# Does the interval (0, 1] satisfy the least upper bound property?

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?

• $E$ does not have a lower bound in $S$.
– bof
Feb 16, 2019 at 13:54

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.

• As I know the theorem says that any nonempty subset of $S$ which is bounded from below (which means it has a lower bound) must have a greatest lower bound. But $E$ is not bounded from below, it doesn't have any lower bounds which are in $S$.
– Mark
Feb 16, 2019 at 13:47
• Yes, I have the book and I just finished reading the proof. He writes "Since $L$ consists of exactly those $y\in S$ which satisfy the inequality $y\leq x$ for every $x\in B$...". So $L$ is the set of lower bounds of $B$ which are in $S$. The set $S$ is our universe, forget that you know there are elements outside of $S$. $S$ is our world. The assumption in the theorem is that $B$ has a lower bound which is in $S$, which means the set of lower bounds $L$ is not empty. But in your example, what lower bounds does the set $E$ you defined have? The set of its lower bounds is empty.
– Mark
Feb 16, 2019 at 14:03
• Ahh, I see now. I had forgotten Definition 1.7: $E$ being bounded above/below implies the existence of an element $\beta \in S$ such that $x \leq \beta$, $\forall x \in E$. Sorry for the confusion and thank you for taking the time to double-check the proof. Feb 16, 2019 at 14:15