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

From: Norbert_Paul <norbertpauls_spambin_at_yahoo.com>
Date: Thu, 17 Dec 2015 12:53:29 +0100
Message-ID: <n4u7j2$tvc$1_at_dont-email.me>

Jan Hidders wrote:
>>>> True or false?
>>>> Proposition. Let σ : 2^X → 2^X be a map (where 2^X is a powerset of X). Then, the following statements are equivalent.
>>>> (1) σ is topological closure operator on X,
>>>> (2) σ satisfies Kuratovski axiom, and there is a Galois connection (K, L) which satisfies σ = LK.
>>>
>>> False, I think. The identity would satisfy always (2) but not necessarily (1).
>> I do not know much about Galois connections, but if identity can be constructed from a
>> Galois conenction in the above manner then it is not a counter-example.
>> Identity is the closure operator of the discrete topological space.
>
>
> Ow, had not realised that. So on any discrete topology, the identity is a closure operator? And in fact the only closure operator?
>
> Jan Hidders

Yes, each topological space has its unique closure operator. Different topologies give different closure operators.

The discrete space: https://nl.wikipedia.org/wiki/Discrete_ruimte When you have an arbitrary set X, then the power set PX is a topology for X:

   Axiom 1: {} in PX, and X in PX
   Axiom 2: Every A, B in PX satisfies A \cap B \in PX.
   Axiom 3: For every subset S of PX the union \bigcap_{A \in S} S is an element of PX.

The space (X, PX) is called the discrete space of X.

The closure operator of the discrete space is

   cl(A) = A for all A subset X ,
hence the identity function id: PX -> PX .

Proof (easy):

   Let p be an arbitrary point in cl A. (Assumption)    Then every open set Up that contains p as an

     element has a non-empty intersection:
     Up \cap A \neq {}                  (Definition of closure)
   Up = {p} is an element of PX         (Definition of PX, Instanciation of Up)
   Hence {p} \cap A \neq {}             (Consequence)
   Hence p \in {p} \cap A               (YA Consequence)

   Hence p in A.
   Therefore cl A \subseteq A
   As A \subseteq cl A always holds we have:

   cl A = A . [qed]

If the closure operator is not the identity function its topological space is not the discrete space.

Other operators of the discrete space:

   Interior: int A = A for all A \subseteq X.    Boundary: bd A = {} for all A \subseteq X.

Norbert Received on Thu Dec 17 2015 - 12:53:29 CET