Blog de Frédéric

Let $P=\{p_{n}:n\in\mathbb{N}\}$ and $\mathbb{Q}=\{q_{n}:n\in\mathbb{N}\}$ be enumerations of $P$ and $\mathbb{Q}$. We shall construct by induction a sequence $f_{0}\subseteq f_{1}\subseteq f_{2}\subseteq...$ of functions such that for all $n\in\mathbb{N}$, $\operatorname{dom}(f_{n})\supseteq\{p_{i}:i<n\}$, $\operatorname{ran}(f_{n})\supseteq\{q_{i}:i<n\}$ and $\forall x,y\in\operatorname{dom}(f_{n}),x<y\Leftrightarrow f(x)<f(y)$. Then $f=\bigcup_{{n\in\mathbb{N}}}f_{n}$ is a function: if $(x,y),(x,z)\in f$ then there is $n$ large enough such that $(x,y),(x,z)\in f_{n}$ and since $f_{n}$ is a function $y=z$. Moreover, $\operatorname{dom}(f)=\bigcup_{{n\in\mathbb{N}}}\operatorname{dom}(f_{n})=P$ and $\operatorname{ran}(f)=\bigcup_{{n\in\mathbb{N}}}\operatorname{ran}(f_{n})=\mathbb{Q}$. Finally, for any $x,y\in P$ there is $n$ large enough such that $x,y\in\operatorname{dom}(f_{n})$ and so $x<y\Leftrightarrow f_{n}(x)<f_{n}(y)\Leftrightarrow f(x)<f(y)$. Hence $f$ is an isomorphism between $(P,<)$ and $(\mathbb{Q},<)$.

Thus let $f_{0}=\emptyset$. If $f_{n}$ is defined, we construct $f_{{n+1}}\supseteq f_{n}$ as follows. Suppose $p_{n}\notin\operatorname{dom}(f_{n})$. If $\forall i<n,p_{n}>p_{i}$ then because $\mathbb{Q}$ is unbounded we can consider the least $n_{0}$ such that $\forall i<n,f(p_{i})<q_{{n_{0}}}$ and set $f_{{n+1}}(p_{n})=q_{{n_{0}}}$. Similarly if $\forall i<n,p_{n}<p_{i}$. Otherwise, let $i_{1},i_{2}<n$ such that $p_{{i_{1}}}<p_{n}<p_{{i_{2}}}$ with $p_{{i_{1}}},p_{{i_{2}}}$ respectively the largest and smallest possible. Because $\mathbb{Q}$ is dense we can consider the least $n_{0}$ such that $f_{{n}}(p_{{i_{1}}})<q_{{n_{0}}}<f_{{n}}(p_{{i_{2}}})$ and again set $f_{{n+1}}(p_{n})=q_{{n_{0}}}$. Similarly, if $n\neq n_{0}$ we use the fact that $P$ is unbounded and dense to find $m_{0}\geq n+1$ that allows to define $f_{{n+1}}(p_{{m_{0}}})=q_{n}$ and ensures $f_{{n+1}}$ is order-preserving.

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$P$ is countable by assumption and since $P\subseteq R$ it is also linearly ordered. If $x,y\in P\subseteq R$ and $x<y$ then by density of $P$ in $R$ there is $z\in P$ such that $x<z<y$. So $P$ is actually dense. Similarly, if $x\in P\subseteq R$ then since $R$ is unbounded there is $y_{1},y_{2}\in R$ such that $y_{2}<x<y_{2}$ and again by density of $P$ in $R$ we can find $z_{1},z_{2}\in P$ such that $y_{2}<z_{2}<x<z_{1}<y_{1}$. So $P$ is unbounded. By the previous lemma, there is an isomorphism of ordered sets $f:P\rightarrow\mathbb{Q}$.

We define for all $x\in R$, $f_{\star}(x)=\sup_{{y\in P;y<x}}f(y)$. Because $P$ is dense in $R$ and $\mathbb{R}$ has the least upper bound property this is well-defined. If $x\in P$ then for all $y<x$ such that $y\in P$ we have $f(y)<f(x)$ and so $f_{\star}(x)\leq f(x)$. If $f_{\star}(x)<f(x)$ we could find (by density of $\mathbb{Q}$ in $\mathbb{R}$) an element $q\in\mathbb{Q}$ such that $f_{\star}(x)<q<f(x)$. For $p=f^{{-1}}(q)$ we get $p<x$ and so $q=f(p)\leq f_{\star}(x)$. A contradiction. So ${f_{\star}}_{{|P}}=f$.

