I am trying to follow this text below, but I am struggling to understand a key component.
Question:
Consider Q where for which A and B are proper subsets of Q, And since A is bounded by B By definition 1.10 suggests that sup of A exists in Q.
But in example 1.9(a) they show it is not the case which is opposite to definition in 1.10. What am I missing?
Then I found the answer to same question here: Least-upper-bound property Rudin book
However, I am curios why do we have the implication in a definition secondly is this correct or is it the case of if and only if as suggested in the link.
By Definition 1.10 An ordered set $S$ is said to have the least-upper-bound property if the following is true:
If $E \subset \mathrm{S}, E$ is not empty, and $E$ is bounded above, then $\sup E$ exists in $S$.
Then looking at:
Example 1.9(a) shows that $Q$ does not have the least-upper-bound property. We shall now show that there is a close relation between greatest lower bounds and least upper bounds, and that every ordered set with the least-upperbound property also has the greatest-lower-bound property.
Example $1.1$ Let $A$ be the set of all positive rationals $p$ such that $p^2<2$ and let $B$ consist of all positive rationals $p$ such that $p^2>2$. We shall show that $A$ contains no largest number and $B$ contains no smallest.
More explicitly, for every $p$ in $A$ we can find a rational $q$ in $A$ such that $p<q$, and for every $p$ in $B$ we can find a rational $q$ in $B$ such that $q<p$. To do this, we associate with each rational $p>0$ the number (3) $$ q=p-\frac{p^2-2}{p+2}=\frac{2 p+2}{p+2} $$ Then (4) $$ q^2-2=\frac{2\left(p^2-2\right)}{(p+2)^2} $$ If $p$ is in $A$ then $p^2-2<0$, (3) shows that $q>p$, and (4) shows that $q^2<2$. Thus $q$ is in $A$
If $p$ is in $B$ then $p^2-2>0$, (3) shows that $0<q<p$, and (4) shows that $q^2>2$. Thus $q$ is in $B$.
1.8 Definition Suppose $S$ is an ordered set, $E \subset S$, and $E$ is bounded above. Suppose there exists an $\alpha \in S$ with the following properties: (i) $\alpha$ is an upper bound of $E$. (ii) If $\gamma<\alpha$ then $\gamma$ is not an upper bound of $E$. Then $\alpha$ is called the least upper bound of $E$ [that there is at most one such $\alpha$ is clear from (ii)] or the supremum of $E$, and we write $$ \alpha=\sup E . $$ The greatest lower bound, or infimum, of a set $E$ which is bounded below is defined in the same manner: The statement $$ \alpha=\inf E $$ means that $\alpha$ is a lower bound of $E$ and that no $\beta$ with $\beta>\alpha$ is a lower bound of $E$.
1.9 Examples (a) Consider the sets $A$ and $B$ of Example $1.1$ as subsets of the ordered set $Q$. The set $A$ is bounded above. In fact, the upper bounds of $A$ are exactly the members of $B$. Since $B$ contains no smallest member, $A$ has no least upper bound in $Q$.
Similarly, $B$ is bounded below: The set of all lower bounds of $B$ consists of $A$ and of all $r \in Q$ with $r \leq 0$. Since $A$ has no lasgest member, $B$ has no greatest lower bound in $Q$. (b) If $\alpha=\sup E$ exists, then $\alpha$ may or may not be a member of $E$. For instance, let $E_1$ be the set of all $r \in Q$ with $r<0$. Let $E_2$ be the set of all $r \in Q$ with $r \leq 0$. Then $$ \sup E_1=\sup E_2=0, $$ and $0 \notin E_1, 0 \in \mathrm{E}_2$. (c) Let $E$ consist of all numbers $1 / n$, where $n=1,2,3, \ldots$ Then $\sup E=1$, which is in $E$, and $\inf E=0$, which is not in $E$.
