Does the Finite Intersection Property Allow Multiple Points in T1 Spaces?

[AKG]

Let $X$ be a space. Let $\mathcal{D}$ be a collection of subsets of $X$ that is maximal with respect to the finite intersection property. Show that if $X$ satisfies the $T_1$ axiom, there is at most one point belonging to:

$$I = \bigcap _{D\in \mathcal{D}}\bar{D}$$

A collection of subsets has the finite intersection property iff the intersection of any finite number of elements of that collection is non-empty.

A collection of subsets is maximal with respect to the finite intersection property iff any collection properly containing it does not have the finite interesection property.

A space satisfies the $T_1$ axiom iff finite point sets are closed in X.

I already know that:

a) Any finite intersection of elements of $\mathcal{D}$ is an element of $\mathcal{D}$
b) If A is a subset of X that intersects every element of $\mathcal{D}$, then A itself is an element of $\mathcal{D}$

and have proved that:

c) If $D \in \mathcal{D}$ and $A \supset D$ then $A \in \mathcal{D}$
d) $x \in \bar{D}$ for every D in $\mathcal{D}$ iff every neighbourhood of x belongs to $\mathcal{D}$.

So now suppose that I has (at least) two distinct points, x and y. I know that by the $T_1$ axiom, {x} and {y} are closed in X. By d) above, I also know that every neighbourhood of x and y are in $\mathcal{D}$, and these include X-{x} and X-{y}. If X were Hausdorff, this would be easy, because I could find a disjoint pair of neighbourhoods U and V for {x} and {y} respectively, and both of these neighbourhoods would have to be in $\mathcal{D}$ by d) but their intersection would be empty since they're disjoint, contradicting the finite intersection property of $\mathcal{D}$. Unfortunately, X is only $T_1$, not Hausdorff.

I can't figure out what to do or how to start.

 

[matt grime]

Well, if the intersection contains no points we are done, if not let x be an element in the intersection, then consider $$\mathcal{D}\cup \{x\}$$ and note via T1 that the closure of x is x and by the other hypotheses we are done. At least to the point where you can fill in the missing detail.

 

[Hurkyl]

Why do we know that $\mathcal{D} \cup \{ x \}$ satisfies the finite intersection property? Mightn't $\mathcal{D}$ contain sets that don't contain x?


When I'm stuck, I often start by looking for a counterexample, and sometimes I can figure out the proof by looking at why I'm having trouble finding a counterexample.

But I think I've found one. (But that's good, because if I figure out why it's not a counterexample, it might help me finish the problem too!)


Suppose we take the real line with a double point at the origin. (That is, take the lines y=0 and y=1 in R², and identify (x, 0) with (x, 1) whenever x is nonzero)

Then, by the axiom of choice, we can generate such a maximal set $\mathcal{D}$ by starting with the collection of sets $\{(0, 1/n) \, | \, n \in \mathbb{Z}^+ \}$

I've surely made a silly mistake, because I've convinced myself that both of the points at the origin must be in the intersection of the closures, yet this is a $T_1$ space.

 

[AKG]

The closure of (0, 1/n) is the intersection of all closed sets containing (0, 1/n), and if the topology of our space is what I think it is (it's generated by a basis consisting of the open intervals in y=1 projected into our space and the open intervals in y=0 projected into our space), then the intersection of all closed sets containing (0, 1/n) is (0, 1/n], since it would basically be

$$[0_0,\, 1/n] \cap [0_1,\, 1/n] = (0,\, 1/n]$$

This closure contains neither zero, so the intersection of all closures will contain neither zero. In fact, the intersection would be contained in:

$$\bigcap _{n \in \mathbb{Z}^+} (0,\, 1/n] = \emptyset$$

so it would be empty. Not exactly sure how this helps.

And yes,

Why do we know that $\mathcal{D} \cup \{x\}$ satisfies the finite intersection property? Mightn't $\mathcal{D}$ contain sets that don't contain x?

$\mathcal{D}$ does not in general contain sets that contain x, so matt grime's argument doesn't work.

 

[Hurkyl]

In the line with a double point at the origin, the closure of (0, 1) includes both of the points at the origin.

 

[AKG]

Hmm, looks like a counterexample to me. Suppose we generate our $\mathcal{D}$ as per your example. Suppose it contains a set D whose closure does not contain one of the zeroes. Then there is an open ball about that zero that doesn't intersect D. This ball has some radius d, so we know that (0, d) doesn't intersect D. This means that for every n such that 1/n < d, (0, 1/n) doesn't intersect D, so $\mathcal{D}$ fails to exhibit the finite intersection property, contradiction. Are you sure you made a silly mistake? Maybe my book did.

 

[matt grime]

And yes,
$\mathcal{D}$ does not in general contain sets that contain x, so matt grime's argument doesn't work.

Yep, I was trying to correct that this morning but failing

 

[AKG]

matt, did you look at Hurkyl's counterexample? Is it really a counterexample, it looks like it to me?

 

[matt grime]

Yeah, I've looked, and I can see a potential problem with it, but can't decide if it's real: appealing to zorn's lemma is fine, but to conclude anything about the behaviour of the maximal one from merely the generating set could lead to problems, though I haven't been able to think of any. (An example would be think of all finite sets in side some infinite set, by appeal to zorn;s lemma there is a maximal one, which is obviously false.)

 

[Hurkyl]

but to conclude anything about the behaviour of the maximal one from merely the generating set could lead to problems, though I haven't been able to think of any.

Yes; but I think AKG's argument is good. It was essentially the idea I had in mind when I constructed this example, but I just haven't sat down with a sheet of paper to verify it myself.