Let $x,y\in R$ such that $x<y$. Because $f$ is increasing we get $f_{\star}(x)\leq f_{\star}(y)$. By density of $P$ in $R$ we can find $p_{1},p_{2}\in P$ such that $x<p_{1}<p_{2}<y$. Again, we get $f_{\star}(x)\leq f_{\star}(p_{1})=p_{1}$ and $p_{2}=f_{\star}(p_{2})\leq f_{\star}(y)$. Hence we actually have $f_{\star}(x)<f_{\star}(y)$. In particular, $f_{\star}$ is one-to-one.

We shall prove that $f_{\star}$ is surjective. Of course $f_{\star}(P)=f(P)=\mathbb{Q}$ so let’s consider $r\in\mathbb{R}\setminus\mathbb{Q}$. Define $x=\sup_{{q<r:q\in\mathbb{Q}}}f^{{-1}}(q)\in R$. This is well-defined because $\mathbb{Q}$ is dense in $\mathbb{R}$ and $R$ has the the least upper bound property. Then for all $q<r$ such that $q\in\mathbb{Q}$ we have $f^{{-1}}(q)\leq x$ by definition. By density of $\mathbb{Q}$ in $\mathbb{R}$ we can actually find $q^{{\prime}}\in\mathbb{Q}$ such that $q<q^{{\prime}}<r$ and so $f^{{-1}}(q)<f^{{-1}}(q^{{\prime}})\leq x$. We have $x>f^{{-1}}(q)\in P$ and so $f_{\star}(x)\geq f(f^{{-1}}(q))=q$. Hence $f_{\star}(x)\geq r$. Suppose that $r<f_{\star}(x)$ and consider (by density of $\mathbb{Q}$ in $\mathbb{R}$) some $q\in\mathbb{Q}$ such that $r<q<f_{\star}(x)$ and let $p=f^{{-1}}(q)$. If $p<x$ then there is $q^{{\prime}}\in\mathbb{Q}$, $q^{{\prime}}<r$, such that $p\leq f^{{-1}}(q^{{\prime}})$. Then $r<q=f(p)\leq q^{{\prime}}<r$ a contradiction. If instead $p\geq x$ then $f_{{\star}}(x)\leq f_{{\star}}(p)=q<f_{\star}(x)$ which is again a contradiction.

Finally, let $x,y\in R$ such that $f_{\star}(x)<f_{\star}(y)$. Because $R$ is totally ordered either $x<y$ or $y<x$. But the latter is impossible since we saw above that it implied $f_{\star}(y)<f_{\star}(x)$. Hence $f_{\star}$ is an isomorphism between $(R,<)$ and $(\mathbb{R},<)$.

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Suppose $(R,<)$ is not isomorphic to $(\mathbb{R},<)$ and in particular does not have any countable dense subset (otherwise we could apply theorem 0.1). We define closed intervals $I_{\alpha}\subseteq R$ by induction on $\alpha<\omega_{1}$. If $I_{\beta}=[a_{\beta},b_{\beta}]$ is defined for all $\beta<\alpha$ then the set $C=\{a_{\beta}:\beta<\alpha\}\cup\{b_{\beta}:\beta<\alpha\}$ is countable and thus is not dense in $R$. Then there is $a_{\alpha}<b_{\alpha}$ such that $I_{\alpha}=[a_{\alpha},b_{\alpha}]$ is disjoint from $C$. We define the set $S=\{I_{\alpha},\alpha<\omega_{1}\}$. Clearly, $|S|=\aleph_{1}$ and $(S,\subsetneq)$ is partially ordered. We note that if $\beta<\alpha<\omega_{1}$ then by construction $a_{\beta},b_{\beta}\notin I_{\alpha}$ and so either $I_{\alpha}\cap I_{\beta}=\emptyset$ or $I_{\alpha}\subsetneq I_{\beta}$. In particular, comparable is the same as compatible in $S$ and any family of pairwise incomparable/incompatible elements of $S$ is a family of pairwise disjoint intervals of $R$ so at most countable.

