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?

  • 1
    $\begingroup$ $E$ does not have a lower bound in $S$. $\endgroup$
    – bof
    Feb 16, 2019 at 13:54

1 Answer 1


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.

  • $\begingroup$ 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 $S$. $\endgroup$
    – Mark
    Feb 16, 2019 at 13:47
  • $\begingroup$ 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. $\endgroup$
    – Mark
    Feb 16, 2019 at 14:03
  • $\begingroup$ 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. $\endgroup$ Feb 16, 2019 at 14:15
  • $\begingroup$ See also: my response to this thread: math.stackexchange.com/questions/2899554/… $\endgroup$ Feb 16, 2019 at 15:16
  • $\begingroup$ Yes, you understood the proof right. I know these things might be confusing at the beginning, but later they become easier. $\endgroup$
    – Mark
    Feb 16, 2019 at 17:07

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