For some sets $S\subseteq T\subseteq U$, when is $\inf_T S=\inf_U S$? I've been struggling with the following problem I found for a while now:
Suppose $(T,\preceq)$ is a partially ordered subset of $(U,\preceq)$ and $S\subseteq T$. If $\inf_T S$ and $u=\inf_U S$ both exist, and there exists a subset $S'\subseteq T$ such that $u=\sup_U S'$, then $\inf_U S=\inf_T S$. 
First I noted that $\inf_T S\preceq\inf_U S$, since $\inf_T S\in U$. 
I also saw that $u=\inf_U([u)\cap T)$ by the following: For $x\in S$, $u\preceq x$ and $x\in T$, so $x\in [u)\cap T$, and so $S\subseteq [u)\cap T$. So $u$ is a lower bound of $[u)\cap T$. Now suppose there exists some $y\in U$ such that $u\prec y$, but $y\preceq x$ for all $x\in [u)\cap T$. Since $S\subseteq [u)\cap T$, $y$ would be a lower bound of $S$ larger than $u$, contradicting the fact that $u=\inf_U S$.
I was hoping to show that $\inf_U S\preceq\inf_T S$ in order to show equality, but I can't piece it together. Can someone explain how to show this?
 A: So, you have shown that $\inf_T(S)\leq \inf_U(S)=u$. You want to show that $u\leq\inf_T(S)$. Clearly, you need to use the existence of $S'$ somehow, because without the existence of $S'$ you can trivially show this is false. 
You know that $\inf_T(S)\leq\inf_U(S)$. You want to show that you do not have $\inf_T(S)\lt\inf_U(S)=u$.
Edit: (added case) If $S'=\emptyset$, then $u=\inf_U(\emptyset)=\sup(U)$, so in particular $\inf_T(S)\leq u$ and we are done. So we may assume that $S'\not= \emptyset$. 
Added later: Actually, this is not needed strictly speaking, but it may be a good point to mention explicitly; if $t\lt u$, then $t$ is not an upper bound for $S'$; the only way that can occur in the first place is if $S'$ is not empty, since everything is an upper bound for the empty set, so this case is covered by the assumption being made.
If $t\in T$ is such that $t\lt u$, then since $t\in U$ and $u=\sup_U(S')$, you know that there exists $s'\in S'$ such that $t\lt s'\leq u$.
Edit: (implicitly assumed certain things were comparable, which may not be the case in general; fixed below). 
Note that for every $s'\in S'$ and $s\in S$, you have $s'\leq u\leq s$ (since $u=\inf_U(S)$), and so in particular $s'\leq \inf_T(S)$, since $S'\subseteq T$. So $\inf_T(S)$ is comparable to every element of $S'$. 
Suppose that $t\in T$ is such that $t\lt u$. Since $u=\sup_U(S')$, it follows that $t$ cannot be an upper bound for $S'$ in $U$. In particular, there exists $s'\in S'$ such that $s'\not\leq t$; so either $t\lt s'$, or else $t$ and $s'$ are incomparable.
In summary, for every $t\in T$, if $t\lt u$ and $t$ is comparable to every element of $S'$, then $t\lt\inf_T(S)$. 
Since $\inf_T(S)\in T$, $\inf_T(S)$ is comparable to every element of $S'$ (as noted above), and $\inf_T(S)\not\lt\inf_T(S)$, you can can conclude by contrapositive that $\inf_T(S)\not\lt u$. Since you already know that $\inf_T(S)\leq u$, you are done.