Let $\alpha<\omega_{1}$ and define $P_{{I_{\alpha}}}=\{I\in S:I\supsetneq I_{\alpha}\}$. Let $X\subseteq P_{{I_{\alpha}}}$ nonempty. Let $\beta$ the least ordinal such that $I_{\beta}\in X$. If $I_{\gamma}\in X$ then $\beta\leq\gamma$ and $I_{\gamma}\cap I_{\beta}\supseteq I_{\alpha}\neq\emptyset$. Hence by the previous remark $I_{\beta}\supseteq I_{\gamma}$. So $P_{{I_{\alpha}}}$ is well-ordered by $\supseteq$ and we define $o(I_{\alpha})$ the order-type of $P_{{I_{\alpha}}}$. We note that the set $P_{{I_{\alpha}}}$ can be enumerated by $I_{{\alpha_{1}}}\supsetneq I_{{\alpha_{2}}}\supsetneq...\supsetneq I_{\alpha}$ for some $\alpha_{\xi}\leq\alpha$ and so $o(I_{\alpha})<\omega_{1}$. Moreover for any $\alpha,\beta<\omega$, if $I_{{\alpha}}\supsetneq I_{{\beta}}$ then $P_{{I_{\alpha}}}$ is an initial segment of $P_{{I_{\beta}}}$ and so $o(I_{\alpha})<o(I_{\beta})$. Hence for each $\alpha<\omega_{1}$ the set $L_{\alpha}=\{I\in S:o(I)=\alpha\}$ has pairwise incomparable elements and so is at most countable.

For any $I\in S$, define $S_{I}=\{J\in S:J\subseteq I\}$ and let $T=\{I\in S:|S_{I}|=\aleph_{1}\}$. Let $\alpha<\omega_{1}$ and suppose that $L_{\alpha}\cap T=\emptyset$. Then $S=\bigcup_{{\beta<\alpha}}L_{\beta}\cup\bigcup_{{I\in L_{\alpha}}}S_{I}$ and since the $L_{\beta}$ are at most countable for $\beta<\omega_{1}$ and the $S_{I}$ are at most countable for $I\in L_{\alpha}$ we would have $S$ at most countable, a contradiction. So for each $\alpha<\omega_{1}$, there is $I\in T\cap L_{\alpha}$ and in particular $|T|=\aleph_{1}$. We note that if $I\in T$ and $J\supseteq I$ then $S_{I}\subseteq S_{J}$ and so $J\in T$. In particular, $P_{I}=\{J\in T:J\supsetneq I\}$ and thus without loss of generality we may assume that $S=T$.

For any $\alpha<\omega_{1}$ let $D_{\alpha}=\{I\in D_{\alpha}:o(I)>\alpha\}$. For any $I\in S$, $|S_{I}|=\aleph_{1}$ and $S_{I}=\left(\bigcup_{{\beta\leq\alpha}}S_{I}\cap L_{\beta}\right)\cup S_{I}\cap D_{\alpha}$. The first term is at most countable and so the second is uncountable and a fortiori nonempty. So $D_{\alpha}$ is a dense subset of $S$.

Using $\mathrm{MA}_{{\aleph_{1}}}$, we find $G\subseteq S$ a filter that intersects each $D_{\alpha}$. By definition, elements of a filter are pairwise compatible and so pairwise comparable. Let us construct by induction on $\alpha<\omega_{1}$, some sets $J_{\alpha}\in G$. If $J_{\beta}$ is constructed for any $\beta<\alpha$ then $\gamma=\sup_{{\beta<\alpha}}o(J_{\beta})<\omega_{1}$ and we can pick $J_{\alpha}\in G\cap D_{\gamma}$. We obtain a decreasing $\omega_{1}$-sequence of intervals $J_{0}\supsetneq J_{1}\supsetneq...\supsetneq J_{\alpha}\supsetneq...$. If $J_{\alpha}=[x_{\alpha},y_{\alpha}]$ then this gives an increasing sequence $x_{0}<x_{1}<x_{2}<...<x_{\alpha}<...$. The sets $(x_{\alpha},x_{{\alpha+1}})$ form an uncountable family of disjoint open intervals. A contradiction.

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Let ${(S_{\alpha})}_{{\alpha<\omega_{1}}}$ be a $\Diamond$-sequence. We first construct a partial ordering $\prec$ of $T=\omega_{1}$. We define for all $1\leq\alpha<\omega_{1}$ an ordering $(T_{\alpha},\prec)$ on initial segments $T_{\alpha}\subseteq\omega_{1}$ and obtain the ordering $\prec$ on $\omega_{1}=T=\bigcup_{{\alpha<\omega_{1}}}T_{\alpha}$.

