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

I have proved that $\sup$≦$\inf$, but dont know how to show that $\inf$≦$\sup$..

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The proof depends on your definition of exponentiation. – Yuval Filmus Jul 6 '12 at 7:01
I haven't defined exponentiation with real index. I have defined exponentiation with a positive base and rational index but this equality seems true and able to be proved with just this info – Katlus Jul 6 '12 at 7:10
up vote 1 down vote accepted

We know that $f(t) = b^t$ is continuous and monotonic. Take two sequences in $\mathbb{Q}$ that converge to $x$, say $(t_n)$ and $(s_n)$. Due to continuity, $\lim b^{t_n} = \lim b^{s_n} = b^x$. Then you only need to show that these limits are really $\inf$ and $\sup$. Use monotony for that.

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Choose once and for all $T$ in $\mathbb Q$ with $T\geqslant x$. For every positive integer $n$, choose $s_n\leqslant x\leqslant t_n$ such that $s_n$ and $t_n$ are in $\mathbb Q$ and $t_n-s_n\leqslant1/n$.

Then, $b\gt1$ hence $b^{t_n}\leqslant b^{s_n}(b^{1/n}-1)+b^{s_n}\leqslant b^T(b^{1/n}-1)+b^{s_n}$. When $n\to\infty$, $b^{1/n}\to1$ hence, for every positive $\varepsilon$ there exists $n$ such that $b^{1/n}\leqslant1+\varepsilon/b^T$. Thus, $b^{t_n}\leqslant \varepsilon+b^{s_n}$.

This shows that $\inf\{b^t\mid t\geqslant x,t\in\mathbb Q\}\leqslant\varepsilon+\sup\{b^s\mid s\leqslant x,t\in\mathbb Q\}$. This holds for every positive $\varepsilon$ hence you are done.

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