First, the set of rationals $(\mathbb{Q},<)$
is well-known to be countable. It is
linearly ordered (for any $x,y\in\mathbb{Q}$ either $x<y$ or $y<x$),
unbounded (for any $x\in\mathbb{Q}$ there is $y_{1},y_{2}\in\mathbb{Q}$ such that
$x<y_{1}$ and $y_{2}<x$) and dense (for any $x,y\in\mathbb{Q}$ if $x<y$ we
can find $z\in\mathbb{Q}$ such that $x<z<y$). It turns out that $\mathbb{Q}$ can
be characterized by these order properties:

###### Proof.

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.

We now notice that $\mathbb{R}$ is linearly ordered, unbounded, dense and has the
least upper-bound property (that is any nonempty bounded subset of $\mathbb{R}$ has
a least upper-bound). Moreover, the subset $\mathbb{Q}$ is countable and dense in
$\mathbb{R}$
(that is for any $x,y\in\mathbb{R}$ such that $x<y$ we can find $z\in\mathbb{Q}$
such that $x<z<y$). Using the previous lemma, we deduce again that these
order properties give a characterization of the set of reals:

###### Proof.

$P$ 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 $(\mathbb{R},<)$.

Now consider a totally ordered set $(R,<)$. For any $a,b$ we define
the open interval $(a,b)=\{x\in R:a<x<b\}$. Suppose
$P=\{p_{n}:n\in\mathbb{N}\}$ is a dense subset of $R$.
If ${\left((a_{i},b_{i})\right)}_{{i\in I}}$
is a family of pairwise disjoint open intervals
then we can associate to any $(a_{i},b_{i})$ the least $n_{i}\in\mathbb{N}$ such that
$p_{{n_{i}}}\in(a_{i},b_{i})$ (by density of $P$ in $R$). Since the family is
disjoint the function $i\mapsto n_{i}$ obtained is one-to-one and so the
family is at most countable. One naturally wonders what happens if we
replace in theorem 0.1
the existence of a countable dense subset by this
weaker property on open intervals:

We have seen how this problem arises from a natural generalization of
a characterization of $\mathbb{R}$. We note that we did not use the Axiom of Choice
in the above analysis and that the problem can be expressed using only
definitions on ordered sets. However, in order to answer Suslin’s Problem
we will need to introduce more concepts. We will assume familiarity with
basic notions of Set Theory like ordinals, cardinals or the Axiom of Choice.
The first five chapters of Thomas Jech’s book ‘‘Set Theory’’ should be enough.
In addition, we will rely on the axiom $\mathrm{MA}_{{\aleph_{1}}}$ and on the
Diamond Principle $\Diamond$, that we define here:

###### Proof.

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.

###### Proof.

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.

The $\Diamond$ principle holds in the model $L$ of constructible sets.
Using iterated forcing, we can construct a model of ZFC in which
$\mathrm{MA}_{{\aleph_{1}}}$ holds. Using theorems
0.2 and 0.3 we deduce that both the positive
and negative answers to Suslin’s problem are consistent with ZFC and
so Suslin’s problem is undecidable in ZFC. It’s remarkable that
a problem that only involves linearly ordered set can be solved using
sophisticated methods from Set Theory.
The above proofs follow chapters 4, 9, 15 and 16 from Thomas Jech’s book
‘‘Set Theory’’. In particular: