The Problem With Closure

Given a triple in the set of all positive properties {p, r, s} the product in the octal of A + B = (A v B)c gives the same product in positive properties r, s with a virtue p as {p,r}+{p,s}={p,(r&s)-1}.

Now here I have a problem: for (r&s)-1immediately references elements outside of the K4 group formed by p, r and s. What is the solution to this? How may I solve for the octal at all when the operation of addition on the positive properties p, r, s are not closed? (They should form a subgroup.)

I am clearly taking the operation of A+B above in the octal group of all positive properties, but I should note that r&s is a conjunction that privates virtue in p. So, given p, our triple {p,r}+{p,s}={p,(r&s)-1} shows that (r&s) most certainly is not a positive property: In fact, it can not be a positive property in the indexing set as it privates a virtue! (And fails the definition of positive.) Then only (r&s)-1 appears as a positive property in the ultrafilter, but the set given p, of {p, r, s} remains open in the octal as I necessitate p and use (A v B)c.

What a quandary! The answer? Simply limit the indexing set for the triple {p, r, s} to the same sets of positive properties. Then it can without fault be said that the K4 group [0, p, r, s] indexes the ultrafilter of ALL positive properties in the union of p, r and s. Then as all the sets represented by singletons in the octal are disjoint I "partition" the octal into singletons, (though subgroups in triples have intersections.) However the subgroups then become acted upon by the triples in their rows from the mapping table (see Chapter 4 of the book), and the disjunction between q and p&q-1permits an ultrafilter on that K4 subgroup of the octal group.

This is quite natural; as the octal has subgroups and they are closed, as are the sets of positive properties together or in triples. It is by construction that from each row of the mapping table, the triples are the same.

So, when I limit the indexing set of p, r and s to just those three sets (closed) I have an ultrafilter and the complement of r&s in the filter is no longer present. So; the operation results in r+s=p.

In fact as (r&s)-1 is always a positive property and disjoint from p also, the third in correspondence to p and (r&s)-1 is not a combination of r or s, rather there is another subgroup altogether, {p, (r&s)-1, (u&v)}.

So, r&s given p is empty of all positivity: Then I simply note that the triple {p, r, s} is closed given p; because r&s is never positive so can not be in the principal ultrafilter over the indexing set of all positive properties! (It privates virtue in p.) Then upon limiting the indexing set I to {p, r, s} the statement that if W is in U (an ultrafilter indexing I) then one of the following is true: W is in U or I \ W is in U - this simply states that r&s is not and cannot be in W; only I \ W. Yet then the complement (r&s)-1 is then not in U if I is modified to {p,r, s} (i.e. if r&s is not in the indexing set either.)

Simply because (r&s)-1 is in the octal, this does not necessitate that r&s or (r&s)-1 are also in our modified indexing set {p, r, s}.

So, if there is any further restriction on U, it may be stated that the greatest difficulty in deriving the octal ultrafilter is showing that the K4 subgroup [0, p, r, s] is closed, and that the symbols p, r and s are "locally closed" as an indexing set with that filter: i.e. they fulfill A + B = (A v B)c becoming p+r=s etc.

As a reminder, I must operate "given p" only. (The virtue p is assumed necessary.)

Can I show that {p,r} + {p,s} = {p} in the octal?

Continue To Next Page

Return To Section Start