Supremum of Subset of Real Numbers/Examples/Example 5
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Example of Supremum of Subset of Real Numbers
The subset $S$ of the real numbers $\R$ defined as:
- $S = \set {x \in \R: x^2 \le 2 x - 1}$
admits a supremum:
- $\sup S = 1$
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 $\sup S \in S$.
$\blacksquare$
Sources
- 1975: W.A. Sutherland: Introduction to Metric and Topological Spaces ... (previous) ... (next): $1$: Review of some real analysis: Exercise $1.5: 4$