Linear-order completion and density

Let $\mathcal{C}(L)$ consist of the subsets $X\subseteq L$ such that (i) $b<a\in X$ implies $b\in X$, and (ii) whenever $X$ has a least upper bound $a$ in $L$, one has $a\in X$. Order $\mathcal{C}(L)$ by inclusion. These are the downward-closed cuts with every already-existing $L$-supremum closed in.

The inclusion order on $\mathcal{C}(L)$ is linear. Indeed, if $X,Y\in \mathcal{C}(L)$ and $a\in X\setminus Y$, then every $b\in Y$ satisfies $b<a$ (otherwise downward closure would put $a$ in $Y$), and hence $b\in X$; thus $Y⊊X$.

Every family $\mathcal{X}\subseteq \mathcal{C}(L)$ has a supremum. Put $U=\bigcup \mathcal{X}$. If $U$ has no least upper bound in $L$, then $U\in \mathcal{C}(L)$ and is the union-supremum. If $a={\sup }_{L}U$ exists, downward closure gives $U\cup \{a\}=(-\mathrm{\infty },a]$, which lies in $\mathcal{C}(L)$ and is the least cut above every member of $\mathcal{X}$. This includes $\mathcal{X}=\mathrm{\varnothing }$. Infima then exist as suprema of sets of lower bounds, so $\mathcal{C}(L)$ is complete.

Send $a\in L$ to $j(a)=(-\mathrm{\infty },a]$. Each $j(a)$ is a cut, and $a<b$ holds exactly when $j(a)⊊j(b)$, so $j$ is an order embedding. Replacing $L$ by its named copy $j[L]$ if necessary gives literal inclusion as required by C1.

Every cut $X$ is $\sup \{j(a):a\in X\}$, including the empty cut. If $a={\sup }_{L}A$ for $A\subseteq L$, then $j(a)$ is an upper bound of $j[A]$; any cut $Y$ above every $j(x)$ for $x\in A$ contains every $b<a$, and it contains $a$ either because $a\in A$ or because clause (ii) closes $Y$ under the supremum $a$. Hence $j(a)=\sup j[A]$. Thus the constructed order satisfies C2-C4.

Let $P$ be any completion satisfying C1-C4 and define ${\tau }_P(x)=\{a\in L:a{\le }_{P}x\}$. This is a cut: downward closure is immediate, while if $b={\sup }_L{\tau }_P(x)$, C4 makes $b$ its supremum in $P$, which also equals $x$ by C3, so $b=x\in {\tau }_P(x)$. If $x<y$, C3 gives an $a\in L$ with $x<a\le y$, so ${\tau }_P(x)⊊{\tau }_P(y)$. Conversely the traces reflect order. For every cut $X$, if $x={\sup }_{P}X$, then ${\tau }_P(x)=X$: an $a<x$ in the trace cannot upper-bound $X$, and if $a=x\in L$, then $a={\sup }_{L}X$ and cut closure puts $a$ in $X$. Thus ${\tau }_P$ is an onto isomorphism from $P$ to $\mathcal{C}(L)$ and fixes $L$. Applying this to two completions gives the unique isomorphism over $L$, since C3 forces any such isomorphism to send each ${\sup }_{P}A$ to ${\sup }_{N}A$.

Suppose now that $L$ is dense and $M$ is a completion. If $x<y$ in $M$, C3 supplies $b\in L$ with $x<b\le y$. If $x\in L$, density in $L$ gives $x<a<b$ for some $a\in L$. If $x\mathrm{\notin }L$ and there were no $a\in L$ with $x<a<b$, then $b$ would be the least upper bound in $L$ of the $L$-points below $x$; C4 would make that supremum equal both $b$ and $x$, a contradiction. Hence in all cases some $a\in L$ satisfies $x<a<y$, so $L$ is order-dense in $M$.

If $(I_{\xi }{)}_{\xi <{\omega }_1}$ were pairwise disjoint nonempty open intervals of $M$, order-density and A1 would choose $a_{\xi }<b_{\xi }$ in $L\cap I_{\xi }$. The nonempty $L$-intervals $(a_{\xi },b_{\xi })$ would remain pairwise disjoint, contradicting the corresponding hypothesis on $L$. Thus the interval ccc passes to $M$.

Suppose a nonempty open interval $I$ of $M$ had a countable dense set $D$. Choose $a<b$ in $L\cap I$. The set $D\cap (a,b)$ is nonempty and countable; list it as $(d_n{)}_{n<\omega }$, repeating entries in the finite case. For every pair $d_i<d_j$, use order-density and A1 to choose $e_{ij}\in L$ with $d_i<e_{ij}<d_j$. The set $E=\{e_{ij}:d_i<d_j\}$ is countable by diagonal enumeration of the pairs of natural indices. Given $u<v$ in $L\cap (a,b)$, density of $D$ first gives $d_i\in (u,v)$ and then $d_j\in (d_i,v)$, so $u<e_{ij}<v$. Therefore $E$ is dense in the nonempty $L$-interval $(a,b)$, contradicting the hypothesis on $L$. No nonempty open interval of $M$ is separable.

Finally assume that $L$ has no endpoints, and delete from $M$ its first and last elements when they exist. Neither deleted point belongs to $L$. The remainder $M^{\circ }$ still contains the order-dense copy of $L$, is dense and has no endpoints, and retains the conclusions of steps 6.1-6.2. If a nonempty $A\subseteq M^{\circ }$ is bounded above there, then ${\sup }_{M}A$ lies above a member of $A$ and below an upper bound in $M^{\circ }$, so it is neither deleted endpoint and belongs to $M^{\circ }$; hence $M^{\circ }$ is boundedly complete. This proves every assertion and records the precise use of AC.