A modified and more direct version of Dedekind cut method to define a completion of $(P, <)$

Question

On the basis of @Daniel Wainfleet's hints. I modify the traditional method using Dedekind cuts to have a more direct proof. It takes me a great amount of time to formalize those ideas, so I'm extremely grateful to anyone to verify my attempt. Does it look fine or contain logical flaws? Thank you for your help!

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Let $(P, <)$ be a dense linearly ordered set without endpoints. Then there exists a complete linearly ordered set $(\mathfrak D,\prec)$ such that

1. $P\subseteq \mathfrak D$.

2. If $p,q\in P$ then $p < q$ if and only if $p \prec q$ ($<$ coincides with $\prec$ on $P$).

3. $P$ is dense in $\mathfrak D$, i.e., for any $a,b\in \mathfrak D$ such that $a\prec b$, there is $p\in P$ with $a\prec p \prec b$.

4. $\mathfrak D$ does not have endpoints.

The linearly ordered set $(\mathfrak D,\prec)$ is called the completion of $(P, <)$.

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My attempt:

A proper Dedekind cut of $(P, <)$ is a pair $(A, B)$ such that

1. $A\neq \emptyset$ and $B\neq \emptyset$.

2. $A\cap B=\emptyset$ and $A\cup B=P$.

3. If $a \in A$ and $b \in B$, then $a < b$.

4. $A$ does not have a greatest element.

5. $B$ does not have a least element.

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Let $\mathfrak C$ be the sets of proper Dedekind cuts of $(P, <)$.

Assuming Axiom of Foundation, it is easy to verify that $\mathfrak C \cap P=\emptyset$

Let $\mathfrak D=\mathfrak C \cup P$. We define an order $\prec$ on $\mathfrak D$ by

\begin{align}\text{For }x,y\in P:&x \prec y \iff x<y \\ \text{For }x\in P, y=(A,B)\in \mathfrak C:&x \prec y \iff x\in A \\ \text{For }x=(A,B)\in \mathfrak C, y\in P:&x \prec y \iff y\in B \\ \text{For }x =(A_1,B_1), y=(A_2,B_2)\in \mathfrak C:&x \prec y \iff A_1 \subsetneq A_2 \end{align}

$\prec$ is a linear order on $\mathfrak D$

It is easy to verify this fact from the definition of $\prec$.

$P \subseteq \mathfrak D$

This fact follows directly from the definition of $\mathfrak D$.

If $p,q\in P$ then $p < q$ if and only if $p \prec q$ ($<$ coincides with $\prec$ on $P$)

This fact follows directly from the definition of $\prec$.

$P$ is dense in $\mathfrak D$

Let $x,y\in\mathfrak D$ and $x\prec y$.

• If $x,y\in P$, $\exists p\in P:x<p<y$ since $P$ is dense $\implies$ $\exists p\in P:x\prec p\prec y$.

• If $x=(A_1,B_1)\in \mathfrak C$ and $y=(A_2,B_2) \in \mathfrak C$, $x\prec y\implies A_1 \subsetneq A_2 \implies \exists p:p\in A_2$ and $p\notin A_1\implies \exists p\in P:p\in A_2$ and $p\in B_1 \implies \exists p\in P: (A_1,B_1) \prec p \prec (A_2,B_2)$ by the definition of $\prec\implies\exists p\in P:x\prec p \prec y$.

• If $x\in P$ and $y=(A,B)\in \mathfrak C$, $x\prec y\implies x\in A\implies \exists p\in A:x<p$ since $\max A$ does not exist $\implies \exists p\in A:x\prec p$. Furthermore, $p\in A \implies p \prec (A,B)\implies \exists p\in P:$ $x \prec p \prec (A,B)\implies\exists p\in P:x \prec p \prec y$.

• If $x=(A,B)\in \mathfrak C$ and $y\in P$, $x\prec y\implies y\in B\implies \exists p\in B:p<y$ since $\min B$ does not exist $\implies \exists p\in B:p\prec y$. Furthermore, $p\in B\implies (A,B) \prec p \implies \exists p\in P:$ $(A,B)\prec p \prec y\implies\exists p\in P:x \prec p \prec y$.

$\mathfrak D$ does not have endpoints

It is easy to verify this fact.

$\mathfrak D$ is complete

Let $S$ be a nonempty subset of $\mathfrak D$ that is bounded from above. Next we prove that $S$ has a supremum.

Let $S_1=\{p\in P\mid p \text{ is an upper bound of }S\}$. It is easy to verify that $S_1 \neq \emptyset$.

For all $s\in S$:

• If $s\in P$, $\exists p\in P:p<s$ since $P$ does not have endpoints $\implies \exists p\in P:p\prec s$.

• If $s=(A,B)\in \mathfrak C$, $\exists p\in A$ since $A \neq \emptyset \implies \exists p\in P:p \prec s$.

$\implies\forall s\in S,\exists p \in P:p \prec s\implies \exists p\in P:p$ is not an upper bound of $S\implies \exists p\in P: p\notin S_1$ $\implies S_1 \neq P$.

Let $S_2=P\setminus S_1 \implies S_2 \neq \emptyset$ since $S_1 \neq P$.

$p \in S_2\implies p\in P$ and $p \notin S_1 \implies p\in P$ and $p$ is not an upper bound of $S \implies p\in P$ and $\exists s\in S:p\prec s \implies \exists p'\in P:p\prec p'\prec s$ since $P$ is dense in $\mathfrak D \implies \exists p'\in P:p\prec p'$ and $p'$ is not an upper bound of $S \implies \exists p'\in P:p\prec p'$ and $p' \notin S_1 \implies \exists p'\in S_2:p \prec p' \implies S_2$ does not a greatest element $\iff \max S_2$ does not exist.

