Suslin Trees

Definition 1.  A tree is a strict poset (T,<) such that set P(x)=\{t:t<x\} of predecessors of x is well-ordered for all x\in T.

This motivates the notion of a root.  We can define the root of x as r(x)=\min\{P(x)\}.  In particular we can say an element r\in T is a root if it is the root of some element.  We define the height of x as h(x)=ord(P(x)), where ord(X) is the order-type (or ordinality) of X.  The ordinality of a toset (totally ordered set) X is the unique ordinal number to which X is isomorphic (as tosets).

Note that ord(X)\geq |X|.  For example |\omega|=|\omega+1|=\aleph_0.  But ord(\omega)\neq ord(\omega+1) since in the bijection f:\omega+1\to\omega defined by f(\omega)=0 and f(n)=n+1 for finite ordinals, we have 0\leq\omega, but f(0)=1\not\leq 0=f(\omega).

Correspondingly we define the height of T to be h(T)=\sup_{x\in T}\{h(x)\}.  An \alphatree is a tree of height \alpha.  We also define the \alphath level of a tree T as the set L_\alpha=\{x\in T:h(x)=\alpha\}.

Definition 2.  A branch in T is a maximal toset/chain in T.  That is, it is not properly contained within any chain in T.

Definition 3.  A tree T is normal if

  1. T has a unique root;
  2. each level of T is countable;
  3. if x\in L_\alpha is not maximal, then it has infinitely many successors in L_{\alpha+1};
  4. all branches have the same height;
  5. if \alpha<h(T) is a limit ordinal and x,y\in L_\alpha such that P(x)=P(y), then x=y.

Note that tosets can be endowed with a topology called an order topology.  It is generated by predecessor and successor sets.  A poset X is dense if for any comparable x,y\in X, there is a z\in X such that x<z<y.  A toset X is complete if every bounded subset has an infimum and supremum.  A toset satisfies the countable chain condition if every collection of disjoint open rays (predecessor/successor sets) is countable.

Suslin’s Problem.  Let X be a dense, complete, and unbounded toset that satisfies the countable chain condition.  Is X isomorphic to \mathbb{R}?

Note that \mathbb{R} is the unique toset which is dense, complete, unbounded and separable (has a countable dense subset).  The separability of \mathbb{R} implies that it satisfies the countable chain condition.  Since any interval in \mathbb{R} has a rational number (from denseness), then any collection of disjoint open rays must be countable.  Hence Suslin’s Problem asks if the converse is true on tosets that are dense, complete, and unbounded.  This is not provable in ZFC, ZFC+CH, ZFC+GCH, or ZFC+\lnotCH.

Definition 4.  A Suslin line is a dense, complete, and unbounded toset that satisfies the countable chain condition but is not separable.

Hence Suslin’s Problem asks whether or not a Suslin line exists.  This turns out to be equivalent to the existence of Suslin trees.  While a chain of a poset is any sub toset, an antichain is a subset in which no elements are comparable.

Definition 5.  A Suslin tree is a tree T such that h(T)=\omega_1, every branch is countable, and every antichain is countable.

Note these assumptions are not contradictory, since the height of T is not equivalent to the supremum of heights of branches (where height of a branch is defined by supremum of heights of its elements).

Lemma 6.  If there exists a Suslin tree, then there exists a normal Suslin tree.

Proposition 7.  There exists a Suslin line if and only if there exists a Suslin tree.

If SH (Suslin’s Hypothesis) is the nonexistence of a Suslin line and MA is Martin’s Axiom, then the proof of independence follows from ZFC[V=L]\vdash\lnot SH and ZFC[MA]\vdash SH where V=L is the Axiom of Constructibility.

[1] Jech, Thomas.  Set Theory.  Third Edition.  Springer Monographs in Mathematics.  Springer-Verlag.  2003.

Advertisements

Leave a Reply

Fill in your details below or click an icon to log in:

WordPress.com Logo

You are commenting using your WordPress.com account. Log Out / Change )

Twitter picture

You are commenting using your Twitter account. Log Out / Change )

Facebook photo

You are commenting using your Facebook account. Log Out / Change )

Google+ photo

You are commenting using your Google+ account. Log Out / Change )

Connecting to %s

%d bloggers like this: