If $x \in \mathbb{R}$, prove that $x = \sup \{r \in \mathbb{Q}: r < x \} = \inf \{s \in \mathbb{Q}: x
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For convenience, let $A = \{r \in \mathbb{Q}: r < x \}$ and $B = \{s \in \mathbb{Q}: x .
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)?