A Mathematician’s Scratchpad

Woo hoo. Blog is back.

Here’s a new silly proof I spotted recently.

Theorem: Let be continuous. Then is eventually constant.

Proof:

This proof assumes the that . The result doesn’t actually need this though, which is one of the main reasons this proof is silly.

So, is continuous. is countable compact, thus so is . But is a metric space, so countably compact subsets are compact. But every compact subspace of has cardinality or . We know that , so it can’t be . Hence it has cardinality . By the pigeon hole principle we must have being constant on some uncountable set. But for any the set is closed, with these being disjoint for distinct values of . You can’t have disjoint uncountable closed sets in , so all but one of these sets must be countable. Thus take an upper bound for all the countable subsets, say . is constant on .

I can’t actually take much credit for this. I made the initial conjecture that sparked this theorem, but my friend John Bytheway was the first person to codify the actual theorem and prove it. My proof of it is pretty independent of his, but I probably wouldn’t have come up with it unless I already knew the theorem was true. (The neccesary part of the theorem is stolen from him, but the sufficiency part is entirely mine).

Note: In this post I take the slightly unconventional approach that a Normal space needn’t be . i.e. a Normal topological space is one in which any two disjoint closed sets have disjoint neighbourhoods, but points needn’t be closed. I could invent a new term for this, like quasinormal, but I really can’t be bothered. Feel free to replace ‘normal’ with ‘quasinormal’ wherever you see it if this makes you more comfortable. It may be readily verified that the usual proof of Urysohn’s lemma in no way depends on the closedness of points, so Urysohn’s lemma holds for this new definition of normal.

First we need some preliminary definitions.

Definition 1:

Let be a topological space and a function, not neccesarily continuous. For we define the oscillation of at to be

where ranges over open sets.

We define the total oscillation of to be

We now prove some preliminary results about

Proposition 2:

Let be an arbitrary topological space and consider the Banach space of bounded functions on and the closed subspace consisting of the continuous bounded functions. Consider

1. iff is continuous.
2. 
3. 
4. 
5. 

These are all perfectly trivial to prove, so I’m not going to bother.

What I noticed is that in almost every case I considered the final inequality was in fact an equality. I could prove a weaker result for Compact hausdorff spaces (I showed that , but I couldn’t do any better, so I passed this over to John to see what he could come up with. He came up with the following theorem, which is the main theorem of this post:

for every iff is normal.

I’ll prove this in two parts. For the right to left implication I’ll in fact prove something stronger:

Theorem 4:

Let be a normal topological space and let . Then there is a (not usually unique) continuous function with .

As is my wont, the proof of this will precede by a couple clever definitions and then drop out as practically a corollary of a Big Theorem.

The big theorem in question is the following, which is due to Tong. I conjectured it, tinkered around with proving it for a bit, went to look up something about semicontinuous functions in Engelking’s general topology book and saw the theorem staring out at me from one of the exercises.

Theorem 5:

Let be a normal topological space. Let be upper semicontinuous and lower semicontinuous respectively with . There is a continuous function with .

Note the direction of the inequality and which are upper and lower semicontinuous respectively. If you reverse this then the theorem becomes false.

I’m not actually going to prove this here - I’ve not yet totally sorted out the details of the proof in my mind. It’s basically a modified version of the standard proof of Urysohn’s lemma, but with additional constraints on the sets constructed. (Update: See here for a proof.)

Anyway, we now make some more definitions:

Definition 6:

Let . Define

Proposition 7:

These satisfying the following properties:

1. 
2. is lower semicontinuous. is upper semicontinuous.
3. If is upper semicontinuous then . Similary if is lower semicontinuous then .
4. The maps and are monotone with respect to the pointwise ordering.
5. If are lower, upper semicontinuous respectively and then
6. 

Again, these are all really very easy to prove (assuming you prove them in order), so I’m not going to do it. I’ll actually not use most of these, but those that I don’t use are of independent interest. i.e. they’re cool.
Now, we have:

Proof of theorem 4:

Note that . Consequently we have that

Now, the left hand side is upper semicontinuous and the right hand side is lower semicontinuous. Thus we have an interpolating continuous function .

