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

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## Example of Bounded Below Subset of Real Numbers

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

- $I := \openint 0 1$

Then $I$ is bounded below by, for example, $0$, $-1$ and $-2$, of which $0$ is the infimum.

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

## Proof

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

Then $0 < \dfrac x 2 < x$ by Mediant is Between.

Thus $\dfrac x 2 \in I$ but $x > \dfrac x 2$.

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

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

$\blacksquare$

## Sources

- 1977: K.G. Binmore:
*Mathematical Analysis: A Straightforward Approach*... (previous) ... (next): $\S 2$: Continuum Property: Exercise $\S 2.10 \ (4) \ \text{(i)}$