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Let $1<b\in \mathbb{R}$ and $x\in \mathbb{R}$. I want to prove that $$\sup\{b^t\in \mathbb{R}\mid x\geq t\in \mathbb{Q}\} = \inf\{b^t\in \mathbb{R}\mid x\leq t\in \mathbb{Q}\}.$$

I have proved that $\sup\leq\inf$, but don't know how to show that $\inf\leq\sup$.

I posted this question a week ago, but I think this could be proved in a more direct way...

  1. I have only defined exponentiation such that base is real and index is rational.
  2. I'm trying to solve this by assuming $\sup<\inf$ and deriving a contradiction such that "there exists $q\in \mathbb{Q}$ such that $\sup < b^q < \inf$".
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The easiest way to me seems to be proving that for $t-t'$ small enough $b^t-b^{t'}$ is very small. –  tomasz Jul 8 '12 at 20:38
@Zev why is this question not a set theory problem? I prefer set-theoretic proof to analysis one.. –  Katlus Jul 8 '12 at 20:38
@Katlus: I'm not sure why it would be a set theory problem. Consider the description of the elementary-set-theory tag (and also the set-theory tag); I don't see how this is related to set theory. Constructing the exponential function seems to me to be something squarely in the realm of analysis. –  Zev Chonoles Jul 8 '12 at 20:44
@tomasz i'm trying to define exponentiation of reals first, then limit, then continuity and etc. –  Katlus Jul 8 '12 at 20:44
@Katlus: So? I meant that for rational t's. The fact I mentioned along with monotonicity of exponentiation should be enough to show the result. –  tomasz Jul 8 '12 at 20:45

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