So

But we have that and .

So

i.e.
But we know that by our very first proposition. Hence we have equality.

QED

Lemma 8:

If for every we have then is normal.

Proof:

This proof is entirely John’s.

Let be disjoint closed sets. Define by , and otherwise.

By considering appropriate neighbourhoods we may see that for every . i.e. . Consequently we have and may find a continuous function with .

Thus we have that and . Composing with some appropriate function from we get a continuous function which separates the two closed sets. Then and are appropriate disjoint neighbourhoods of and .

QED

Here begins a new series, akin to the “Silly Proofs” series. Obscure results which are cool, but which you probably haven’t heard of.

Suppose we’ve got a pair of measurable spaces (sets with a sigma algebra on them) . We make by taking the sigma algebra generated by sets of the form . In the case where we have topologies on and and are giving them their Borel algebras, we might suppose this agrees with the Borel algebra of the product. Alas, ’tis not so! It does in the second countable case, but in general not:

Theorem (Nedoma’s Pathology): Let be a measurable space with . Then the product algebra on does not contain the diagonal.

In particular, if is Hausdorff then the diagonal is a closed set in the product topology which is not contained in the product algebra.

The proof proceeds by way of two lemmas:

Lemma: Let be a set, and . There exist with

Proof: The set of satisfying the conclusion of the theorem is a algebra containing .

Lemma: Let be measurable. Then is the union of at most sets of the form .

Proof:

By the preceding lemma we can find with

For , define where if and otherwise.

Sets of the form then form a partition of of . Thus the sets which can be written as a union of sets of the form form a algebra. This contains each of the , and so contains . Thus . There are at most sets in this union. Hence the desired result.

Finally we have the proof of the theorem:

Let be the diagonal. Suppose is measurable. Then is the union of at most sets of the form . Because at least one of these sets must contain two points. Say and . But then it also contains . This is a contradiction.

QED

To be honest, this theorem doesn’t seem that useful to me. But knowing about it lets you avoid a potential pitfall - when you’re dealing with measures on large spaces (e.g. on , which it’s really important to be able to do when you’re playing with certain forcing constructions) things are significantly less well behaved than you might hope.

I went to an interesting maths conference this week (Set theory and its neighbours meets the Cameleon), and I’m probably going to write a report on it at some point. This isn’t that report.

This is just a quick note to say that the stuff about Boolean algebras and operator algebras is at least known, if not well known, and has been for a good few decades. Oh well. The noncommutative stuff probably hasn’t - I’m going to email the guy whose talk was on a related subject and ask him about it.

David

This one is much more speculative, as I still haven’t worked out the details. I think I worked out a few more back when I was actively investigating this stuff, but if so then I’ve forgotten them all.

The question is, what are the natural analogue of Boolean algebras in the noncommutative case?

The answer is “Err, well, I’m not quite sure!”

I’ve tried to see how far I can extend the construction to the noncommutative case and see what happens. The results are interesting, but perhaps not very informative.
First of all, we can’t just look at arbitrary idempotents. These turn out to not be very well behaved. Instead we’ll look at self-adjoint idempotents - the projections. (In the commutative case every idempotent is self adjoint). We’ll call the set of these as before.
Second of all, we run into a problem. The product of two projections is only a projection if they commute. So, in the noncommutative case, we don’t have the multiplication operation making it into a Boolean algebra. We can however still make it into a partial order:

Define if .

Thie definition may seem like there should also be a right-multiplication version, but in fact there isn’t. If then (one of the many reasons it was important that these things are self adjoint).

So, if and then . So the relation is antisymmetric. It’s evidently transitive and reflexive, so we have a partial order.

It has top and bottom elements, and .

We have an operator which maps into itself. It has the following properties:

• It’s order reversing.
• 
• 
• 

Note that it is not determined by these latter two properties. is not a distributive lattice!

Is even a lattice? I don’t know. It looks like it.

In the case of we have is isomorphic to the lattice of closed subspaces of , with being orthogonal complement and being . This is a lattice, with and being the closed linear span of .

