# The n+2 Management Policy

I think this is a decent way to operate in a hierarchy–particularly when it comes to big decisions like hiring/firing/approving new policies. We first assume the hierarchy is a tree (i.e. each member has only one immediate supervisor) and that there is a maximal element, called an “administrator”. The setup is that if something happens at level in the hierarchy (say hiring/firing someone at level (presuming the action is consistent with other policy)), then the relevant supervisor makes the nomination/recommendation for the action, and his/her supervisor confirms the action.

Hence let be a tree, and denote the supervisor of Typically in trees, minimal elements are considered level Here we reverse the ordering and call the maximal element the **administrator**, denoted by and say We define a **rank** **policy** as a policy that affects all successors (subordinates) of an element such that Hence we have:

**The n+2 Operational Policy**. Let be a rank policy and denote the member whose subordinates are affected. The policy then becomes activated provided *proposes* it and *approves* it.

Hence the Operational Policy itself is a rank policy as it affects everything in the hierarchy. But we arrive at a problem in continuing to execute this policy at levels and So we adjoin another set to the hierarchy called the **board**, denoted and redefine our hierarchy as where denotes the initial tree with unique maximal element We then define and

The main positive of this method is the minimizing of micromanagement. The only big negative that stands out is the possibility of actions diverging from intentions of a level member as one goes down the ladder. But I’d argue that this just means more level policy needs to be implemented in order to prevent that.

# The Calculus Tree

We first start with a tree structure on the set of real coefficient polynomials with domain and codomain , and call it the **polynomial tree**. We will say iff for some

**Proposition 1.** This gives us a partial order on .

*Proof.* since If and then and thus so And lastly if and then and Thus which implies and thus so

The fact that differentiation yields a unique function gives us the tree structure: the chain of derivatives (predecessors) is well ordered with minimal element By the Weierstrass approximation theorem we can attempt to extend this to the closure of the class of smooth functions with compact support. We extend our indexing set to the ordinal and define the th (and th) derivative linearly by the rule

Recall integration is a set valued operation–sending a function to its set of antiderivatives which all differ by some constant.

**Example 2.** Let We have

Hence for finite since the “last term” is and If we begin to integrate, we will add polynomials trailing after Continuing this denumerably we then obtain

where is the conumber of in the sense that it satisfies in the th term of the th partial sum. The denumerable integration converges if the sequence of constants chosen satisfies

Suppose the sequence satisfies this condition and let us assume the sum converges to some then we can write

In particular we thus have since

We thus have and We also define the **degree** of the functions which are not polynomials as Let be a polynomial with degree and have infinite degree. Then we clearly have that and have degree

**Definition 3.** A function is **cyclic** if for some nonzero finite number The number is called the **differential order** of and is denoted If a function is not cyclic we say (note in this definition ).

**Example 4**. The functions and are cyclic with orders and respectively. The only function with finite degree which is cyclic is with order

In fact being cyclic occurs iff

for at least some where and Thus we have

With this presentation it is clear that the order of the th term in the above term is if Otherwise the order is

Let us now define the equivalence relation where iff and are cyclic with same order and where such that Reflexivity is clear, and symmetry/transitivity involve simple arithmetic.

**Example 5**. Hence some equivalence classes are and for any polynomial

**Proposition 6**. gives a tree structure on which we call the **calculus tree**.

*Proof.* This is trivial considering the relation we modded out by was precisely loops in the poset–together with the fact that the poset has a unique minimal element.

Also see update.

# Suslin Trees

**Definition 1.** A **tree** is a strict poset such that set of predecessors of is well-ordered for all .

This motivates the notion of a root. We can define the **root of ** as . In particular we can say an element is a **root** if it is the root of some element. We define the **height** of as , where is the order-type (or ordinality) of . The ordinality of a toset (totally ordered set) is the unique ordinal number to which is isomorphic (as tosets).

Note that . For example . But since in the bijection defined by and for finite ordinals, we have , but .

Correspondingly we define the **height** of to be . An –**tree** is a tree of height . We also define the **th level** of a tree as the set .

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

**Definition 3.** A tree is **normal** if

- has a unique root;
- each level of is countable;
- if is not maximal, then it has infinitely many successors in ;
- all branches have the same height;
- if is a limit ordinal and such that , then .

Note that tosets can be endowed with a topology called an **order topology**. It is generated by predecessor and successor sets. A poset is **dense** if for any comparable , there is a such that . A toset 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 be a dense, complete, and unbounded toset that satisfies the countable chain condition. Is isomorphic to ?

Note that is the unique toset which is dense, complete, unbounded and separable (has a countable dense subset). The separability of implies that it satisfies the countable chain condition. Since any interval in 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+CH.

**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 such that , every branch is countable, and every antichain is countable.

Note these assumptions are not contradictory, since the height of 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 and where is the Axiom of Constructibility.

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