Talk:Kuratowski closure axioms

From Wikipedia, the free encyclopedia

You have new messages (last change).

How do you get axiom 4 (empty set is fixed point) from the single statement you have (pres. of finitary unions)? I don't see it. Unless you didn't mean that it was equivalent, that's what it sounds like. Revolver

It would be preservation of a nullary union. Not a big deal.

It just means that axioms 3) and 4) together are equivalent to
\forall n: c(A_{1} \cup \cdots \cup A_{n}) = c(A_{1}) \cup \cdots \cup c(A_{n}) \!.
One way it is just induction. The other way round you apply the formula with n = 0 and :n = 2 respectively. Cthulhu.mythos 08:30, 16 May 2006 (UTC)

Charles Matthews 19:18, 1 Dec 2003 (UTC)

I changed this back to the plural article name, for basically the same reasons that the article is called Orthogonal polynomials, not "orthogonal polynomial", i.e. it doesn't make sense to talk about a "kuratowski closure axiom" simply by itself, it is the SET of all 4 axioms together that constitute the object we're talking about, just one or two of the axioms by themselves is not the same thing. Revolver 19:03, 2 Feb 2004 (UTC)

was first restricted to T1 not T2 (Hausdorff).

Moore requires 3) monotone this gives

c(A) \cup c(B) \subseteq c(A \cup B)\!;

[edit] Rewrote article

I rewrote and extended the article to make clear how other basic topological notions (like continuous function) can be defined using only the closure operator. MathMartin 17:35, 12 Mar 2005 (UTC)

[edit] intersection of closed sets

how to prove \bigcap_{i\in I} cl(A_{i}) = cl(\bigcap_{i\in I} cl(A_{i})) ? --itaj 03:46, 12 May 2006 (UTC)

Isn't there an axiom missing? This is a property of normally defined closure operators, but I don't think you can derive it from the other axioms:
cl(\bigcap_{i\in I}A_i) \subseteq \bigcap_{i\in I}cl(A_i) 165.146.66.94 19:24, 7 June 2006 (UTC)
oh sorry, you can show it once you have monotonicity from axiom 3. Copying and pasting from [1]:
Well, for any j:
  • ∩ cl A ⊆ cl A_j (essential property of intersections)
  • cl ∩ cl A ⊆ cl cl A_j (axiom 3 implies monotonic)
  • cl ∩ cl A ⊆ cl A_j (axiom 2)
  • cl ∩ cl A ⊆ ∩ cl A (essential property of intersections)
  • cl ∩ cl A = ∩ cl A (axiom 1)
User:Melchoir 03:51, 16 May 2006 (UTC)

[edit] Moore closure

As remarked above the conditions claimed to give Moore closure are insufficient to establish monotonicity. From the definitions I find by Google search I have the impression that X can be more abstract than a set; it further appears that cl(Ø) = X (the top of the lattice). I propose to delete this part as being dubious (and in any case of dubious value). -LambiamTalk 08:48, 16 May 2006 (UTC)

Retrieved from "http://en.wikipedia.org../../../k/u/r/Talk%7EKuratowski_closure_axioms_79a4.html"