Further, fix . Then for . So one might expect that . Unfortunately this isn’t true. Usual problem with exchanging limits. It’s true in some sort of weak topology, so if can be represented as a closed subspace of in this topology then this limit is in and we have the desired result. I imagine this only works if is closed in the strong operator topology. i.e. is a von Neumann algebra. I’ll need to learn more about these before I can say for sure.
I’m not yet really sure if this works. I doubt it, and even if it does then it’s probably far too much work and there’s a better proof.

Some negative results:

The norm is not uniquely determined on the projections. Rather, every projection has norm , but projections can have taking any value in .

Also, these posets are very very large. In the simplest example, take . Then is an antichain of size plus a top and bottom element.

Note this is also an example of how badly complementation fails to be unique. For any distinct with we have and .

I’ll post more as I think of it.

I’m going to be gradually migrating posts over from the old version of the site ( http://mathsscratchpad.blogspot.com ), and setting them to their old time stamps. So if you’re wondering why the archives are gradually filling up, that’s the reason.

This isn’t going to happen that quickly though, because upgrading them to LaTeX is really tedious.

This is something I noticed a while ago, and I don’t seem to be going anywhere with it recently, so I thought I’d blog about it. It’s a cool factoid which may be new (At least one person has muttered something along the lines of ‘I vaguely recall having heard about something like that’, but was unable to give any specifics), but hasn’t really got enough content to be publishable.

What motivated the question was the following:

We have full and faithful functors

The categories of Boolean algebras, Compact Hausdorff spaces and C* algebras respectively. S is the Stone space functor, C is the ‘continuous functions on X’ functor.

Now, in a slight abuse of notation, these combine to give a functor

In particular, this functor is covariant! It’s an equivalence of categories between the category of Boolean algebras and some subcategory of the category of commutative C* algebras.

So, I asked two questions: What’s the image of this functor, and is there a natural way of realising it? (The Stone functor is fairly nonconstructive, so is in some sense not ‘natural’).

The image of the Stone functor is precisely the class of zero-dimensional compact Hausdorff spaces, and the C functor is an isomorphism of categories, so rephrasing the first question we ask how can we tell if is zero-dimensional from looking at ?

This turns out to have a nice answer. is zero-dimensional iff is generated by its idempotents (i.e. they generate a dense sub-algebra).

The way to see this is an easy application of the following lemma:

Lemma: Let be compact Hausdorff. generates a dense subalgebra iff the usual topology on is the topology generated by (i.e. the coarsest topology with respect to which each element of is continuous). Equivalently, if the set , where is restricted to range over the open sets, is a basis.

I won’t bother proving this lemma here. It’s just a vaguely clever application of Stone-Weierstrass + the fact that compact Hausdorff spaces are minimally Hausdorff.

Theorem: Let be compact Hausdorff. is zero-dimensional iff is generated by its idempotents.

Proof:

If is zero-dimensional then the set of characteristic functions of clopen sets generates a dense subalgebra, by the above.

Any idempotent is in fact the characteristic function of a clopen set, so again by the above lemma if the idempotents generate a dense subalgebra then the clopen sets form a basis.

QED

Right. So that’s that sorted.

Some definitions. We’ll (confusingly) call a C* algebra a Boolean C* algebra if it is commutative and generated by its idempotents. The above shows that the category of Boolean algebras is equivalent to the category of Boolean C* algebras. Now we’d just like to construct a more direct equivalence.

It turns out to be much easier to go from the Boolean C* algebras to the Boolean algebras, so we’ll do that first.

Let be a commutative C* algebra. Define

Now, if then we have and . So, define the following:

So, is closed under these operations. Easy check: They make I(A) into a Boolean algebra.

It’s easy to see that if is a homomorphism then . Further this is a Boolean algebra homomorphism. Hence is functorial in a natural way.

So, what we want to do is clear. We want to realise an arbitrary Boolean algebra as of some C* algebra.

We’ll study the structure of a little more before we set off to do so.

First of all, note that if (i.e. ) then . This means that we can write any as with non-zero and for .

This is important for a number of reasons. Among them, because it means that the C* condition fixes the norm on (Of course, the norm on is uniquely determined, but < I(A) \right> will almost never be complete).