1. $\max S$ exists

$\implies \sup S=\max S$.

2. $\max S$ does not exist

$\implies$ $(u$ is an upper bound of $S$ and $s\in S\implies s\prec u) \implies S\cap S_1=\emptyset$.

a. $\min S_1$ exists

It's clear that $\min S_1$ is an upper bound of $S$. Let $u$ be an upper bound of $S$.

• If $u\in P$, $u\in S_1 \implies \min S_1 \preccurlyeq u$.

• If $u \notin P$, $u=(A,B)\in \mathfrak C \implies \min S_1\in A$. If not, $\min S_1\in B \implies \exists p\in B:p\prec \min S_1$ since $\min B$ does not exist. Furthermore, $p\in B\implies (A,B)\prec p\implies p$ is an upper bound of $S \implies$ $p\in S_1 \implies \min S_1 \preccurlyeq p$. This is a contradiction $\implies \min S_1\in A \implies$ $\min S_1 \prec (A,B)$ $\implies \min S_1 \prec u$.

As a result, $u$ is an upper bound of $S\implies \min S_1 \preccurlyeq u \implies\sup S=\min S_1$.

b. $\min S_1$ does not exist

• $S_2 \neq \emptyset$ and $S_1 \neq \emptyset$.

• $S_2=P\setminus S_1 \implies S_1\cap S_2=\emptyset$ and $S_1\cup S_2=P$.

• If $a\in S_2$ and $b\in S_1$, $a\notin S_1$ and $b\in S_1\implies a$ is not an upper bound of $S$ and $b$ is an upper bound of $S \implies \exists s\in S:a \prec s$ and $s\prec b\implies a \prec b$.

• $\max S_2$ does not exist $\implies$ $S_2$ does not have a greatest element.

• $\min S_1$ does not exist $\implies$ $S_1$ does not have a least element.

As a result, $(S_2,S_1)$ is a proper Dedekind cut of $(P,<)$. Next we prove $(S_2,S_1)$ is a supremum of $S$.

For $s\in S$:

• If $s\in P$, $s\in S$ and $s\in P \implies s\notin S_1$ and $s\in P\implies s\in S_2 \implies s\prec (S_2,S_1)$.

• If $s=(A,B) \in \mathfrak C$, $p\in A\implies p\prec (A,B) \implies p\in P$ and $p$ is not an upper bound of $S \implies p\in P$ and $p\notin S_1 \implies p\in S_2 \implies A\subseteq S_2 \implies (A,B) \subseteq (S_2,S_1)\implies$ $s\preccurlyeq (S_2,S_1)$.

As a result, $(S_2,S_1)$ is an upper bound of $S$.

Let $u$ be an upper bound of $S$.

• If $u\in P$, $u\in S_1\implies (S_2,S_1)\prec u$.

• If $u\notin P$, $u=(A,B)\in \mathfrak C$. $p\in B\implies p\in P$ and $(A,B)\prec p \implies p\in P$ and $p$ is an upper bound of $S \implies p\in S_1 \implies B\subseteq S_1 \implies S_2\subseteq A\implies (S_2,S_1)\preccurlyeq (A,B) \implies$ $(S_2,S_1) \preccurlyeq u$.$$\tag*{$\blacksquare$}$$

Answer 1

Overall it is pretty nice but there is still a room for improvement:

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$...\implies A_1 \subsetneq A_2 \implies \exists p:p\in A_2\land p\notin A_1$

why? need justification.

For all $s\in S$:

• If $s\in P$, $\exists p\in P:p<s$ since $P$ does not have endpoints $\implies \exists p\in P:p\prec s$.

• If $s=(A,B)\in \mathfrak C$, $\exists p\in A$ since $A \neq \emptyset \implies \exists p\in P:p \prec s$.

$\implies\forall s\in S,\exists p \in P:p \prec s\implies \exists p\in P:p$ is not an upper bound of $S$

Correct, but way too much work, look on the following:

If $|S|=1$ then, by definition $x\in S$ is the maximum and hence the supremum, otherwise there exists $x,y\in S$ such that $x\prec y\implies x$ not an upper bound $\implies\exists z\in P(z\prec x)$(no endpoints+dense).

$S_2 \neq \emptyset$ since $S_1 \neq P$

Wrong reasoning, or rather, part of the explanation. $\{1\}\setminus\{\mbox{cat},1\}=\emptyset$ yet $\{1\}\overset*\ne\{\mbox{cat},1\}$.

• If $u \notin P$, $u=(A,B)\in \mathfrak C \implies \min S_1\in A$. If not, $\min S_1\in B \implies \exists p\in B:p\prec \min S_1$ since $\min B$ does not exist. $p\in B\implies (A,B)\prec p\implies p$ is an upper bound of $S \implies p\in S_1 \implies \min S_1 \preccurlyeq p$. This is a contradiction $\implies \min S_1\in A \implies\min S_1 \prec (A,B) \implies \min S_1 \prec u$.

First, it is important deal with the case $u=\min S_1$, second, you again working way to hard:

If $u\mathfrak C$ is an upper bound, then either $u=\min S_1$ or $\min S_1\preccurlyeq u$, because, by $P$ being dense, there exists $q$ between $u$ and the minimum, this $q$ is clearly in $S_1$ thus it has to be greater than $\min S_1$ thus lesser than $u$.

$...\implies A\subseteq S_2 \implies (A,B) \subseteq (S_2,S_1)\implies...$

It should be

$...\implies A\subseteq S_2 \implies (A,B) \preccurlyeq (S_2,S_1)\implies...$

If $u\notin P$, $u=(A,B)\in \mathfrak C$...

You should add that you are assuming $u$ is an upper bound.

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*As far as I know at least.