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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

${\displaystyle \forall n:c(A_{1}\cup \cdots \cup A_{n})=c(A_{1})\cup \cdots \cup c(A_{n})\!}$.

One way it is just induction. The other way round you apply the formula with ${\displaystyle n=0}$ and :${\displaystyle n=2}$ respectively. Cthulhu.mythos 08:30, 16 May 2006 (UTC)[reply]

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

${\displaystyle c(A)\cup c(B)\subseteq c(A\cup B)\!}$;

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)

how to prove ${\displaystyle \bigcap _{i\in I}cl(A_{i})=cl(\bigcap _{i\in I}cl(A_{i}))}$ ? --itaj 03:46, 12 May 2006 (UTC)[reply]

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:

${\displaystyle cl(\bigcap _{i\in I}A_{i})\subseteq \bigcap _{i\in I}cl(A_{i})}$ 165.146.66.94 19:24, 7 June 2006 (UTC)[reply]

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)[reply]

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). -Lambiam^Talk 08:48, 16 May 2006 (UTC)[reply]

"The space ${\displaystyle X}$ is connected if no two subsets are separated" is wrong. I can easily find two separated subsets of ${\displaystyle \mathbb {R} }$. — Preceding unsigned comment added by 223.229.158.209 (talk) 04:44, 22 August 2019 (UTC)[reply]

Fixed.129.234.21.188 (talk) 15:52, 31 October 2019 (UTC)[reply]

The discussion of weakenings and alternative axioms seems somewhat poorly construed. For example

"Because of extensivity [K2], it is possible to weaken the equality in [K3] to a simple inclusion" suggests that one could replace [K3] by "for all ${\displaystyle A\subseteq X}$, ${\displaystyle \mathbf {c} (A)\subseteq \mathbf {c} (\mathbf {c} (A))}$". Of course this is merely a consequence of [K2], and the 'simple inclusion' meant is ${\displaystyle \mathbf {c} (A)\supseteq \mathbf {c} (\mathbf {c} (A))}$. Is anything gained by changing the equality to an inequality is this way? I have removed this.

The sentence "It is straightforward to show that weakening the equality in [K4] to a simple inclusion (subadditivity) and assuming [K4'] yields back to [K4]" is also somewhat tortuous. Given what we have just seen, in what direction can we imagine our 'simple inclusion' to run? I have rephrased but it is perhaps still not perfect?129.234.21.188 (talk) 17:18, 31 October 2019 (UTC)[reply]

The article of Monteiro (and thus the axiom [M]) is not about Kuratowski closure, it is about Moore closure (no preservation of unions), although the article calls it Sierpinski closure.

There is a remark about the axiom [BT] (which does axiomatise Kuratowski closure, given that the empty set is a fixed point) in the article. It is remarked that it would axiomatise Moore closure if the equality in [BT] is replaced by an inclusion. 164.15.254.98 (talk) 13:32, 12 June 2023 (UTC)[reply]

This page doesn't render well in dark mode https://en.wikipedia.org/wiki/Kuratowski_closure_axioms?vectornightmode=1&useskin=vector-2022#Induction_of_topology_from_closure

All math equations in the collapsible box are not visible. I suggest using a new template or working with the existing template editors to make it compatible with sark mode.

mw:Recommendations for night mode compatibility on Wikimedia wikis should have all the information you need to get this working. Let me know if I can help in any waY! Jon (WMF) (talk) 18:10, 19 July 2024 (UTC)[reply]