Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

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 $\inf$≦$\sup$..

share|improve this question
1  
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

2 Answers 2

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.

share|improve this answer

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.

share|improve this answer

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.