Another way of looking at it is that the fact any set in $\mathcal{D}$ must have a nonempty intersection (0, 1/n) (because (0, 1/n) is in $\mathcal{D}$), and therefore must contain a sequence of points whose limit points are the origins.

I've also been worried about the claim that this is a T₁ space (since it's the kind of thing I would expect to screw up), but I've convinced myself several times that it really is T₁.


It strikes me that this would be a good counterexample to some of the things I was wondering about when I was playing with filters... if only I could remember the questions I had!

 

[matt grime]

This is does appear false: have done a quick google for 'T1 property finite intersection' and been presented with a counter examples such as:

http://at.yorku.ca/cgi-bin/bbqa?forum=ask_a_topologist_2005;task=show_msg;msg=3168.0001

perhaps this is from the same source as yours.

The example is the cofinite topology on a countable discrete set.

I was playing around with the zariski topology on R (which is the cofinite topology as well) but not getting anywhere since that maximal problem kept coming up

Out of curiosity, what would happen there if we took all open sets which has the FIP then took the maximal one via zorn's lemma?

 

[AKG]

My book states as a lemma that given a collection $\mathcal{A}$ of subsets of some space X with the finite intersection property, there exists a collection $\mathcal{D}$ containing $\mathcal{A}$ which is maximal with respect to the finite intersection property. The proof in the book does appeal explicitly to Zorn's lemma.

but to conclude anything about the behaviour of the maximal one from merely the generating set could lead to problems

I don't see anything in my post #6 that does anything that could lead to problems.

--------

To show this space is T1, finite point sets must be closed. Let x and y be the two origins. A finite point set is some finite subset S of R - {x,y}, possibly unioned with {x} or {y}. It's complement will be the complement of S, possibly intersected with {x}^c or {y}^c. It suffices to show that S^c, {x}^c and {y}^c are open. {x}^c and {y}^c are obviously open. S^c is the union of (S^c - {x}) and (S^c - {y}). Each of these are just the projections of the complements of S in y=0 and y=1 respectively, and each of those spaces are T1, so the complement of S in each of those spaces is open, hence their projections are open in our space, hence so is their union, and so S^c is closed in our space, as desired. This might not have been the most efficient proof, I don't think that matters.

 

[Hurkyl]

Out of curiosity, what would happen there if we took all open sets which has the FIP then took the maximal one via zorn's lemma?

Oh wow; the intersection of the closures would have to be the entire space. (Since the only sets whose closure is not the entire space are finite, and no finite set can be in the maximal FIP set generated by the nonempty open sets!)


In any case, I think the question does become interesting if you make $\mathcal{D}$ a collection of closed subsets. (In the fact that it leads to interesting theorems, not because it is interesting to prove!)

 

[AKG]

In any case, I think the question does become interesting if you make $\mathcal{D}$ a collection of closed subsets. (In the fact that it leads to interesting theorems, not because it is interesting to prove!)

If $\mathcal{D}$ consisted of only closed sets, then it would only contain $\mathbb{N}$ and some collection of nonempty finite sets. We want no nontrivial open sets to be in this collection, thus for every finite subset S of $\mathb{N}$, $\mathcal{D}$ must contain a nonempty subset of S. In order to keep $\mathbb{N} - \{x\}$ out of $\mathcal{D}$, we need to have {x} in $\mathcal{D}$, so in general, {x} must be in $\mathcal{D}$ for each natural x, but then we fail to meet the finite intersection property, so no collection that is maximal with respect to the finite intersection property consists only of closed sets.

 

[Hurkyl]

If restricted to closed sets, why must $\mathcal{D}$ contain $\mathbb{N}$, or any finite set?

And you would require $\mathcal{D}$ to be maximal amongst collections of closed sets that satisfy the FIP, of course.

 

[matt grime]

I think AKG assumes you mean take this question in the specifc case of N with the cofinite topology and D a collection of closed sets.

I have some questions/observations that you might have more insight about.

What kinds of maximal collections are there? We could take D to be the collection of all sets that contain some point x. But what other things can occur? SUre we can appeal to Zorn's Lemma but that doesn't give anything that has another description that I can think of.

 

[AKG]

If it is to contain only closed sets, then the complement of any element must be an open set, and the open sets in the cofinite topology are {} and those sets with finite complement. Therefore, any element of a collection of closed sets is either the complement of {}, or the complement of the complement of a finite set. Thus any element of this collection is either N or some finite set S. Clearly, S cannot be empty otherwise the collection fails to exhibit the finite intersection property.

I don't think you can speak of a single collection that is maximal among collections of closed sets, because you can start with {1} and create a maximal collection of closed sets D such that {1} is in D and D has the finite intersection property. You could also do the same with {2}. But there is no collection that will contain both {1} and {2}.

So you want to investigate:

Let $\mathcal{D}$ be a collection of closed sets with the finite intersection property such that if $\mathcal{E} \supseteq \mathcal{D}$ is also a collection of closed sets with the finite intersection property, then $\mathcal{E} = \mathcal{D}$. What can be said about

$$\bigcap _{D \in \mathcal{D}}D?$$

Well

$$(\forall D \in \mathcal{D})((\exists E \in \mathcal{D})(E \subset D)\ \vee \ (\forall C \in \mathcal{D})(D \subseteq C))$$

Since each D is finite, and {} can't be in the collection, then the collections of the type that $\mathcal{D}$ can be enumerated by $\mathbb{N}$, where

$$\mathcal{D}_n = \{\mathbb{N}\} \cup \{S | n \in S\ \wedge \ |S| \in \mathbb{N}\}$$

 

[matt grime]

But AKG, i do'nt think Halls was referring to the cofinite topology on N at all hence his confusion.