# Supremum of Subset of Real Numbers/Examples/Example 6

## Example of Supremum of Subset of Real Numbers

The subset $S$ of the real numbers $\R$ defined as:

$S = \set {x \in \R: x^2 + 2 x \le 1}$

$\sup S = -1 + \sqrt 2$

such that $\sup S \in S$.

## Proof

 $\ds x^2 + 2 x$ $\le$ $\ds 1$ $\ds \leadsto \ \$ $\ds x^2 + 2 x - 1$ $\le$ $\ds 0$ $\ds \leadsto \ \$ $\ds \paren {x + 1 - \sqrt 2} \paren {x + 1 + \sqrt 2}$ $\le$ $\ds 0$

When $x < -1 - \sqrt 2$ we have that both $x + 1 - \sqrt 2 < 0$ and $x + 1 + \sqrt 2 < 0$, and so $x^2 + 2 x - 1 > 0$.

When $x > -1 + \sqrt 2$ we have that both $x + 1 - \sqrt 2 > 0$ and $x + 1 + \sqrt 2 > 0$, and so $x^2 + 2 x - 1 > 0$.

When $-1 - \sqrt x \le x \le -1 + \sqrt 2$, we have that $x + 1 - \sqrt 2 \ge 0$ and $x + 1 + \sqrt 2 \le 0$, and so $x^2 + 2 x - 1 < 0$

Hence:

$S = \set {x \in \R: -1 - \sqrt x \le x \le -1 + \sqrt 2}$

and it follows that:

$\sup \set {x \in \R: x^2 + 2 x \le 1} = -1 + \sqrt 2$

and it follows that $\sup S \in S$.

$\blacksquare$