I saw saw the statement I am trying to prove in "a proof of lub property" in a Mathematics Stack Exchange post. The statement I am trying to prove is :
$∃\ b ∈ \mathbb {R},(b < a\ \text { AND }$
$\text {Set of all rationals to the left of}\ b = A)$
where $A$ is the set of all rationals to the left of $a ∈ Q$
Is this statement indeed true ?
I get the Theorem is false
Below is a proof of lub property I got online:
The author uses $\subset$ as $\subseteq$. Also notice the difference by font that $\mathscr{C}$ is a class, while $C$ a set.
Definition. A cut in $\Bbb{Q}$ is a pair of subsets $A,B$ of $\Bbb{Q}$ such that
- $A\cup B=\Bbb{Q},\quad A\ne\emptyset,\quad B\ne\emptyset,\quad A\cap B=\emptyset.$
- $a\in A \land b \in B \implies a\lt b.$
- $A$ contains no largest element.
Definition. The cut $x=A|B$ is less than or equal to the cut $y=C|D$ if $A\subset C$.
Theorem. If $S$ is a non-empty subset of $\Bbb{R}$ and is bounded above then in $\Bbb{R}$ there exists a least upper bound for $S$.
Proof: Let $\mathscr{C}\subset\Bbb{R}$ be any non-empty collection of cuts which is bounded above, say by the cut $X|Y$. Define $$C=\{a\in\Bbb{Q}: \stackrel{?_1}{\text{for some cut }}A|B\in\mathscr{C},a\in A\},D=\text{the rest of }\Bbb{Q}.$$ It is easy to see that $z=C|D$ is a cut. Clearly, it is an upper bound for $\mathscr{C}$ since the "$A$" for every element of $\mathscr{C}$ is contained in $C$. Let $z'=C'|D'$ be any upper bound for $\mathscr{C}$. By the assumption that $A|B\le C'|D'$ for all $A|B\in\mathscr{C}$, we see that the "$A$" for every member of $\mathscr{C}$ is contained in $C'$. Hence $C\subset C'$, so $z\le z'$. That is, among all upper bounds for $\mathscr{C}$, $z$ is $\stackrel{?_2}{\text{least}}$.
What I don't get is, it is written right after the definition of $C$ :
It is easy to see that
$z=C|D$ is a cut.
Why is it easy to see...