Each $(T_{\alpha},\prec)$ will be a tree i.e. for any $x$ in the tree the set $P_{x}=\{y:y\prec x\}$ is well-ordered. As in the proof of 0.2 we can define $o(x)$ the order-type of $P_{x}$. The level $\alpha$ is the set of elements such that $o(x)=\alpha$. The height of a tree is defined as the supremium of the $o(x)+1$. In a tree, a branch is a maximal linearly ordered subset and an antichain a subset of pairwise incomparable elements. A branch is also well-ordered and so we can define its length as its order-type. $T_{\alpha}$ is constructed such that its height is $\alpha$ and for each $x\in T_{\alpha}$ there is some $y\succ x$ at each higher level less than $\alpha$.

We let $T_{1}=\{0\}$. If $\alpha$ is a limit ordinal then $(T_{\alpha},\prec)$ is the union of $(T_{\beta},\prec)$ for $\beta<\alpha$. If $\alpha=\beta+1$ is a successor ordinal, then the highest level of $T_{\alpha}$ is $L_{\beta}=\{x\in T_{\alpha}:o(x)=\beta\}$. $T_{{\alpha+1}}$ is obtained by adding $\aleph_{0}$ immediate successors to each element of $L_{\beta}$. These successors are taken from $\omega_{1}$ in a way that $T_{{\alpha+1}}$ is an initial segment of $\omega_{1}$.

Let $\alpha$ is a limit ordinal. Let $A=S_{\alpha}$ if it is a maximal antichain in $(T_{\alpha},\prec)$ and take $A$ an arbitrary maximal antichain of $T_{\alpha}$ otherwise. Then for each $t\in T_{\alpha}$ there is $a\in A$ such that either $a\prec t$ or $t\prec a$. Let $b_{t}$ be a branch that contains $a,t$. We construct $(T_{{\alpha+1}},\prec)$ by adding for each branch $b_{t}$ some $x_{{b_{t}}}$ that is greater than all the elements of $b_{t}$. We can choose $b_{t}$ in way that it contains an element of each level less than $\alpha$ and so $o(x_{{b_{t}}})=\alpha$ and the height of $T_{{\alpha+1}}$ is $\alpha+1$. We note that each $t\in T_{{\alpha+1}}$ is either in $T_{\alpha}$ (so comparable with some element of $A$) or greater than (a fortiori comparable with) one element of $A$. So $A$ is an antichain in $(T_{{\alpha+1}},\prec)$.

Now consider a maximal antichain $A\subseteq T=\omega_{1}$ and $C$ the set of ordinals $\alpha<\omega_{1}$ such that ‘‘$A\cap T_{\alpha}$ is a maximal antichain in $T_{\alpha}$ and $T_{\alpha}=\alpha$’’. Let $\alpha_{0}<\alpha_{1}<...<\alpha_{\xi}<...$ ($\xi<\gamma<\omega_{1}$) be a sequence of elements in $C$ and consider the limit ordinal $\lambda=\lim_{{\xi<\gamma}}\alpha_{\xi}$. By construction, $T_{\lambda}=\bigcup_{{\alpha<\lambda}}T_{\alpha}=\bigcup_{{\xi<\gamma}}T_{{\alpha_{\xi}}}=\bigcup_{{\xi<\gamma}}\alpha_{\xi}=\lambda$. If $x\in T_{\lambda}$, there is $\xi<\gamma$ such that $x\in T_{{\alpha_{\xi}}}$ and so $x$ is comparable with some $y\in A\cap T_{{\alpha_{\xi}}}\subseteq A\cap T_{\lambda}$. So $A\cap T_{\lambda}$ is a maximal antichain in $T_{\lambda}$. Finally $\lambda\in C$ and $C$ is closed.

