Supremum of Subset of Real Numbers/Examples/Example 3
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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: $\text{(iii)}$