For $S\subseteq T\subseteq\mathbb{Q}$, it is possible for $\sup_T S$ to exist, but $\sup_\mathbb{Q} S$ does not? I've been reading a bit about how the set of bounds changes for a set depending on what superset one works with. I considered the sets $S\subseteq T\subseteq\mathbb{Q}$ and worked out a few contrived examples:
If $S=T=$ {$x\in\mathbb{Q}\ | \ x^2\lt 2$}, so here $S$ is not bounded above in $T$, but it is bounded about in $\mathbb{Q}$, with $2$ being a possibility.
Also, if $S=$ {$x\in\mathbb{Q}\ | \ x^2\lt 1$} and $T=$ {$x\in\mathbb{Q}\ | \ x\lt 2 \ \text{and}\ x\neq 1$}, then $S$ is bounded in $T$ and $\sup_\mathbb{Q} S=1$ exists, but $\sup_T S$ does not exist.
My question is, is it possible for $S$ to be bounded in $T$ where $\sup_T S$ exists, but $\sup_\mathbb{Q} S$ does not? And moreover, can both $\sup_T S$ and $\sup_\mathbb{Q} S$ exist, but not be equal? Any example of this would be much appreciated.
 A: Yes, it is possible for $\sup_T(S)$ to exist, but $\sup_\mathbb{Q}(S)$ not to exist. Yes, it's possible for both to exist and be distinct.
Here is an example of the latter. Take $S=\{x\in\mathbb{Q}|0\leq x\lt 1\}$. Let $T=S\cup{{2}}$. Then $S$ is bounded above in $T$, and has a supremum, namely $2$. Indeed, for every $t\in T$, if $t\lt 2$, then there exists $s\in S$ such that $t\lt s\leq 2$, so $2$ is the supremum of $S$. And of course, the supremum of $S$ in $\mathbb{Q}$ is $1$.
For the former, take $S=\{x\in\mathbb{Q}|0\leq x \lt \sqrt{2}\}$ (or any irrational number you please), and take $T=S\cup\{2\}$ (or any number greater than $\sqrt{2}$). 
In general, if you have partially ordered sets $P\subseteq Q$, and a subset $A$ of $P$, you can have that $A$ has a supremum in $P$ but not in $Q$; has a supremum in $Q$ but not in $P$; has suprema in neither; has the same supremum in both; or has suprema in both but they are distinct. The one thing you can say is that if both suprema exist, then $\sup_Q(A)\leq\sup_P(A)$. So in your situation, you do know that if they both exist you will have $\sup_{\mathbb{Q}}(S)\leq\sup_{T}(S)$, but you can have strict inequality.
A: For the first question, let $S$ be the set of all rationals less than $\pi$.  Then $S$ has no least upper bound in $\mathbb{Q}$.  On the other hand, if $T = S\cup {4}$, then 4 is the least upper bound of $S$ inside $T$.
For the second question, let $S$ be the set of all rationals strictly less than 1, and let $T = \mathbb{Q}\cap \left([-\infty, 1)\cup [2, \infty)\right)$.  Then the least upper bound of $S$ inside $T$ is 2.  But of course the least upper bound of $S$ in $\mathbb{Q}$ is 1.  
This shows the answer to your second question is "Yes."
