Proof of Supremum of a set

Question

Prove that $$\sup\{1−1/n^2\}=1.$$ I have tried to prove it by Archimedean property as let $a>1$ is the Supremum and tried to find contradiction but can't .

Answer 1

if a<1, then a is not a supremum: let $n$ bu such that $\frac{1}{n^2}<1-a$ then $1-\frac{1}{n^2}<a$.

if a = 1 then $a > 1-\frac{1}{n^2}$ for any n. Hence a is the supremum

Answer 2

Let $$A=\left\{1-\frac1{n^2}|n\in\mathbb N\right\}$$

Clearly, if $a>1$, $a$ cannot be the supremum because $1$ is also the upper bound for the set.

So, let $a<1$. What you need to prove is that $a$ is not an upper bound for the set. You can do that by finding $x\in A$ such that $x>a$. Since each $x\in A$ can be written as $1-\frac{1}{n^2}$, this means you must find some $n\in\mathbb N$ such that

$$1-\frac{1}{n^2}>a$$

The inequality can be rewritten as $$\frac{1}{n^2}<1-a$$

Can you continue from here?

Answer 3

for every $\epsilon > 0$, if $s $ is supremum, then $s- \epsilon$ is not Supremum. then following must hold

$s-\epsilon < x \leq s$ where x is 1 element in given sequence of form $(1-\frac{1}{n^2})$

$s-\epsilon < (1-\frac{1}{n^2}) \leq s < 1$ (1 is upper bound which should be greater than least upper bound)

applying $n \to \infty$ we get $1 \leq s < 1$ . By squeeze theorem s = 1

Answer 4

Hint:

Show that $x > y \ge 1 \implies x^{-1} < y^{-1} \le 1$.

Then use Archimedean property to show that $\forall \epsilon \exists n \in \Bbb N : \frac 1 {n^2} \le \frac 1 n < \epsilon$