Re: Comments on Norbert's topological extension of relational algebra

vadimtro_at_gmail.com wrote:
> On Tuesday, December 22, 2015 at 1:49:12 AM UTC-8, Jan Hidders wrote:
>> I don't follow you here. In your construction for Heath's theorem you are taking a set of
>> polynomial equations and in them you substitute a variable with a function over the other
>> variables. Are you now claiming that there is always an equivalent set of polynomial equations
>> for this new system, no matter what that function is?
>>
>
> When we substitute a polynomial expression into polynomial expression we get yet another
> polynomial expression. Initially we had a system over x,y,z. We have eliminated y, and have
> gotten the system over x,z. It corresponds to projection of original relation

To get your point it would be helpful to write down the polynomil expression and to /explicitly/ specify the transformation rules between polynomial expressions and their "corresponding" relations.

> [x y z]
> 1 1 1
> 2 1 1
> 3 2 1
> 3 2 2

What does that mean?

   1x + 1y + 1z      (1 1 1)
+ 2x + 1y + 1z      (2 1 1)
+ 3x + 2y + 1z      (3 2 1)
+ 3x + 2y + 2z      (3 2 2)

or else

   x^1 + y^1 + z^1      (1 1 1)
+ x^2 + y^1 + z^1      (2 1 1)
+ x^3 + y^2 + z^1      (3 2 1)
+ x^3 + y^2 + z^2      (3 2 2)

   (x^1 + y^1 + z^1)      (1 1 1)
* (x^2 + y^1 + z^1)      (2 1 1)
* (x^3 + y^2 + z^1)      (3 2 1)
* (x^3 + y^2 + z^2)      (3 2 2)

or something completetly different?

> The second system consists of the two equations, the one constraining the domain of x, and
> another is functional dependency itself (in explicit analytical form). It corresponds to
> relation/finite variety
>
> x=1,y=1 x=2,y=1 x=3,y=2

So merely a set of three discrete points? What is so interesting about storing a finite point set S \subset R^n (or Q^n, or double^n) into an n-ary relation? Received on Wed Dec 23 2015 - 11:49:19 CET