### 6. First subexponential bounds.

**Proposition 1:**

**How to prove it:** This is easy to prove: Given two sets and in our family , we first find a path of the form where, and . We let with and consider the family . This is a family of -subsets of an set () . It follows that we can have a path from to in of length at most . Putting all these paths together gives us the required result. (We remind the notations at the end of this post.)

**How to use it:** It is not obvious how to use Proposition 1. Barnette’s argument from part 3 was about , and it used something a bit more sophisticated. Applying Proposition 1 directly for does not give anything non trivial. However, when is small compared to and is a small fraction of we can say something.

Let us start with an example: . let us choose . Consider a path in from two sets and . Suppose also that in this path

(*) , for every .

Let be the last set in the path whose intersection with has at least elements. Let be the last set in the path whose intersection with has at least elements. I claim that is a path in .

To see this note that must contain new elements not already in . Next must contain at least d/2 elements not already in and . Together the three sets must therefore contain at least elements. This means that their union has at least elements, hence their union contains at least elements from and by (*) and share at least elements. **Sababa.**

This argument extends to the following proposition:

**Proposition 2:** .

**So what? ****How to use these propositions:** Remember that the bound to beat was (actually, Larman improved it to , but in any case, it is exponential in .) Applying the two propositions and the trivial bound we can get

**Proposition 3:** .

**Can we automatize it?**Let me return to the question of whether such arguments can be automatized. The above argument for Proposition 2 was simple but somewhat ad hoc. You can get a slightly worse upper bound by noting that if is a family of -subsets of an n-set and , then does not contain an independent set of size . Just use the fact that . This implies that its diameter is not larger than . The bound on the independence set based on a standard estimetes for union of sets and the connection between the diameter and the independence number both look rather automatable. So is “thinking about” and proving Proposition 1 and deducing Proposition 3. (But I must admit that overall I am less optimistic about the ability to make these very elementary attacks on the problem automatic.)

**What else? **There is a little more to be said. The problem we face using Propositions 1 and 2 is that the ratio between and may deteriorate. Once is large compared with the situation is hopeless. But if we force the ratio between and to be bounded also for families we can get better (polynomial!!) bounds. I will state these bounds for polytopes keeping in mind the simple connection between the abstract problem and the diameter problem for graphs of polytopes.

**Proposition:** Let be a simple -polytope and suppose that for every face of the number of facets of is at most . Then the diameter of the graph of is at most . Here .

For it is not hard to see that has a diameter at most 2 and to then deduce that the graph of the polytope has diameter at most .

### Reminder: Our Diameter problem for families of sets and some notations and basic observations.

Consider a family of subsets of size d of the set N={1,2,…,n}.

Associate to a graph as follows: The vertices of are simply the sets in . Two vertices and are adjacent if .

For a subset let denote the subfamily of all subsets of which contain .

**MAIN ASSUMPTION**: Suppose that for every for which is not empty is **connected.**

**MAIN QUESTION: **How large can the diameter of be in terms of and .

**Let us denote the answer by . **

### More notations and a basic observation from previous parts:

Let be the family obtained from by removing the elements of A from every set. Note that . Therefore, the diameter of is at most , where and is the number of elements in the union of all the sets in .

We associated more general graphs to as follows: For an integer define as follows: The vertices of are simply the sets in . Two vertices and are adjacent if . Our original problem dealt with the case . Thus, .

Let be the maximum diameter of in terms of and , for all families of -subsets of satisfying our connectivity relations.

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