# 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?

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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.

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Thanks for this response! One question though, you mention you know that there exists $s'\in S'$ such that $t\lt s'\leq u$. Is this acceptable in the case where $S'$ is empty? And furthermore, how do you know that $t$ is strictly smaller than $s'$? Would you not only know that $t\lt u$ and $s'\leq u$? –  yunone Sep 11 '10 at 21:37
@yunone: No, it's not; you are correct that it should be treated separately. If $S'$ is empty, then $\sup_U(S')=u = \sup_U(\emptyset)=\inf(U)$. But If $U$ is the infimum of $U$, then certainly we have $u= \inf_U(S)\leq\inf_T(S)$ and you are done. –  Arturo Magidin Sep 11 '10 at 21:59
@yunone: So, it wasn't just one question, then... There is small argument I skipped for the second part. I'll edit the reply. Sorry. –  Arturo Magidin Sep 11 '10 at 22:05
I hope I am not asking too many small questions, I just want to see all the detail. Thank you so far for your helpful explanations. Sorry to beat a dead horse, but we are guaranteed the existence of $\inf(U)$ since it is given that there exists $u=\sup_U(S')$, and that is why we may assume that $\sup_U(\emptyset)=\inf(U)$ truly does hold, correct? –  yunone Sep 11 '10 at 22:09
Yes: We know that if $\sup_U(\emptyset)$ exists, it must be the infimum of $U$ (it is the least upper bound of $\emptyset$ by definition, which implies comparability, and everything in $U$ is an upper bound for $\emptyset$). Since here we are told it exists, we know it must the infimum of $U$. –  Arturo Magidin Sep 11 '10 at 22:18