I'm having difficulty understanding one of the very early theorems proved in Rudin, namely that the least upper bound property implies the greatest lower bound property.
Why is the following not a counterexample?
The interval $S = (0, 1] = \{x | x \in \mathbb{R}, 0 < x, x \leq 1\}$ seems to satisfy the least upper bound property, but not the greatest lower bound property (let $E$ be (0, 0.5]. Then $E$ is non-empty, $E \subset S$, and $E$ has a lower bound, but inf $E = 0,$ and $0 \not \in S$).
But why does $S$ not nevertheless satisfy the least upper bound property? Is there a non-empty, upper-bounded subset $B$ whose supremum is not in $S$?
Edit: a little more context, because the initial replies seem to be off the mark.
Definition: The Least Upper Bound Property
An ordered set $S$ is said to have the least upper bound property if the following is true:
if $E \subset S$, $E$ is not empty, and $E$ is bounded above, then sup $E$ exists in $S$.
The interval (0,1] seems to satisfy this property. At any rate, I cannot think of a subset $E$ meeting the above criterion whose supremum does not exist in $S$.
As far as I know, the definition of the greatest lower bound property is symmetric. And yet, (0,1] clearly does not satisfy that property, because if $E = (0,0.5]$, then inf $E \not \in S$
But according to the Principles Of Mathematical Analysis, the first property implies the second!
So why is my counterexample not a true counterexample?
Best Answer
But the set $E$ you defined has no lower bound which belongs to $S$. Take any number in $S$ and it is not a lower bound of $E$. Hence the set is not bounded from below, so it doesn't need to have a greatest lower bound.