Bounded Above Subset of Real Numbers/Examples/Open Interval from 0 to 1

Example of Bounded Above Subset of Real Numbers

Let $I$ be the open real interval defined as:

$I := \openint 0 1$

Then $I$ is bounded above by, for example, $1$, $2$ and $3$, of which $1$ is the supremum.

However, $I$ does not have a greatest element.

Proof

Aiming for a contradiction, suppose $x \in I$ is the greatest element of $I$.

Then $x < \dfrac {x + 1} 2 < 1$ by Mediant is Between.

Thus $\dfrac {x + 1} 2 \in I$ but $x < \dfrac {x + 1} 2$.

So $x$ is not the greatest element of $I$ after all.

So by Proof by Contradiction it follows that $I$ has no greatest element.

$\blacksquare$

Sources

• 1977: K.G. Binmore: Mathematical Analysis: A Straightforward Approach ... (previous) ... (next): $\S 2$: Continuum Property: Exercise $\S 2.10 \ (3) \ \text{(i)}$