# Supremum of Subset of Real Numbers/Examples/Example 5

## Example of Supremum of Subset of Real Numbers

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

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

$\sup S = 1$

such that $\sup S \in S$.

## Proof

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

But as $\forall x \in \R: \paren {x - 1}^2 \ge 0$ it follows that:

$\paren {x - 1}^2 \le 0 \implies x - 1 = 0$

and so:

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

and so trivially:

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

and as $1 \in \set 1$ it follows that $\sup S \in S$.

$\blacksquare$