We have .

So

This formula for the norm is as important as the fact that the norm is determined uniquely. The following consequences in particular will be of crucial importance:

Corollary : Let be distinct. Then .

Proof: . These are disjoint, hence .
Theorem: Let be a Boolean subalgebra which generates a dense subalgebra of (remember, that’s C* subalgebra, not Boolean). Then .

Proof:

Suppose not, then we can find some idempotent . Let . We can find such that for . with . Then for perpendicular. So . In particular .

So, for every we have . Now, choose small enough so that this requires that or . Let . Then is an idempotent and . We may choose to be as small as we like, so let . Then .

But and are both idempotents, and by our previous corollary if they were distinct then we'd have . Hence , and so .

Hence
QED

You may not believe it, but we’re now essentially done!

Start with some Boolean algebra . Generate the free algebra with as its generators. For we’ll denote the corresponding element of the algebra by . Quotient out by the requirement that and , that and that .

Now, for define .

Again, we can write as a disjoint sum of the . Having done so define .

Easy checks show that this gives an algebra norm which satisfies the C* condition. Thus its completion is a C* algebra. Call this algebra
You can quickly see that morphisms between Boolean algebras extend to morphisms between the generated algebras, and so extend to the completion. Thus this is functorial.

Our previous theorem shows that , as is a Boolean subalgebra of which generates a dense C* subalgebra of .

So, we’re basically done. Some trivial checking of details remains, but I hope I’ve now convinced you that these functors give an equivalence of categories.

But do they do what we originally wanted?

Sure. Requiring that is isomorphic to (sorry for the overloading of labels) is the same as showing there’s a natural isomorphism from the character space of C(B) to the Stone space of B.

But elements of the character space are just homomorphisms . is where is the 2-element Boolean algebra . So these homomorphisms naturally biject with the Boolean algebra homomorphisms . i.e. the elements of the Stone space. It’s then easy to check that this bijection is a homeomorphism.

Phew. So, that’s all over and done with.

Now, why do we actually care about this? I’m afraid it’s not because this gives us tons of exciting new examples of Boolean algebras or C* algebras - intriguingly it seems like natural examples of one correspond to natural examples of others. This is actually quite surprising, and more than a little cool! It gives us multiple ways to look at existing objects. So far I’ve only found one case where this was actually useful (and it could be done without it), but I’m sure there are more. I’ve not been looking very hard.

Here are some examples:

• If is the Boolean algebra of finite and cofinite subsets of then is the C* algebra of convergent complex valued sequences.
• 
• If Meas is the Boolean algebra of measurable subsets of modulo the ideal of null sets, then

This last example is the one I found useful - it’s much more obvious what the Stone space of Meas is than what the character space of L^{\infty}[0, 1] is (although once you know what it is it isn’t that hard to prove directly). I needed an example of a commutative C* algebra with an inseparable character space but a separable Hilbert space representation, and this was it.

Questions I’d like to go on to answer:

What’s the noncommutative generalisation of this? You can certainly get a poset out of it, and the poset has an order reversing operator analagous to . Unfortunately I can’t seem to prove that is always a lattice here - it certainly seems to be in the cases I’ve seen. The problem is that it isn’t closed under multiplication of noncommuting elements, so the obvious candidate for doesn’t work.

For topological spaces we have an additional way of getting a Boolean algebra out - the regular open sets. This then gives us the notion of the absolute of a space (the stone space of the algebra of regular open sets, which is also the maximal irreducible preimage of ). Does this correspond to anything nice in the C* algebra setting?

Other stuff I haven’t yet thought of.

This is potentially another rambling post. It relates what I’ve been doing on a cute but rather strange corner of point-set topology. For now, when I say ’space’ I mean ‘Hausdorff space’. Most of what I’m doing here isn’t very interesting in the non-Hausdorff case. Some of it is, but in those cases the Hausdorff assumption makes no real difference either way.
I’ve been playing with the structure of compact spaces which aren’t sequentially compact for a while. Primarily with the aim of showing that in the absence of the full axiom of choice these don’t neccesarily exist (specifically I wanted to show it was consistent with ZF + DC that they didn’t exist). I’ve got to the point where if I really put a lot of effort into it I could probably get a solution, but I’ve rather lost steam on that. It’s not all that interesting a result even if it is true.