We note that $T_{1}=\{0\}\geq 1$. If $T_{\alpha}\geq\alpha$ then $T_{{\alpha+1}}$ is obtained by adding at least one element at the end of the initial segment $T_{\alpha}$ and so $T_{{\alpha+1}}\geq\alpha+1$. Finally if $\lambda>0$ is limit and $T_{\alpha}\geq\alpha$ for each $\alpha<\lambda$ then $T_{\lambda}=\bigcup_{{\alpha<\lambda}}T_{\alpha}\geq\sup_{{\alpha<\lambda}}\alpha=\lambda$. Moreover by definition, each $T_{\alpha}$ is at most countable. Let’s come back to the closed set $C$ above. Let $\alpha_{0}<\omega_{1}$ be arbitrary. For each $n<\omega$, we let $\alpha_{{2n+1}}$ be the limit of the sequence $\alpha_{0}\leq T_{{\alpha_{0}}}\leq T_{{T_{{\alpha_{0}}}}}\leq T_{{T_{{T_{{\alpha_{0}}}}}}}...$. By definition, $T_{{\alpha_{{2n+1}}}}=\bigcup_{{\xi<\alpha_{{2n+1}}}}T_{\xi}=\alpha_{0}\cup T_{{\alpha_{0}}}\cup T_{{T_{{\alpha_{0}}}}}...=\alpha_{{2n+1}}$. Because $A$ is a maximal antichain in $T$, for each $x\in T_{{\alpha_{{2n+1}}}}$ we can find some $\alpha_{x}\geq\alpha_{{2n+1}}$ and $a_{x}\in A_{{\alpha_{x}}}$ that is comparable with $x$. Because $T_{{\alpha_{{2n+1}}}}$ is countable we can define $\alpha_{{2n+2}}=\sup_{{x\in T_{{\alpha_{{2n+1}}}}}}\alpha_{x}<\omega_{1}$. Then any $x\in T_{{\alpha_{{2n+1}}}}$ is comparable with some element of $a_{x}\in A\cap T_{{\alpha_{{2n+2}}}}$. Let $\lambda=\lim_{{n<\omega}}\alpha_{n}$. With the same method as to prove the fact that $C$ is closed, the equality $\lambda=\lim_{{n<\omega}}\alpha_{{2n+1}}$ shows that $T_{\lambda}=\lambda$ while the equality $\lambda=\lim_{{n<\omega}}\alpha_{{2n+2}}$ shows that $A\cap T_{\lambda}$ is a maximal antichain in $T_{\lambda}$. So $\lambda\in C$ and $C$ is closed unbounded.

Using the Diamond principle, $\{\alpha<\omega_{1}:A\cap\alpha=S_{\alpha}\}\cap C\neq\emptyset$. If $\alpha$ is in the intersection, then $S_{\alpha}=A\cap T_{\alpha}$ is a maximal antichain in $T_{\alpha}$. By construction, it is also a maximal antichain in $T_{{\alpha+1}}$. Each element of $a\in A\cap T_{{\alpha+1}}$ is at level $o(a)\leq\alpha$. Any $t\in T_{{\alpha+1}}$ is comparable with some element of $A\cap T_{{\alpha+1}}$. Moreover, by contruction any $t^{{\prime}}\in T\setminus T_{{\alpha+1}}$ has some predecessor $t\in T_{{\alpha+1}}$ at level $\alpha$ and there is $a\in A\cap T_{{\alpha+1}}$ that is comparable with $t$. Necessarily, $o(a)<o(t)=\alpha$ and so $a\prec t\preceq t^{{\prime}}$. Thus $A\cap T_{{\alpha+1}}$ is maximal in $T$ and $A=A\cap T_{{\alpha+1}}\subseteq T_{{\alpha+1}}$ is at most countable.

Let $B$ be a branch in $T$. It is clear that $B$ is nonempty. Actually it is infinite: otherwise there is some limit ordinal $\alpha>0$ such that $B\subseteq T_{\alpha}$ and by construction we can find some $y\succ\max B$ contradicting the maximality of $B$. By construction, each $x\in T$ has infinitely many successors at the next level and so these successors are pairwise incomparable. Hence if $x\in B$ then we can pick one $z_{x}\notin B$ among these succcessors. Let $x\prec y$ be two elements of $B$. Then $o(z_{x})=o(x)+1>o(y)+1=o(z_{y})$ so $z_{x}\preceq z_{y}$ is impossible. Suppose $z_{y}\prec z_{x}$. We also have $x\prec z_{x}$ by definition. If $z_{y}\preceq x$ then we would have $z_{y}\in B$ because $B$ is a branch. If $x\prec z_{y}$ we would get $o(x)<o(z_{y})=o(y)+1\leq o(x)$. Hence $z_{x},z_{y}$ are incomparable (in particular distinct) and the set $\{z_{x}:x\in B\}$ is an antichain in $T$ and so $B$ is countable.

