Re: 14 sets, 14 modalities, epistemic logic, Cantor spaces and an Euler tour

From: William Elliot (marsh_at_hevanet.remove.com)
Date: 03/29/05 

Date: Mon, 28 Mar 2005 22:38:00 -0800

On Tue, 29 Mar 2005, mitch wrote:

> One solution to Kuratowski's 14 set problem is obtained
> using an isolated point, a punctuated half-open interval,
> and an interval of rationales,
>
> x o----o----x x****x
>
> The 14 sets are then obtained as shown
>
> ~cl(~cl(~cl(X))) <---------o o----o o---->
> cl(~cl(~cl(X)) x---------x x----x
> ~cl(~cl(X)) o---------o o----o
> cl(~cl(X)) <---------x x----x x---->
> ~cl(X) <----o----o o----o o---->
> cl(X) x x---------x x----x
> X x o----o----x x****x
> ~X <----o----x x o----o####o---->
> cl(~X) <---------x x x-------------->
> ~cl(~X) o----o----o
> cl(~cl(~X)) x---------x
> ~cl(~cl(~X)) <---------o o-------------->
> cl(~cl(~cl(~X))) <---------x x-------------->
> ~cl(~cl(~cl(~X))) o---------o
>
Oh groan, a huge accounting problem. Instead
let's use regular open and regular closed sets.
A set U is regular open when int cl U = U
A set K is regular closed when cl int K = K

Now with complementation and regular closure, ie the
operator cl int, how many different sets can be produced?

> As previously noted, for the given set, one direction
> yields 6 sets and the other direction yields 7 sets.

> One may observe that the capacity to generate distinct
> sets fails in both directions in the same way. There is
> an "invariant" four cycle,
>
> Note that my use of "A" and "B" here is not a naming function.
>
[snip][snip][snip][snip][snip][snip][snip][snip]

Oh no, a long lengthy prolonged stream of thought soliloquy. Rather that
keep us in suspenseful boredom, get to the shaggy dog punch line. Give an
abstract, ie 300 words or less, stating the important results you've made
or the most significant vein you've uncovered for exploration.

Now an operator that may amuse you is the boundary operator with
        bd bd bd A = bd bd A
How many different sets can be produced with the boundary operator
and complementation? Hint, find a set for which bd bd A /= bd A.
BTW, such a set is neither open nor closed.

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