# Proof of least upper bound and greatest lower bound property of an ordered set

I am currently self-studying topology using the book of Munkres and since I am completely new to this topic (engineering grad) I need your opinion/verification regarding the proof that I came up with for the exercise 3.13. It seems to me a little bit simpler (though maybe not correct) comparing to this one.

Prove the following:

Theorem. If an ordered set A has the least upper bound property, then it has the greatest lower bound property.

If A has the least upper bound property then there is $$A_0\subset A$$ with $$A_0$$ bounded above and $$c = \sup \{A_0\} \in A$$, with $$x\leq c$$ for any $$x \in A_0 \subset A$$. Now consider the singleton subset $$A_1 = \{c\} \subset A$$, consisting of only the element $$c$$, which is bounded below. Assuming that $$A$$ does not have the greatest lower bound property, then $$\inf \{A_1\}\notin A$$, but $$\inf\{A_1\}=c \in A$$. Thus, by contradiction, the set $$A$$ must also have the greatest lower bound property. $$\square$$

Edit: Since the above attempt to prove the theorem is clearly false, I post a revised one, according to the hint provided by @fleablood.

Let $$B_0$$ be any subset of $$A$$ that is bounded below and $$A_0$$ the set of all lower bounds of $$B_0$$ in A, i.e. $$A_0 = \{ x \in A\ |\ x \leq x_0, \forall x_0 \in B_0 \}$$. Then the set $$A_0$$ is bounded above since there is $$x_0 \in B_0 \subset A$$ such that $$x \leq x_0$$ for all $$x \in A_0$$. Assuming now that the set $$A$$ has the l.u.b. property, we can write $$c = \sup \{ A_0\} \in A$$ and $$x \leq c \leq x_0$$, $$\forall x \in A_0$$, $$\forall x_0 \in B_0$$, because if there is $$x_0 \in B_0$$ such that $$x_0 < c$$ then $$c$$ would not be the l.u.b. of $$A_0$$. That also means that $$c\in A_0$$ (from the definition of $$A_0$$) and thus $$c$$ is the g.l.b. of $$B_0$$ which exists in $$A$$. Conversely, let $$B_0$$ be any subset of $$A$$ that is bounded above and $$A_0$$ the set of all upper bounds of $$B_0$$ in $$A$$. In a similar way we conclude that if the set $$A$$ has the g.l.b. property it must also have the l.u.b. property. $$\square$$

Could you please confirm me if it's correct or if there are still holes or false arguments?

• Poorly written. There is $A_0$... is the wrong start. Let $A_0$ be any subset... is getting off on the right foot. The rest was bumpy at best. Commented Aug 1, 2018 at 4:39
• Let $A_0$ be any subset... is another way of saying that there is arbitrary $A_0$, but I am missing the word `arbitrary', that's your point? By bumpy you mean badly written or incorrect?
– ares
Commented Aug 1, 2018 at 4:53
• No. Not having the greatest lower bound property doesn't mean that greatest lower bound never exists. It just means it doesn't always have to exist. And there is nothing important about $c=A_1$. You could make the exact same argument with any $\{d\}: d\in A$. Does that mean every ordered set has the greatest lower bound property? If so there really wasn't any point in making a definition for it. Commented Aug 1, 2018 at 4:55
• "By bumpy you mean badly written or incorrect?" I suspect he meant badly written because he didn't bother reading it. It started out okay (albeit badly written) but... it is incorrect. You seem to thing l.u.b/g.l.b property means either i) sets can be bounded above/below or that it means ii) l.u.b/g.l.b might exist. The l.u.b/g.l.b property does not mean either of those. It mean l.u.b/g.l.b must always exist. Commented Aug 1, 2018 at 5:17
• You misunderstood what that means. It doesn't mean that some $A_0$ exist. It means that ever every nonempty set, $A_0$ that is bounded above then there will be a $c$ that is $\sup A_0$. Which means that for any $x\in A_0$ you will have $x \le c$ but ALSO that for any $c_0 < c$ that will not be the case... note: $\sup (0,2) = 2$ but $2.25$ is ALSO an upper bound. for all $x \in (0,2)$ we have $x \le 2.25$. But $2.25$ isn't the LEAST number with this property. Commented Aug 1, 2018 at 5:31

If A has the least upper bound property then there is A0⊂A with A0 bounded above

That'd be true if $A$ didn't have the least upper bound property. You can always find a set that is bounded above or below. The question is whether every such set will always have a least upper bound.

Example: Let $A_0 = \{q| q^2 < 2\} \subset \mathbb Q$. That IS bounded above by $2$. But $2$ is not the least upper bound. There is no least upper bound for that set.

Also not having the least upper bound property doesn't mean that a least upper bound doesn't exist.

Example: Let $B = \{q| q < 3\}\subset \mathbb Q$. $\sup B$ does exist and $\sup B = 3$. Even though $\mathbb Q$ doesn't have the least upper bound property. In $\mathbb Q$, $B$ has a least upper bound.... and $A$ does not..

c=sup{A0}∈A, with x≤c for any x∈A0⊂A

Okay. $A$ does have the least upper bound property so $c = \sup A_0$ does exist. That's true.

Now consider the singleton subset A1={c}⊂A, consisting of only the element c, which is bounded below.

That'd be true of any singleton set, whether $A$ had the least upper bound principal or not.

Consider $A = \mathbb Q$ and $A_0=\{q| q < 3\}$ and $\sup A_0 = c$.

Now let $A_1 = \{c\} = \{3\}$.

Nothing interesting is going to come from this.....

Assuming that A does not have the greatest lower bound property, then inf{A1}∉A

Nonsense. $\mathbb Q$ does not have the greatest lower bound property but $\inf \{3\} = 3$.

Not having the greatest lower bound doesn't mean a greatest lower bound can't exist. It just means it doesn't have to exist.

$B=\{q| q^2 > 3\} \subset Q$ is such that $\inf B$ does not exist.

But $\inf A_1 = \inf \{c\} = c$ does.