In playing with this in general I’ve noticed something a bit more interesting, and related to my original reasons for forming this conjecture.

We’ve basically got two ‘classic’ examples of compact non-sequentially compact spaces. and . The first is not sequentially compact for any one of a number of reasons, the second is not sequentially compact because the sequence has no convergent subsequence.

One thing I noticed recently: This is actually only one example. We can identify with the set of ultrafilters on . i.e. a subset of , which is in turn identified in a natural way with . Under this identification, the subset topology and the normal topology on agree. Further, the defined above are the principal ultrafilters at . So their closure is precisely a copy of .

So, our two classic examples are really only one example!

I’ve searched the literature for examples of compact spaces which are not sequentially compact, and turned up surprisingly short. Other than these two, there seems to only be one other class of examples (I’ll come to that in a minute). At this point I started getting suspicious - does every compact space which isn’t sequentially compact contain a copy of ?

Well, no. At least, not quite. So far the only counterexampe I have depends on the value of the reaping number, . It’s fairly standard (I’ll write a post on it at some point) that is not sequentially compact. If then this is an example of a compact non-sequentially compact space of weight , so can't contain a copy of , which has weight .

So, if the answer isn't simply no then the problem depends on the underlying combinatorial structure of the universe (or, if you prefer, is independent of ZFC). I suspect that if , or possibly if diamond holds, then the answer will be yes.

You can at least pare down any example into a compactification of for which the sequence does not have any convergent subsequences. Suppose is some non-convergent sequence. We can find a subsequence which has the discrete topology on it. The closure of this subsequence together with the map is our desired compactification of .
Allow me to phrase the problem in what may be a slightly more suggestive way.

Let be topological spaces. Say is a minor of if it is the continuous image of a closed subset of , and write . Also say is a major of . Note that forms a preorder on topological spaces. This is not a partial order on isomorphism classes. For example, and are both minors of eachother and are not isomorphic. Also, and are minors of eachother which are not isomorphic.

A standard result is that every compact hausdorff space of weight is a minor of . So given a minor closed class in order to show that it contains all spaces of weight it suffices to show that it includes . e.g. in order to show all compact spaces of weight are sequentially compact, it suffices to show that is for every .

Note: The 'minor' notation is totally nonstandard as far as I'm aware. I've stolen it from graph theory in order to suggest certain analogies (but I know very little graph theory, so these analogies may be totally wrong).
We'll be interested in looking at classes of spaces which are closed under taking minors. Examples of these include:

• The class of compact spaces.
• The class of sequentially compact spaces.
• The compact spaces of weight (note: The compactness assumption is neccesary here. In general weight is not neccesarily non-decreasing on continuous images).
• For any space , the class of minors of .

A class is closed under taking minors iff its complement is closed under taking majors. This is important!

has the following interesting property. If is compact (Hausdorff) and is a continuous surjection then there is an embedding such that . The proof is fairly straightforward, but the important part to take home is that if is a minor of then it is a subspace of . So the class of spaces which contain is major closed.

Final point: Every compact space which is not sequentially compact contains a separable subspace which is not (indeed a compactification of . There are only separable Hausdorff spaces, so there is a set of representatives for the isomorphism classes of such subspaces. Thus we can find some set such that is not sequentially compact iff it has a minor in . So far we’ve shown we can choose , and we want to show we can choose . Oh well, a little way to go…
Maybe this isn’t that suggestive, but I think it’s a neat way of looking at it and may prove useful. Certainly I’d like to study results about minor closed spaces some more - obviously I’ve been motivated by the excluded minor theorem, that a minor closed set of graphs is precisely the set of graphs which exclude some finite set of forbidden minors. I really doubt this is true in topological version, even restricting ourself to compact Hausdorff spaces, (I suspect the succesor ordinals form a counterexample), but some analogue of it might be.

In fact, for any finite topological space we have that .

