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