# One more fun proof1

• Nov 12th 2006, 06:37 PM
Swamifez
One more fun proof1
Thank you in advance for any more help!

Prove that S = { n−1/n |n ∈ N} is bounded above and that its supremumn
is equal to 1.
(Hint: First show that 1 is an upper bound for S. Then suppose, for a
contradiction, that sup S = α < 1. Using a theorem from the lecture
we can ﬁnd a rational number alpha < p < 1. Deduce that we can ﬁnd an
n ∈ N with αn infiniti alpha< n/n-1 . Contradiction to infiniti alpha being an upper bound.)
• Nov 12th 2006, 07:03 PM
Soltras
Quote:

Originally Posted by Swamifez
Thank you in advance for any more help!

Prove that S = { n−1/n |n ∈ N} is bounded above and that its supremumn
is equal to 1.
(Hint: First show that 1 is an upper bound for S. Then suppose, for a
contradiction, that sup S = α < 1. Using a theorem from the lecture
we can ﬁnd a rational number alpha < p < 1. Deduce that we can ﬁnd an
n ∈ N with αn infiniti alpha< n/n-1 . Contradiction to infiniti alpha being an upper bound.)

Suppose 1 weren't an upper bound.
Then there exists some $\displaystyle n \in \mathbb{N}$ such that $\displaystyle \frac{n-1}{n} > 1$.
Thus, $\displaystyle n - 1 > n$, particularly $\displaystyle -1 > 0$.
So 1 must be an upper bound of $\displaystyle S$.

So $\displaystyle S$ is a bounded set. We still do not know if there is a supremum in the rationals because $\displaystyle S \subset \mathbb{Q}$, and $\displaystyle \mathbb{Q}$ does not carry the least upper bound property.

However, we know since the reals do have the least upper bound property that $\displaystyle \sup S$ at least lies in $\displaystyle \mathbb{R}$.

Suppose 1 were not the least upper bound of $\displaystyle S$.

Let $\displaystyle \sup S = \alpha \in \mathbb{R}$.
Then $\displaystyle \alpha < 1$.

By density of the rationals in the reals, there exists $\displaystyle \beta \in \mathbb{Q}$ such that $\displaystyle \alpha < \beta < 1$.
Since $\displaystyle \beta > \alpha$, then $\displaystyle \beta$ is an upper bound of $\displaystyle S$.

Since $\displaystyle \beta \in \mathbb{Q}$, write $\displaystyle \beta = \frac{\gamma}{\delta}$, for $\displaystyle \gamma, \delta \in \mathbb{N}, \delta \ne 0$. Note that $\displaystyle \gamma < \delta$, for if not then $\displaystyle \beta \ge 1$ which cannot be true.

Then $\displaystyle \frac{\gamma}{\delta} \le \frac{\delta - 1}{\delta}$.

However, if this were true, then $\displaystyle \beta$ would not be an upper bound of $\displaystyle S$ because for $\displaystyle n \ge \delta+1$, then $\displaystyle \frac{n-1}{n} > \beta$.

So therefore $\displaystyle \alpha$ cannot be an upper bound of $\displaystyle S$ (and certainly not its least upper bound).

Therefore, the least upper bound of $\displaystyle S$ is 1 as desired. QED.