Please help to solve the following:
The set is partially ordered with respect to the “less than or equal to” relation, ≤, for real numbers. In following case, determine whether the set has a greatest or least element.
{x ∈ R| 0 < x < 1}
Since R is rational I assume the least element is 0+R and the greatest is 1-R. For better visibility I change the form a bit:
0+R <= x <= 1-R
In this case we have $X = \{x \in R \mid 0 < x < 1 \}$. No there is no greatest or least element in $ X$ since the given set is an open set. Consider the least element, this should be the smallest element in the set. For each element $ y \in X $ there exist an $\epsilon > 0$ such that $y- \epsilon \in X$. The same reasoning holds for the greatest element.
Note that the standard ordering $\leq$ is not a well-ordering for the positive real numbers since the above example does not contain a least element.
Note if we have $X = \{x \in R \mid 0 \leq x \leq 1 \}$ then there is a least and a greatest element. The least element is $x=0$ and the greatest element is $x=1$. This holds true since each value smaller than 0 is outside the set and each value larger than 1 is outside the set.