Limit-preserving point-wise maximum

Question

Let $F$ denote the set of monotonic functions in $N^{\infty} \rightarrow N^{\infty}$ (where $N^{\infty}$ includes 0 and $\infty$) ordered by the point-wise application of $\le$ i.e., for $f,g \in F$ $$ f \le g \iff \forall n \in N^{\infty} :~ f(n) \le g(n) $$

then I'd like to show that the point-wise maximum ($\vee$), defined such that $(f \vee g)(n) = \max(f(n), g(n))$ is limit-preserving i.e. $$ \inf_{(f,g) \in C} (f \vee g) = (\inf_{(f,g) \in C} f) \vee (\inf_{(f,g) \in C} g) $$ for a chain (totally ordered subset) $C \subseteq F \times F$ ordered by the element-wise comparison.

Answer 1

In short: since these functions take values in $\mathbb{N}^\infty$, pointwise, these inf are attained min by the well-ordering principle. The point is that over a chain $C$, the inf on the right are attained jointly for the same element $(f^*,g^*)\in C$.

I assume your definition of monotonic is $n\leq m\Rightarrow f(n)\leq f(m)$. But it could be strictly monotonic with the same conclusions.

First, it is clear that $f\vee g\in F$ if $(f,g)\in F\times F$. Now for any non-empty set $S\subseteq F$, and for every $n$, $\{f(n)\,;\,f\in S\}$ is nonempty in $\mathbb{N}^\infty$ whence $\inf_S f(n)$ exists and is a least element $\min_S f(n)\in \mathbb{N}^\infty$. So the function $\inf_S f=\min_S f$ is well-defined. Then for $n\leq m$, and for every fixed $g\in S$ we have $$ \min_S f(n)\leq g(n)\leq g(m). $$ Taking the inf over $g$ shows that $\min_S f$ is motonic and belongs to $F$.

So what you want to prove is a min-max theorem: for every $n\in\mathbb{N}^\infty$ $$ \min_{(f,g)\in C} \max (f(n),g(n))=\max \left(\min_{(f,g)\in C} f(n), \min_{(f,g)\in C} g(n) \right). $$

Proof: note that $n$ is fixed. For every $(f_0,g_0)\in C$ fixed, it is clear that $$ \max (f_0(n),g_0(n))\geq \max \left(\min_{(f,g)\in C} f(n), \min_{(f,g)\in C} g(n) \right). $$ Hence the inequality $\geq $ holds by taking the inf over all $(f_0,g_0)\in C$.

The other direction is where the fact that $C$ is a chain is useful. Take $(f_0,g_0)\in C$ such that $\min_{(f,g)\in C} f(n)=f_0(n)$ and $(f_1,g_1)\in C$ such that $\min_{(f,g)\in C} g(n)=g_1(n)$. Then either $(f_0,g_0)\leq (f_1,g_1)$, or $(f_0,g_0)\geq (f_1,g_1)$ by chain condition. In the former case, we have $$ g_0(n)\leq g_1(n)=\min_{(f,g)\in C} g(n)\leq g_0(n)\quad\Rightarrow \quad g_0(n)=\min_{(f,g)\in C}g(n). $$ In the latter case, we would conclude that $f_1(n)=\min_{(f,g)\in C} f(n)$. So in any case, there exists $(f^*,g^*)\in C$ such that $$ f^*(n)=\min_{(f,g)\in C} f(n)\quad\mbox{ and}\quad g^*(n)=\min_{(f,g)\in C} g(n). $$ Finally $$ \min_{(f,g)\in C}\max(f(n),g(n))\leq \max(f^*(n),g^*(n)) $$ which proves the desired reverse inequality. $\Box$