Finally, we define $S$ the set of all branches of $T$. By construction, each $x\in T$ has countably many immediate successors and we order them as $\mathbb{Q}$. Let $B_{1},B_{2}$ be two branches and $\alpha$ the least level where they differ. The level 0 is $T_{1}=\{0\}$ so $\alpha>0$. If $\alpha$ is limit then the restriction of the branches $B_{1},B_{2}$ to $T_{\alpha}$ is the same branch $b$ and $T_{{\alpha+1}}$ has been contructed in a way that there is only one possible element at level $\alpha$ to extend this branch and this is a contradiction. So $\alpha$ is actually a successor ordinal $\beta+1$. Hence if $B_{1}(\alpha)\in B_{1}$ and $B_{2}(\alpha)\in B_{2}$ are the immediate successors of the point in $B_{1}\cap B_{2}$ at level $\beta$, we order $B_{1},B_{2}$ according to whether $B_{1}(\alpha)<B_{2}(\alpha)$ or $B_{1}(\alpha)>B_{2}(\alpha)$, using the $\mathbb{Q}$-isomorphic order we just defined. Clearly, this gives a linear ordering $<_{S}$. It is also unbounded: for any $B\in S$, if $B(1)\in B$ is the element at level 1 pick $x$ greater (or smaller) than for the $\mathbb{Q}$-isomorphic order on successors of $B(0)=0$ and consider a branch extending $0\prec x$.

Now consider two branches $B_{1}<_{S}B_{2}$. Let $\alpha=\beta+1$ be as above. We can find $x$ an immediate successor of $B_{1}(\beta)$ such that $B_{1}(\alpha)<x<B_{2}(\alpha)$ for the $\mathbb{Q}$-isomorphic order on immediate successors of $B_{1}(\beta)$. Let $I_{x}=\{B\in S:x\in B\}$. Any $B\in I_{x}$ contains $\{y\in T:y\prec x\}=\{y\in T:y\prec B_{1}(\beta)\}=\{y\in T:y\prec B_{2}(\beta)\}$. Moreover $x=B(\alpha)$ and so $B_{1}<B<B_{2}$. $I_{x}$ is nonempty (we can extend $\{y\in T:y\preceq x\}$ to a maximal branch) and so $S$ is dense. If $I_{x}\cap I_{y}=\emptyset$ then $x,y$ are incomparable. So from any collection of disjoint open intervals ${(B_{{1i}},B_{{2i}})}_{{i\in I}}$ we get an antichain $\{x_{i}:i\in I\}$ and so $I$ is at most countable.

Let $C$ be a countable set of branches in $T$. Since these branches are countable, we can find $\alpha<\omega_{1}$ larger than the length of any branches in $C$. If $x\in T$ is at level greater than $\alpha$ then for all $B\in C$, $x\notin B$ so $B\notin I_{x}$. Finally $C\cap I_{x}=\emptyset$ and $C$ is not dense.

Now let $R$ be the Dedekind-MacNeille completion of $S$. It is unbounded, linearly ordered, has the least upper-bound property and $S$ is dense in $R$. Using the fact that $S$ is dense in $R$ we deduce that $R$ is dense. Similarly, if ${(a_{i},b_{i})}_{{i\in I}}$ is any collection of disjoint open intervals in $R$ we can find $c_{i},d_{i}\in S$ such that $a_{i}<c_{i}<d_{i}<b_{i}$. Then ${(c_{i},d_{i})}_{{i\in I}}$ is a collection of disjoint open intervals in $S$ and so $I$ is at most countable.

Finally, it remains to prove that $R$ is not isomorphic to $\mathbb{R}$ and it suffices to show that $R$ does not have any countable dense subset $C$. If $C$ is a countable subset of $R$, then for any elements $a<b$ in $C$ we pick $c\in S$ such that $a<c<b$. This gives a countable subset $C^{{\prime}}$ of $S$. If $a<b$ are elements of $S$, we can find $c,d\in C$ such that $a<c<d<b$ and so $e\in C^{{\prime}}$ such that $a<c<e<d<b$. Thus $C^{{\prime}}$ would be a countable dense subset of $S$. A contradiction.

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