# Confusion about the least upper bound property

We say an ordered set $$S$$ has the least upper bound property if every nonempty subset of $$S_0$$ of $$S$$ that is bounded above has a least upper bound.

Let $$A = (-1,1) \subset \mathbb{R}$$. Does $$A$$ have the least upper bound property? Well, let $$A_0$$ be any nonempty subset of $$A$$. If $$A_0$$ is bounded above, then $$\exists a \in A$$ such that $$\forall x \in A_0, x \le a$$. Now clearly, $$x \le 1$$, but $$1 \not\in A$$. So how do we we deal with this? We assume (or prove) $$\mathbb{R}$$ has the least upper bound property. So assuming that it does, then every bounded above subset of $$\mathbb{R}$$ has a least upper bound in $$\mathbb{R}$$. Clearly $$(-1,1)$$ is a bounded above subset of $$\mathbb{R}$$, since $$\forall x \in (-1,1), x \le 1$$, so it has a least upper bound in $$\mathbb{R}$$, but not in $$A$$?

Similarly, why doesn't $$(-1,0)\cup(0,1)$$ have the least upper bound property?

• Be careful about the precise statement of a 'least upper bound property"! If a set of real numbers has an upper bound then it has a least upper bound. It does NOT say the least upper bound must be IN that set! Commented Aug 13, 2022 at 18:27
• @GeorgeIvey I understand that. The set $(-1,0)\cup(0,1)$ should have a least upper bound $1$. Why doesn't that work? Commented Aug 13, 2022 at 18:33
• The least upper bound needs to be in $\mathbb{R}$. $1$ is a real number. But the least upper bound does not need to belong to $(-1,0) \cup (0,1)$. Have you learnt about Cauchy sequences and all that stuff? Commented Aug 13, 2022 at 18:34
• @MordeusMorgenstern No. I'm just wondering why $(-1,1)$ has the property, but why $(-1,0)\cup(0,1)$ doesn't. Commented Aug 13, 2022 at 18:35
• But $(-1,1)$ doesn't have that property. It has least upper bound $1$ but $1 \notin (-1,1)$. However, $1$ is a real number. Commented Aug 13, 2022 at 18:35

Considered as a subset of $$(-1, 1)$$, the set $$(-1, 1)$$ is not bounded.

The precise statement we are looking for is as follows:

Consider a partially ordered set $$(A, \leq)$$. A subset $$B \subseteq A$$ is said to be $$(A, \leq)$$-bounded from above iff there exists some $$a \in A$$ such that $$\forall b \in B (b \leq a)$$. When $$(A, \leq)$$ is clear from context, we just use the term “bounded from above”.

In particular, we see that $$(-1, 1)$$ is $$(\mathbb{R}, \leq)$$-bounded from above, since $$1$$ is an upper bound. However, $$(-1, 1)$$ is not $$((-1, 1), \leq)$$-bounded from above since there is no upper bound of $$(-1, 1)$$ in $$(-1, 1)$$.

The least-upper-bound property states that for all $$B \subseteq A$$, if $$B$$ is non-empty and $$(A, \leq)$$-bounded from above, then there is some $$a \in A$$ which is the least upper bound of $$B$$. In particular, $$\mathbb{R}$$ has the least upper bound property. Since $$(-1, 1)$$ and $$\mathbb{R}$$ are order-isomorphic, $$(-1, 1)$$ also has the least upper bound property.

• Then why doesn't $(-1,0)\cup(0,1)$ have the least upper bound property? Commented Aug 13, 2022 at 19:06
• Because $(-1,0)$ does not have a least upper bound within $(-1,0)\cup(0,1)$ but it does have upper bounds like $\frac 12$ Commented Aug 13, 2022 at 19:07
• @RossMillikan “$(-1,0)\cup(0,1)$ does have a least upper bound in $\mathbb{R}$, so why can't we use the same argument as we did for $(-1,1)$? Commented Aug 13, 2022 at 19:12
• because $(-1,1)$ does not have an upper bound in the set, so it does not need to have a least upper bound. Commented Aug 13, 2022 at 19:21
• @RossMillikan This is not a valid deduction. $(-1, 0)$ has a least upper bound in $(-1, 0) \cup \{1\}$ even though $0$, its supremum in $\mathbb{R}$, is not in $(-1, 0) \cup \{1\}$. Commented Aug 14, 2022 at 0:42