Supremum of Subset of Real Numbers/Examples/Example 6
Jump to navigation
Jump to search
Example of Supremum of Subset of Real Numbers
The subset $S$ of the real numbers $\R$ defined as:
- $S = \set {x \in \R: x^2 + 2 x \le 1}$
admits a supremum:
- $\sup S = -1 + \sqrt 2$
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 $\sup S \in S$.
$\blacksquare$
Sources
- 1975: W.A. Sutherland: Introduction to Metric and Topological Spaces ... (previous) ... (next): $1$: Review of some real analysis: Exercise $1.5: 4$