1. ## Question on boundedness

Let $\displaystyle a\in\mathbb{R}$ and consider the set $\displaystyle S_a=\{a^n:n\in\mathbb{N},n>0\}$. Prove carefully that a is not bounded above when a>1.

My first thoughts are to use a contradiction, by using the definition of supremum, but I can't seem to find one. Could someone help?

2. ## Re: Question on boundedness

Originally Posted by worc3247
Let $\displaystyle a\in\mathbb{R}$ and consider the set $\displaystyle S_a=\{a^n:n\in\mathbb{N},n>0\}$. Prove carefully that a is not bounded above when a>1.

My first thoughts are to use a contradiction, by using the definition of supremum, but I can't seem to find one. Could someone help?
Yeah, if you want to do it that way. Maybe say something like if it were bounded above it would have a supremum $\displaystyle \alpha$ but by definition of supremum and the fact that $\displaystyle a>1$ we have that there is some $\displaystyle m\in\mathbb{N}$ for which $\displaystyle \displaystyle \frac{\alpha}{a}<a^m$ and so $\displaystyle \alpha< a^{m+1}$ contradiction.

3. ## Re: Question on boundedness

just out of curiosity: how do we know there is such an m? because, to me, this seems like a shell game: we are implcitly assuming that we can make a^m bigger than any real number (in particular, α/a), which is to say, {a^n} is unbounded (circular reasoning).

i mean, it's clear {a^n} is increasing, since a > 1, but increasing without bound? i don't see it in the proof (that is to say, how exactly do we choose m?).

4. ## Re: Question on boundedness

Originally Posted by Deveno
just out of curiosity: how do we know there is such an m? because, to me, this seems like a shell game: we are implcitly assuming that we can make a^m bigger than any real number (in particular, α/a), which is to say, {a^n} is unbounded (circular reasoning).

i mean, it's clear {a^n} is increasing, since a > 1, but increasing without bound? i don't see it in the proof (that is to say, how exactly do we choose m?).
Because by definition, since $\displaystyle \displaystyle \frac{\alpha}{a}<\alpha$ it cannot be an upper bound for $\displaystyle S_a$ and thus there must exist some element of $\displaystyle S_a$ which is strictly bigger than $\displaystyle \displaystyle \frac{\alpha}{a}$--I just generically called that element $\displaystyle a^m$.

5. ## Re: Question on boundedness

thanks. it is important (from a logical standpoint) that one note that α/a < α (this is what uses the condition that a > 1).

this actually tells us "which" m, it is the first one for which a^m > α/a, that is, the proof becomes constructive.