# Question on boundedness

• Jun 18th 2011, 10:50 AM
worc3247
Question on boundedness
Let $a\in\mathbb{R}$ and consider the set $S_a=\{a^n:n\in\mathbb{N},n>0\}$. 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?
• Jun 18th 2011, 11:33 AM
Drexel28
Re: Question on boundedness
Quote:

Originally Posted by worc3247
Let $a\in\mathbb{R}$ and consider the set $S_a=\{a^n:n\in\mathbb{N},n>0\}$. 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 $\alpha$ but by definition of supremum and the fact that $a>1$ we have that there is some $m\in\mathbb{N}$ for which $\displaystyle \frac{\alpha}{a} and so $\alpha< a^{m+1}$ contradiction.
• Jun 18th 2011, 11:51 AM
Deveno
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?).
• Jun 18th 2011, 12:40 PM
Drexel28
Re: Question on boundedness
Quote:

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 $\displaystyle \frac{\alpha}{a}<\alpha$ it cannot be an upper bound for $S_a$ and thus there must exist some element of $S_a$ which is strictly bigger than $\displaystyle \frac{\alpha}{a}$--I just generically called that element $a^m$.
• Jun 18th 2011, 12:59 PM
Deveno
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.