I am currently trying to understand R. C. Flagg's "Quantales and continuity spaces".
However I am struggling a bit with his definitions and would like to have a good simple (but not too simple) example of a value quantale. Hence my question:
Let $$P=\{f:[0,1]\rightarrow [0,\infty]\}$$ be equipped with a usual pointwise order: $$f\leq_P g\quad\text{ iff }\quad \forall x \ \ \ f(x)\leq g(x),$$ and pointwise addition: $$(f+_Pg)(x):=f(x)+g(x).$$ ($[0,1]$ and $[0,\infty]$ are considered with their usual orders and addition operations.)
Is $(P,\leq_P,+_P)$ a value quantale?
$\bf{Here\ is\ a\ more\ general\ question:}$ suppose that we have a family $\{P_j\}_{j\in J}$ of value quantales. Is their Cartesian product $\prod_{j\in J}P_j$ equipped with usual coordinate-wise order and coordinate-wise operation a value quantale?
It is not, though it meets most of the requirements. Indeed, most of the requirements are clear; only those involving the well above relation relation are not. It is true that $\langle P,\le_P\rangle$ is completely distributive.
Let $f,g\in P$ with $f<_Pg$. If there is an $x\in[0,1]$ such that $f(x)=g(x)$, for each $n\in\Bbb N$ define
$$h_n:[0,1]\to[0,\infty]:t\mapsto\begin{cases} f(t)+2^{-n},&\text{if }t=x\\ f(t),&\text{otherwise}\;, \end{cases}$$
and let $S=\{h_n:n\in\Bbb N\}$; then $\bigwedge S=f$, but $g(x)=f(x)<f(x)+2^{-n}=h_n(x)$ for each $n\in\Bbb N$, so $h_n\not\le_P g$, and hence $f\not\prec g$.
If $g(x)>f(x)$ for all $x\in[0,1]$, let $A=\{x\in[0,1]:g(x)\ne\infty\}$. Assume for now that $|A|\ge 2$, and let $\varphi:[0,1]\times\Bbb N\to A$ be any surjection such that for each $\langle x,n\rangle\in[0,1]\times\Bbb N$, $\varphi(\langle x,n\rangle)\ne x$. For $\langle x,n\rangle\in[0,1]\times\Bbb N$ define
$$h_{\langle x,n\rangle}:[0,1]\to[0,\infty]:t\mapsto\begin{cases} f(x)+2^{-n},&\text{if }t=x\\ g(t)+1,&\text{if }t=\varphi(\langle x,n\rangle)\\ f(t),&\text{otherwise}\;, \end{cases}$$
and let $S=\{h_{\langle x,n\rangle}:\langle x,n\rangle\in[0,1]\times\Bbb N\}$. Then $\bigwedge S=f$, and $g\big(\varphi(\langle x,n\rangle)\big)<h_{\langle x,n\rangle}\big(\varphi(\langle x,n\rangle)\big)$ for each $\langle x,n\rangle\in[0,1]\times\Bbb N$, so again $f\not\prec g$.
Thus, if $f\prec g$, there can be at most one $x\in[0,1]$ such that $g(x)\ne\infty$, and if there is such a point $x$, then $g(x)>f(x)$.
Conversely, it’s not hard to check that
$f\prec g$ if there is an $x\in[0,1]$ such that $g(t)=\infty$ for $t\in[0,1]\setminus\{x\}$, and $f(x)<g(x)$; and
if $\mathbf{1}_P\in P$ is defined by $\mathbf{1}_P(t)=\infty$ for all $t\in[0,1]$, then $\mathbf{1}_P\prec\mathbf{1}_P$.
From this it follows easily that $f=\bigwedge\{g\in P:f\prec g\}$ for all $f\in P$ and hence that $\langle P,\le_P\rangle$ is completely distributive.
Now let $\mathbf{0}_P$ be defined by $\mathbf{0}_P(t)=0$ for all $t\in[0,1]$; to show that $\langle P,\le_P,+_P\rangle$ is a value quantale we would have to show that if $\mathbf{0}_P\prec f,g\in P$, then $\mathbf{0}_P\prec f\land g$. Unfortunately, this need not be true. Let
$$f:[0,1]\to[0,\infty]:t\mapsto\begin{cases} 1,&\text{if }t=0\\ \infty,&\text{otherwise} \end{cases}$$
and
$$g:[0,1]\to[0,\infty]:t\mapsto\begin{cases} 1,&\text{if }t=1\\ \infty,&\text{otherwise}\;. \end{cases}$$
We’ve just seen that $\mathbf{0}_P\prec f,g$, but $(f\land g)(t)<\infty$ for both $t=0$ and $t=1$, so $\mathbf{0}_P\not\prec f\land g$.