей Мадонны на вершине по проекту архитектора  Карло Мадерна. Колонну назвали «Колонной мира» (Colonna della Pace), поскольку базилику Максенция-Константина называли в то время «Храмом Мира» (Tempio della Pace).

Создавая в 1506—1514 годах новый проект собора Св. Петра в Ватикане, архитектор  римского классицизма Донато Браманте утверждал, что стремился водрузить купол Пантеона на своды «Храма Мира» (базилики Максенция-Константина). Проект ватиканского храма, разработанный Браманте, следовал центрическому типу крестово-купольных зданий (одна из основных идей эпохи итальянского Возрождения), но также опирался на достижения архитектуры императорского Рима.

См. также
 Колосс Константина
 Базилика Константина (Трир)

Примечания

Ссылки

Дохристианские храмы Рима
Римский форум
Появились в IV веке

redpajama_set_name: RedPajamaWikipedia


In order to do the job properly, you need the right tools. At Ashby Precision Engineering we have invested heavily in acquiring the latest technology, including around 14 different CNC machines. With almost 40 years' experience we are skilled in a broad range of materials, enabling us to take on any request.
To see our full list of precision engineering equipment, click here [PDF].
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Get in touch with us today to receive your quote! With years of experience in the industry, we are able to work expertly with a broad range of materials and handle any request you may have.