Supremum of Subset of Real Numbers/Examples/Example 8

Example of Supremum of Subset of Real Numbers

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

$S = \set {x \in \R: x \sin x < 1}$

does not admit a supremum.

Proof

Let $x = n \pi$.

From Sine of Multiple of Pi, we have that:

$\sin x = 0$

and so:

$x \sin x = 0$

That is:

$\forall n \in \Z: n \pi \sin n \pi < 1$

and $n \pi$ is not bounded above.

$\blacksquare$

Sources

• 1975: W.A. Sutherland: Introduction to Metric and Topological Spaces ... (previous) ... (next): $1$: Review of some real analysis: Exercise $1.5: 4$