Talk:Finite intersection property

Finite intersction axiom

Some definitions of compactness invoke the finite intersection axiom, related to axiom of choice by Tychonoff's theorem; the relationship is not clear to me just right now. I guess this is an article request. linas 02:16, 21 November 2005 (UTC)

Merge with centered system of sets

Yesterday I wrote an article about the definition of centered system of sets. Today I saw that it has been proposed to be merged with this article, since the two notions are the same.

I don't know how the two terms emerged. The term centered system of sets seams to be uncommon in English literature and my guess is that it originated in Russian and found it's way into English books through translations.

I am willing to do the merging. How should the merger of two definitions in mathematics go? Create a section in the destination article about the alternative naming?

Sogartar (talk) 07:42, 4 February 2010 (UTC)

There wasn't any useful information at centered system of sets, so I've redirected it and added a note at the top of this article mentioning the definition—so perhaps technically a merge, but nothing was copied over. xnn (talk) 22:28, 24 February 2010 (UTC)

Grammar

Is "has non-empty intersection" proper English, or should it be "has a non-empty intersection"?

Why not the empty family?

At multiple points in the article we define or require that the family (or collection) A of sets must have at least one set in order to try to satisfy the finite intersection property. But I don't see any statements that fail to be true if that condition is removed. Occam's razor would seem to tell us that if we don't need to require that the family A be non-empty then we should remove that condition. —Quantling (talk | contribs) 18:18, 27 August 2025 (UTC)

Uncountability of perfect spaces

We show that there exists a counterexample where a space satisfying all the criteria except, say, the Hausdorff condition, can fail to be uncountable. But we haven't disproved that there might be a condition weaker than Hausdorff (perhaps the T₁ condition or something along those lines) that would be sufficient as a replacement for the Hausdorff condition. In other words, our assertion that We cannot eliminate the Hausdorff condition isn't proved by the existence of our counterexample. Can we improve the article language to remove that ambiguity? ... and likewise with the counterexamples provided for the other conditions of the theorem. —Quantling (talk | contribs) 23:30, 2 December 2025 (UTC)

For a $T_{1}$ counterexample consider $\mathbb {N}$ with the cofinite topology. Jean Abou Samra (talk) 00:08, 3 December 2025 (UTC)

Right -- I wasn't suggesting that T₁ would do as a substitute for Hausdorff / T₂. I am suggesting that the existence of a non-Hausdorff counterexample does not prove that the Hausdorff condition within this theorem can't be weakened.

If happens to be that we can't prove in any sensible way that the Hausdorff condition within this theorem can't be weakened then let's remove the sentence that says We cannot eliminate the Hausdorff condition and replace it with something that is unambiguously true. —Quantling (talk | contribs) 20:38, 3 December 2025 (UTC)

In my eyes, "We cannot eliminate the Hausdorff condition" is unambiguous. It just means that the same theorem with the word "Hausdorff" removed is false. But yes, it could be nice to put examples with weaker separation conditions. Jean Abou Samra (talk) 20:56, 3 December 2025 (UTC)

Maybe it is easier to see if we turn the theorem around a little bit. The theorem is exactly equivalent to Let X be a non-empty, countable, compact space that satisfies the property that no one-point set is open. Then X is not Hausdorff. (I "went contrapositive" with the Hausdorff and uncountable conditions.)

In this formulation, it begs the question, what else can we say about X? Sure, it is not Hausdorff, but is there anything more we can say about it? The answer is yes, there are some non-Hausdorff spaces that X can't turn out to be; we can conclude more than non-Hausdorff.

So, that tells us that the Hausdorf condition of the original theorem statement is not as weak as it could be. It is my fear that some readers might consider that truth to be inconsistent with the statement that We cannot eliminate the Huasdorff condition. —Quantling (talk | contribs) 21:47, 3 December 2025 (UTC)