Thread: Question on boundedness

Let and consider the set . 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?

Originally Posted by worc3247

Let and consider the set . 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 but by definition of supremum and the fact that we have that there is some for which and so contradiction.

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?).

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 it cannot be an upper bound for and thus there must exist some element of which is strictly bigger than --I just generically called that element .

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.