combinatorial uniqueness of Hesse Configuration
If is an abstract Hesse configuration, then has 12 elements.
If is an abstract Hesse configuration then, for every , there exist exactly four elements such that .
To every such that , there exists exactly one such that and . Furthermore, for every such that , there will be exactly two elements of other than . Hence, there exist elements such that . ∎
If is an abstract Hesse configuration and , then there exist such that .
By the foregoing result, given , there are four elements of to which belongs. One of these, of course, is itself, and the other three are distinct from . Since has three elements, this means that there are at most elements such that . Because has 12 elements. there must exist such that .
It remains to show that . Suppose to the contrary that there exists a such that and . Since , it follows that , hence there will exist three distinct elements of containing and an element of . Because , these three elements are distinct from and . That makes for a total of five distinct elements of containing , which contradicts the previous theorem, hence . ∎
If are elements of such that and , then has exactly one element in common with each of .
Since each element of is a subset of with three elements and are pairwise disjoint but only has nine elements, it follows that every element of must belong to exactly one of . In particular, this means that every element of must belong to one of . Were two elements of to belong to the same element of then, by the third defining property, that element would have to equal , contrary to its definition. Hence, each element of must belong to a distinct element of . ∎
If is an abstract Hesse configuration, then we can label the elements of as A,B,C,D,E,F,G,H,I in such a way that the elements of are
By theorem 3, there exist such that . Since has twelve elements, there must exist an a elemetn of distinct from . Pick such an element and call it . By another application of theorem 3, there must exist such that .
By theorem 4, must have exactly one element in common with each of ; let the element it has in common with , be the element it has in common with and be the element it has in common with . Likewise, must have exactly one element in common with each of , as must . Let be the element has in common with , be the element has in common with , be the element has in common with , be the element has in common with , be the element has in common with and be the element has in common with .
Summarizing what we just said another way, we have assigned labels to the elements of in such a way that
That is half of what we set out to do; we must still label the remaining six elements of .
By theorem 4, if , then must have exactly one element in common with each of and exactly one element in common with each of .
Suppose that . It could not be the case that because then would have two elements in common with . Since must have one element in common with , that means that either or . If , then the element has in common with cannot be because would have both and in common with and it cannot be because and would have both and in common, hence the only possibility is to have , i.e. . Likewise, if , it follows that .
Summarrizing the last few sentences, if , then either or . By a similar line of reasoning, if , then either or and, if , then either or . Since must contain one of , it follows that there are omnly the following six possibilities for :
However, since has cardinality six, all these possibilities must be actual members of the set. ∎
|Title||combinatorial uniqueness of Hesse Configuration|
|Date of creation||2014-02-28 16:18:25|
|Last modified on||2014-02-28 16:18:25|
|Last modified by||rspuzio (6075)|