I'm having a little bit of a problem with this; any help would be handy!

Find, with proof the least upper bound of this set of real numbers!

E = {6 - (2/(3 n + 1)) : n is a element of N}

Where N is the natural Numbers

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- Jan 24th 2010, 08:03 AMSuperFuzzyGremlinProof of the Least Upper Bound(Supremum) of a Set
I'm having a little bit of a problem with this; any help would be handy!

Find, with proof the least upper bound of this set of real numbers!

E = {6 - (2/(3 n + 1)) : n is a element of N}

Where N is the natural Numbers - Jan 24th 2010, 08:28 AMPlato
- Jan 25th 2010, 12:58 PMSuperFuzzyGremlinSoloution
Claim Sup(E) = 6

To prove this, first 6 must be found to be a valid upper bound for E. Then we must ascertain that their is no smaller upper bound then 6. Where y < 6, where y might be a smaller upper bound for E.

Proof:

Let x ∈ E with x>y

$\displaystyle \left(6-\frac{2}{3n+1}\right)$>y

$\displaystyle \left(-\frac{2}{3n+1}\right)$>y-6

$\displaystyle \left(-\frac{2}{3n+1}\right)$>y-6

-2>(y-6)(3n+1)

n>$\displaystyle \frac{1}{3}\left(\frac{2}{y-6}+1\right)$

Then pick $\displaystyle n \in N$

such that n>$\displaystyle \frac{1}{3}\left(\frac{2}{y-6}+1\right)$

This means that $\displaystyle \left(6-\frac{2}{3n+1}\right)$>y

Hence y can not be an upper bound for E!!!

Finally got it after pouring through a book for a very long while!!!

Thanks for the help! - Jan 25th 2010, 01:15 PMJhevon
this is incorrect. you assumed x > y. that is, y is not an upper bound. this is what you want to show, not assume. your choosing of n seems to be inconsequential. moreover, you ended up with exactly what you assumed, which means you've shown nothing, just went in a circle.

I would proceed this way:

Clearly 6 is an upper bound for E, since (i leave it to you to show this).

To show that $\displaystyle 6 = \sup E$, we need to show that 6 is*the least*upper bound. Assume, to the contrary, that it isn't. Rather, let $\displaystyle y = \sup E$ where $\displaystyle y \in \mathbb R,~y < 6$. Then, for all $\displaystyle x \in E$, we have $\displaystyle x \le y < 6$.

Now, let $\displaystyle y = 6 - \epsilon$ for some $\displaystyle \epsilon > 0$. Then we have, for all $\displaystyle n \in \mathbb N$, that

$\displaystyle 6 - \frac 2{3n + 1} \le 6 - \epsilon < 6$

$\displaystyle \Leftrightarrow 0 < \epsilon \le \frac 2{3n + 1}$ for all $\displaystyle n$.

Which is absurd (why?). Hence our original assumption must have been in error and, in fact, $\displaystyle \sup E = 6$