= Solution
This is the <pointwise order> on a <function space>. Each of the three <partial order> axioms follows from the corresponding axiom in $B$.
For <reflexivity>, $f(x)\preceq_B f(x)$ for every $x$, so $f\preceq_{\mathcal F}f$. For <antisymmetry>, if $f\preceq_{\mathcal F}g$ and $g\preceq_{\mathcal F}f$, then $f(x)=g(x)$ for every $x\in A$; equality of <functions> therefore gives $f=g$. For <transitivity>, if $f\preceq_{\mathcal F}g\preceq_{\mathcal F}h$, then $f(x)\preceq_B g(x)\preceq_B h(x)$ for every $x$, so $f\preceq_{\mathcal F}h$.
Thus \b[$(\mathcal F,\preceq_{\mathcal F})$ is a <partially ordered set>]. The argument also covers an empty domain $A$, since then there is exactly one <function> $A\to B$.
Back to article page
