Supremum of Subset of Real Numbers/Examples/Example 3

Example of Supremum of Subset of Real Numbers

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

$V := \set {x \in \R: x > 0}$

does not admit a supremum.

Proof

Aiming for a contradiction, suppose $x \in \R$ is a supremum for $V$.

Then we have that:

$x + 1 \in V$

and it is seen that $x$ is not a supremum after all.

$\blacksquare$

Sources

• 1977: K.G. Binmore: Mathematical Analysis: A Straightforward Approach ... (previous) ... (next): $\S 2$: Continuum Property: $\S 2.5$: Examples: $\text{(iii)}$