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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.2012-11-01
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    @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.2012-11-01
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    $\bf Q$ 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 $\bf R$?2012-11-01

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