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If $x \in \mathbb{R}$, prove that $x = \sup \{r \in \mathbb{Q}: r < x \} = \inf \{s \in \mathbb{Q}: x <s \}$.

For convenience, let $A = \{r \in \mathbb{Q}: r < x \}$ and $B = \{s \in \mathbb{Q}: x <s \}$.

I think both of these sets are non-empty by the denseness of the rationals. The elements of $B$ are upper bounds of $A$, and the elements of $A$ are lower bounds of $B$. Hence $\sup A$ and $\inf B$ exist. To actually show that these are equal to $x$, would you suppose $x_1$ was an arbitrary upper bound and show that $x_1 > x$ (similar thing with lower bounds)?

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up vote 1 down vote accepted

The fact that they are not empty is a consequence of the Archimedean property, rather than of the denseness of the rationals (it would be hard to use the denseness of the rationals to establish that both $A$ and $B$ are nonempty for $x\in\mathbb{Q}$, for example...)

Note that since every element of $A$ is a lower bound for $B$, then $a\leq\mathrm{inf}(B)$ for all $a\in A$; therefore, $\mathrm{inf}(B)$ is an upper bound for $A$, so $\mathrm{sup}(A)\leq\mathrm{inf}(B)$. So you only need to show that they are equal.

Moreover, it is easy to check that $\mathrm{sup}(A)\leq x\leq \mathrm{inf}(B)$.

If you want to proceed by contradiction, one possible way would be to assume that $\mathrm{sup}(A)\lt\mathrm{inf}(B)$. In that case, at least one of $\mathrm{sup}(A)$ and $\mathrm{inf}(B)$ is not equal to $x$. Say $\mathrm{sup}(A)\neq x$; then $\mathrm{sup}(A)\lt x$. Can you come up with some $q\in\mathbb{Q}$ that would satisfy $\mathrm{sup}(A)\lt q\lt x$ in that situation? Then you can try something similar under the assumption that $x\lt \mathrm{inf}(B)$.

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Put $q := (\sup A+x)/2$? So in this case we use the denseness of the rationals? – Damien Jun 21 '11 at 5:48
@Damien: That $q$ doesn't necessarily work, because you don't whether it is a rational. However, it is indeed a good time to use denseness of the rationals (or the archimedean property again, but denseness will do here). – Arturo Magidin Jun 21 '11 at 6:00
@Didier: Thanks! – Arturo Magidin Jun 21 '11 at 6:02
For example, $\frac{1+\pi}{2}$ is not rational. Proof: in fact, if $x$ is irrational and $y$ is rational, then $\frac{x+y}{2}$ is irrational. If it were rational, then we could substract the rational number $\frac{y}{2}$ and obtain another rational number; in particular, $\frac{x}{2}$ would be rational. However, this implies that $x$ is rational which is a contradiction. (Hence the average value of a rational number and an irrational number is always irrational.) – Amitesh Datta Jun 21 '11 at 8:05
@Arturo: Can I just say that there exists a $q \in \mathbb{Q}$ such that $\sup(A) < q < x$ by the denseness of the rationals? Or would I have to explicitly state what $q$ is? So $q$ is not an upper bound but we know that $\sup(A)$ is. Contradiction? – Damien Jun 21 '11 at 16:19

Maybe a bit different solution:
Wlog we can assume $x>0$ and we only need to show it for the supremum, as $\sup (M)=-\inf(-M)$. As $x\in \mathbb{R}$ we can express it as a sequence of it's digits.
So $$x=\sum_{k=-m}^\infty a_{k} \cdot 10^{-k},$$ with $a_k\in\{0,1,\dots,9\}$.
Now we define $x_n$ as $$x_n=\sum_{k=-m}^n a_{k} \cdot 10^{-k}$$ Obviously $x_n\to x$ so the upper bound is the lowest upper bound and hence the $\sup$

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