This is another clarifying question; alas, I find myself confused once again by a seemingly innocuous statement in my lecture notes.
Let $S$ be a subset of a partially ordered set $(T, \preceq)$, and let $E$ be a subset of $S$, and let $\sup_{T}(E)$ denote the supremum of $E$ when $E$ is viewed as a subset of $T$, $\sup_{S}(E)$ when $E$ is viewed as a subset of $S$. Now, I have written down here that "For a subset $E \subseteq S$, the sets of upper and lower bounds will generally depend on whether one considers $E$ as a subset of $S$ or $T$". This all makes much sense to me. However, I then have an exercise asking me to "Show that, for $E \subseteq S$, one has $\sup_{T}(E) \preceq \sup_{S}(E)$ whenever both suprema exist". See, intuitively I would think that it has to be the other way around (that is, $\sup_{S}(E) \preceq \sup_{T}(E)$ whenever both suprema exist) simply because if $T \supseteq S$ then $\sup_{T}(E)$ has to "proceed" or "be at" $\sup_{S}(E)$ since either $\sup_{S}(E) \in T$ or $\sup_{S}(E)$ is somewhere "higher up" in the set of all upper bounds of $T$. I mean, I can definitely see how $\sup_{T}(E)$ = $\sup_{S}(E)$, but how could we ever have $\sup_{T}(E) \prec \sup_{S}(E)$?
Denote by $\def\ub{\mathop{\rm ub}\nolimits}\ub_S(E)$ the set of upper bounds of $E$ in $S$ (analogusly write $\ub_T(E)$). As obviously for $s\in \ub_S(E)$ we have $E \le s$ and $s \in T$, it follows that $\ub_S(E) \subseteq \ub_T(E)$. Now the suprema are the minima of the upper bounds, so $$ \sup_T(E) = \min \ub_T(E) \le \min \ub_S(E) = \sup_S(E) $$ Strict inequality can occur, if the supremum $\sup_T(E)$ is missing in $S$. For example, let $T = \omega + 2$, $S = T \setminus \{\omega\}$ and $E = \omega$. Then $\sup_T E = \omega$, but $\sup_S E = \omega + 1$.