To prove that the set $M = \{ x \in \mathbb{R} \mid 0 < x < 1 \}$ has neither a minimum nor a maximum, we need to examine the nature of open intervals and the absence of boundary points within the set.

Proof

1. No Minimum:

   • Suppose, for the sake of contradiction, that $M$ has a minimum element $m$. By the definition of $M$, $0 < m < 1$.
   • However, for any $m \in M$, we can find a smaller number $m' = \frac{m}{2}$ (since $m > 0$, $m' > 0$ as well) such that $0 < m' < m$.
   • This contradicts the assumption that $m$ is the minimum element because there exists an element $m' \in M$ smaller than $m$.
   • Therefore, $M$ cannot have a minimum.

2. No Maximum:

   • Suppose, again for contradiction, that $M$ has a maximum element $M$.
   • For any $M \in M$, we can find a larger number $M' = \frac{M + 1}{2}$ such that $M < M' < 1$.
   • This contradicts the assumption that $M$ is the maximum element because there exists an element $M' \in M$ larger than $M$.
   • Therefore, $M$ cannot have a maximum.

Conclusion

Since $M$ lacks both a minimum and a maximum element, we have proven that $M$ has neither.

Would you like further details or have any questions?

Here are some related questions:

1. What are the properties of open intervals in real analysis?
2. How do we formally define a minimum and maximum of a set?
3. How does this proof differ for closed intervals?
4. Can we apply similar reasoning to bounded but infinite sets?
5. How do minimum and maximum relate to infimum and supremum?

Tip: When analyzing bounded intervals, remember that open intervals do not include their endpoints, which is often the key to proving the absence of minimum or maximum elements.