# Show that a bounded set of rational numbers has a supremum.

For a problem set, I have to show that the set $\mathbb{Q}_{x} :=$ set of all rational numbers for which $q \leq x$ has a supremum in $x$.

My attempt is to suppose that there is a $y < x$ that is an upper bound of $\mathbb{Q}_{x}$ and then find a $q \in \mathbb{Q}_{x} > y$. Unfortunately, I am not allowed to use the fact in between any arbitrary real numbers lies a rational one (as this is part of a proof to show that the rational numbers are dense in the real ones...)

I would appreciate any hints greatly, as I find myself banging my head against a wall with this simple problem for some time now...

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This question can't be properly answered if you don't provide the definition of real numbers we're working with. With some definitions, what you are trying to prove is actually a part of the definition. – tomasz Nov 1 '12 at 16:12
@tomasz Thanks for your reply. We defined $\mathbb{R}$ as an ordered field, with R being closed under multiplication and addition, and both operation satisfy the associative and commutative property. Also $\alpha, \beta, \gamma \in \mathbb{R}, \gamma = \alpha*\gamma + \beta*\gamma$ is said to hold. – padrino Nov 1 '12 at 16:22
$\bf Q$ itself satisfies your definition, so you might want to say that it is a complete ordered field (but then, complete in what sense?). But then: I assume that $\bf Q$ is defined as the smallest subfield of $\bf R$? – tomasz Nov 1 '12 at 16:27

## 1 Answer

Consider $x - y$ and use the Archimedean property. Of course, I'm assuming that you've already built up $\mathbb{R}$ axiomatically, which might not be the case. If you haven't already developed the real numbers in this way, you should probably include more information about what has been covered and what tools are at your disposal.

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