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Given a complete lattice $(A,\preceq)$

$ \ \bullet \text{Prove: if}\ S,T \subset A, \text{then} \ \forall x \in S.\ \exists y\in T.x\preceq y \implies \bigvee S \preceq \bigvee T. $ (1)

$ \ \bullet \text{Given a monoton function}$ $f: \mathbb{N} \times \mathbb{N}\rightarrow A \ \text{such that} \ \forall \ i,i',j,j'\in \mathbb{N} $ $$i \le i' \wedge j \le j' \implies f(i,j)\preceq f(i',j'). $$ Prove: $$ \bigvee_{i,j\in \mathbb{N}\times \mathbb{N}} f(i,j)= \bigvee_{k\in \mathbb{N}} f(k,k)$$

  1. The first point seems pretty obvious and follows (almost) from the definition of a complete lattice, we know that for any subsets $S, T \subset A$ there exists a supremum and infimum and form (1) it follows that $ \vee S \preceq \vee T.$ Note: (its just an idea I'm not sure if its correct)
  2. I'm not quite sure how to do the second part, I know that $f$ is monotone $\implies f$ is injective if I can prove $|A|=|\mathbb{N}| \implies f $ is surjective, the proof follows. But I don't know if $f$ is surjective so how can I be sure there isn't a bigger$ \ \vee f(k,k) $
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  • $\begingroup$ 1 does need a small argument, give it! $\endgroup$ – Henno Brandsma Jun 9 '19 at 21:49
  • $\begingroup$ constant maps are also monotone, so don't jump to conclusions... $\endgroup$ – Henno Brandsma Jun 9 '19 at 21:50
  • $\begingroup$ You are right I didn't think of constant functions in this context, and your proof is quite sleek, thanks helped a lot $\endgroup$ – Nejc.Z Jun 10 '19 at 5:38
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Suppose the condition $\forall x \in S: \exists y \in T: x \preceq y$ holds for $S$ and $T$. The completeness of the lattice ensures that $\bigvee S$ and $\bigvee T$ both exist in $A$.

If $x\in S$ then we have $y \in T$ with $x \preceq y$. Also $y \preceq \bigvee T$ as the latter is an upperbound for $T$. So for all $x \in S$ we know (by transitivity) that $x \preceq \bigvee T$, so $\bigvee T$ is an upperbound for $S$ and $\bigvee S$ is the smallest one of those, so by definition $\bigvee S \preceq \bigvee T$.

The second is a consequence: First, if $(i,j) \in \Bbb N \times \Bbb N$ then let $k=\max(i,j)$ and $i \le k \land j \le k$ is obvious. So $f$ being monotone tells us that $f(i,j) \preceq f(k,k)$.

This means that $S=\{f(i,j): (i,j) \in \Bbb N \times \Bbb N\}$ and $T=\{f(k,k): k \in \Bbb N\}$ satisfy the condition of 1. So $\bigvee S \preceq \bigvee T$ but as $T \subseteq S$ trivially, we also have $\bigvee T \preceq \bigvee S$ and hence equality (antisymmetry of $\preceq$), which is what was asked to prove.

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