Re: Proposal: 6NF

From: JOG <jog_at_cs.nott.ac.uk>
Date: 19 Oct 2006 18:08:24 -0700
Message-ID: <1161306504.907916.306400_at_k70g2000cwa.googlegroups.com>


Thanks for the clarification guys. My confusion in this discussion stems from this quote on set closure from the same site:

"Set Closure:

A set S and a binary operator * are said to exhibit closure if applying the binary operator to two elements S returns a value which is itself a member of S.

The closure of a set A is the smallest closed set containing A. Closed sets are closed under arbitrary intersection, so it is also the intersection of all closed sets containing A."

The first statement, from my reading, support's vc's argument wholeheartedly - a binary operator returns a value from the same set if it is to exhibit closure.However the second statement indicates that the closure of a set is the smallest closed set /containing A/ - i.e. the closure can be a superset of A. This seems to support Jan's point of view.

In relation to the example under discussion I read it as: * A = {0, 1, 2, 3, 4} and 'plusMod4' exhibit closure. * But the closure of B = {2,3} under 'plusMod4' , is the smallest /closed/ set containing B - which would be A.

There seems a very subtle point here which I am currently missing. Or perhaps the terminology is being used in two subtly different ways? Either way I am vexed, as from a certain angle you both appear to be right.

On Oct 20, 1:18 am, Gene Wirchenko <g..._at_ocis.net> wrote:
> "vc" <boston..._at_hotmail.com> wrote:
> >JOG wrote:[snip]
>
> >> As far as I understood from my school years, an operation just mapped
> >> one set of values to another. I remain unconvinced of the need for an
> >> operation upon a set to map to itself.
>
> >f:SxS -> S is not "a set to map to itself". We are talking here about The page you suggested sure seems to say that though!
>
> >very very basic stuff that's known if not from the primary school then
> >at least from the secondary school agebra (or at least should be).
>
> >>Where is such a definition? As I
> >> said, I am open to convincing, but I was not aware of such a
> >> pre-requisite for closure?
>
> >If books are not read any more nowadays, then this might help:
>
> >http://mathworld.wolfram.com/BinaryOperation.html Look at clause 2.
>
> "Binary Operation
>
> A binary operation f(x,y) is an operation that applies to two
> quantities or expressions x and y.
>
> A binary operation on a nonempty set A is a map f:AxA->A such that
>
> 1. f is defined for every pair of elements in A, and
>
> 2. f uniquely associates each pair of elements in A to some element of
> A.
>
> Examples of binary operation on A from AxA to A include addition (+),
> subtraction (-), multiplication (x) and division (÷)."
>
> Sincerely,
>
> Gene Wirchenko
>
> Computerese Irregular Verb Conjugation:
> I have preferences.
> You have biases.
> He/She has prejudices.
Received on Fri Oct 20 2006 - 03:08:24 CEST

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