Proving supremum for non-empty, bounded subsets of Q iff supremum in R is rational Let E be a nonempty bounded subset of ℚ. Prove that E has a supremum in ℚ if and only if its supremum in ℝ is rational and that in this case, the two are equal.
This seems intuitive enough, and I know that not all nonempty, bounded subsets of R have a supremum in Q, but how do I prove this?
I can proceed by proving that if E's sup in R is irrational, E does not have sup in Q - but this I have only though one example, not in a general case. How do I say this generally?
I also have to prove the reverse, which is that if sup in R is rational, then E has supremum in Q. Again, I get this intuitively, but how do I prove it?
Help appreciated!
 A: HINTS:


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*Suppose that $\sup_{\Bbb R}E=\alpha$ is irrational, but that $E$ has a (necessarily rational and therefore necessarily different) supremum $s$ in $\Bbb Q$. Either $s<\alpha$ or $s>\alpha$, and you can get a contradiction either way.

*Suppose now that $\sup_{\Bbb R}E=\alpha\in\Bbb Q$. Clearly $e\le\alpha$ for each $e\in E$, so $\alpha$ is an upper bound for $E$ in $\Bbb Q$. Show that if $q\in\Bbb Q$, and $q<\alpha$, then $q$ is not an upper bound for $E$: there is some $e\in E$ such that $e>q$. This will show that $\alpha$ is the least upper bound of $E$ in $\Bbb Q$, i.e., that $\alpha=\sup_{\Bbb Q}E$.
A: I find it easier to prove directly.
Suppose $E$ has a supremum in $\mathbb{Q}$. Let $\alpha = \sup_\mathbb{Q} E$. Then $\alpha \in \mathbb{Q}$, $\alpha$ is an upper bound for $E$, and if $\beta \in \mathbb{Q}$ is an upper bound for $E$, then $\alpha \le \beta$.
Now suppose $\gamma \in \mathbb{R}$ is an upper bound for $E$, and let $q_n \in \mathbb{Q}$ be such that $q_n \downarrow \gamma$. Since $\gamma$ is an upper bound, and $q_n \ge \gamma$, $q_n$ is an upper bound for $E$, and hence $q_n \ge \alpha$. Hence we have $\gamma \ge \alpha$. Hence $\alpha$ is the supremum of $E$ in $\mathbb{R}$.
Now suppose $E$ has a supremum in $\mathbb{R}$, and the supremum is rational. That is, $\alpha = \sup_\mathbb{R} E \in \mathbb{Q}$. Then $\alpha$ is an upper bound for $E$. Suppose $\beta \in \mathbb{Q}$ is an upper bound for $E$, then since $\beta \in \mathbb{R}$, we must have $\alpha \le \beta$. Since $\alpha \in \mathbb{Q}$, it follows that $\alpha = \sup_\mathbb{Q} E$.
A: Suppose $b$ is the smallest rational upper bound of $E\subseteq\mathbb Q$.  Then for every $c\in\mathbb Q$, if $c<b$ then some member of $E$ is $>c$ (otherwise $c$ would be a rational upper bound of $E$ despite $c$ being $<b$).  Therefore the smallest upper bound of $E$ in $\mathbb R$ would have to be bigger than every rational number $<b$, but could not be bigger than $b$ since $b$ is an upper bound.  Thus the smallest real upper bound of $E$ must be either $b$ or a real number less than $b$ but bigger than all rational numbers less than $b$.  It therefore suffices to show there can be no real number $x$. less than $b$ but bigger than all rationals less than $b$.  This can easily be shown to be equivalent to saying there can be no real $\varepsilon\ (=b-x)$ that is positive but smaller than all positive rationals.  If there were, then it would be smaller than $1/n$ for all $n\in\mathbb N$, so its reciprocal would be an upper bound of $\mathbb N$.  Then $\mathbb N$ would have a least upper bound $d$.  But then some integer would be in the interval $(d-1,d\,]$ and that integer plus $1$ would be in $\mathbb N$ but bigger than $d$, which cannot happen.
That shows that if $E$ has a sup in $\mathbb Q$, then that is equal to its sup in $\mathbb R$.
The converse is that if the sup of $E$ in $\mathbb R$ is rational, than that's also the sup of $E$ in $\mathbb Q$.  Let $b=\sup E$ (meaning the sup in $\mathbb R$) and suppose $b$ is rational.  Since every member of $E$ is $\le b$ and $b\in\mathbb Q$, the number $b$ is certainly an upper bound of $E$ in $\mathbb Q$.  The problem then is showing it's the smallest one.  Suppose there were a smaller one: a rational upper bound of $E$ that is $< b$.  Then that would also be a real upper bound of $E$ that is $<b$ ---- hence a contradiction.
The nonexistence of a real that is bigger than all rationals less than $b$ but smaller than $b$ would be a triviality if one were actually to define reals as Dedekind cuts.
