Why does sup-dense imply inf-continuous? I am confused. The question is as follows:
Let $(S, \preceq$), be a sup-dense subset of $(T, \preceq)$, both partially ordered sets. In other words, for every $t \in T$, there exists a subset $E \subseteq S$ such that $t = \sup_T E$. Show that the canonical "inclusion" (an order embedding such that $\iota(s) = s \forall s \in S$) is inf-continuous, or in other words, that for every subset $E \subseteq S$ such that $\inf_S E$ exists, $\inf_S E = \inf_T E$.
I can't figure out how to prove that $\inf_T E$ exists in the first place. How might I go about doing that?
 A: You prove that $\inf_TE$ exists by showing that $\inf_SE$ is still the infimum of $E$ in the full partial order $T$.
Let $e_0=\inf_SE$; you’re assuming that $e_0$ exists, and you know that you want to be able to conclude that $\inf_TE=e_0$, so the most natural thing to try is to show directly that $e_0$ is the infimum of $E$ in $T$. It’s clear that $e_0$ is a lower bound for $E$ in $T$ as well as in $S$, since $e_0\preceq e$ for each $e\in E$. Thus, you need only show that if $t\in T$, and $t\preceq e$ for each $e\in E$, then $t\preceq e_0$.
Suppose that $t\in T$, and $t\preceq e$ for each $e\in E$. There’s only one reasonable way to begin: your hypothesis that $S$ is sup-dense in $T$ gives you a set $A_t\subseteq S$ such that $t=\sup_TA_t$. Clearly $s\preceq e$ for every $s\in A_t$ and $e\in E$, so every element of $A_t$ is a lower bound for $E$ in $S$. Recall that $\inf_SE=e_0$; can you finish it from there?
A: Let $t \in T$ be an lower bound for $E$. By assumption, there is an $E_t \subseteq S$, such that $\sup_T E_t = t$. Now $E_t \le t \le E$, so every $f\in E_t$ is a lower bound for $E$, hence $f \le \inf_S E$ for every $f \in E_t$. So $\inf_S E$ is an upper bound for $E_t$, hence $t = \sup_T E_t \le \inf_S E$. Does this help?
