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A poset $P$ is called Bourbakian if every order-preserving map $P\rightarrow P$ has a LEAST fixed point. Let $P, Q$ be Bourbakian. Let $P\times Q$ be ordered pointwise, that is $(a,b)\le (c,d)$ if and only if $a\le c$ and $b\le d $. Let $h:P\times Q\rightarrow P\times Q$ be order-preserving. Let $h(x,y)=(h_1(x,y), h_2(x,y))$.

For each $x\in P$ we can define $g_x:Q\rightarrow Q$ by $g_x(y)=h_2(x,y)$. If $y\le y'$ then we see $h(x,y)\le h(x,y')$ so $g_x(y)\le g_x(y')$ so $g$ is order preserving, hence $g_x$ has a least fixed point in $Q$, which we shall call $m(x)$.

The first question is: if $x_1\le x_2\in P$, by considering the effect of $g_{x_1}$ on $\{y\in Q:y\le m(x_2)\}$, show that $m:P\rightarrow Q$ is order-preserving.

Added: Next, define $f:P\rightarrow P$ by $f(x)=h_1(x, m(x))$. This is order preserving, since $id$ and $m$ are both order preserving. So let $x_0$ be the least fixed point of $f$ on $P$. Now the second question is to show that $(x_0, m(x_0))$ is the least fixed point of $h$.

There is no difficulty in showing $h(x_0, m(x_0))=(x_0, m(x_0))$. I need to show if $h(a, b)=(a,b)$, then $a\ge x_0$. If I can show that, then because $b\ge m(a)$, $m(a)\ge m(x_0)$, $(a,b)\ge (x_0, m(x_0))$. Any ideas?

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HINT: For any $y\in Q$ we have $h(x_1,y)\le h(x_2,y)$, so $h_2(x_1,y)\le h_2(x_2,y)$, i.e., $g_{x_1}(y)\le g_{x_2}(y)$. If $y\le m(x_2)$, then $g_{x_1}(y)\le g_{x_2}(y)\le g_{x_2}(m(x_2))=m(x_2)$. If $Q_0=\{y\in Q:y\le m(x_2)\}$, we’ve just shown that $g_{x_1}$ maps $Q_0$ into itself.

Define a map

$\varphi:Q\to Q:y\mapsto\begin{cases} g_{x_1}(y),&\text{if }y\in Q_0\\\\ m(x_2),&\text{otherwise}\;. \end{cases}$

Show that $\varphi$ is order-preserving and therefore has a fixed point $y_0$. Clearly $y_0\le m(x_2)$; finish by showing that $y_0$ is a fixed point of $g_{x_1}$.