One more fun proof1

**Swamifez** · November 12th 2006, 06:37 PM

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.)

**Soltras** · November 12th 2006, 07:03 PM

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 $n \in \mathbb{N}$ such that $\frac{n-1}{n} > 1$ .
Thus, $n - 1 > n$ , particularly $-1 > 0$ .
So 1 must be an upper bound of $S$ .

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

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

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

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

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

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

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

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

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

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

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