Proof: For any point define to be the intersection of all open sets containing . The set is a basis for the topology of of cardinality .

See here for far more details.

Warning: This post isn’t really of an expository nature. It’s more me rambling for my own benefit about something I find interesting. It isn’t in any especially coherent order. You might find this interesting, or you might not.

A while ago I claimed that all interesting topological spaces were Hausdorff. Even then I knew this wasn’t really the case - for example the algebraic geometers’ Spec is almost never Hausdorff, or even (a prime ideal is closed iff it’s maximal). What I really mean of course is interesting for my purposes - Hausdorff is precisely the requirement that limits are unique (in an appropriate sense), and given that most of the stuff studied in analysis is concerned with limits, this seemed to me to be a fairly compelling argument that Hausdorff was an extremely important condition.

One seperation axiom that really is required in order for a topological space to be ‘interesting’ is that of being . Two points x and y are said to be indistinguishable iff they’re contained in precisely the same open sets. Equivalently they’re indistinguishable iff . Topologically speaking, you can’t tell the difference between indistinguishable points. A space is if it has no distinct indistinguishable points.

Of course, one might argue that there are examples of spaces with additional structure which might distinguish these points. So from that perspective there is certainly worth in studying spaces which are . But, from a topological perspective, this is essentially the same as studying their Kolmogorov quotient, which is .

The Kolmogorov quotient is defined in an obvious way. Indistinguishability of points is an equivalence relation. Put the quotient topology on the set of equivalence classes. This is the Kolmogorov quotient of the space. It is universal in the sense that if is a topological space and is its Kolmogorov quotient then any map , where Y is , factors uniquely through the quotient map .
Usually if we have some algebraic structure compatible with the topology then the Kolmogorov quotient will inherit this.

For example, suppose we have a topological group which may not be . Some minor faffing will show that x and y are equivalent iff . In particular the group is iff the identity (and so every point) is closed.

The closure of the identity is a closed normal subgroup of , and so we can quotient by it to get a topological group in which the identity is closed. Further, this quotient is precisely the Kolmogorov quotient of . So, as promised, the Kolmogorov quotient inherits a natural algebraic structure.

As an aside note, if a topological group is then it is Hausdorff, as the diagonal is the preimage of the identity under the map and so is closed. Another reason why I thought Hausdorff spaces were really the appropriate setting for what I was doing (and while I still maintain that Hausdorff is part of the definition of topological group).

One thing that seems to be the case is that the theories of Hausdorff and non-Hausdorff topological spaces seem to be very different - definitions important in one either don’t work or become essentially trivial in the other. Almost everything you’d normally file under ‘point-set topology’ is a bit uninteresting or downright wrong when you start weakening separation conditions.

All that being said, some rather cool stuff does happen in the case where your spaces fail to be .

Let be some topological space. For define if . Then this is a preorder on , and it’s a partial order precisely when is .

Now, in the familiar cases where the spaces are this tells you precisely bugger all about your underlying topologies - the resulting partial order is just a great big antichain.

But in the case where is finite, this tells you everything about the topological space. Remember that there’s only one topology on a finite set - the discrete topology.

Suppose you know the preorder on . Then for each point you know what is. So for any you have that . So you know the closure operator, and so the topology, precisely.

Another way of looking at this is that given a preorder on a set you can form a topology on by taking to be closed iff it is downward closed (i.e. if and then ), or the open sets to be the upwards closed sets. In the finite case composing this with the above construction gives you back the original topology on .

Note that both of these constructions are functorial. I suspect they form an adjoint pair of some sort, but have yet to work out the details. In the finite case they are an isomorphism of categories.

This is cool. Finite topological spaces are precisely the same thing as finite preorders (and finite spaces precisely the same as finite partial orders). This is related to why we have such a hard time of enumerating the finite topological spaces - because enumerating the finite partial orders is hard (and I think determining if two of them are isomorphic is NP-complete - it is after all essentially a not very special case of graph isomorphism).

Note that this construction, in the infinite case, gives us a topology whose open sets are closed under arbitrary intersection, as a consequence of which holds in general. These are called Alexandroff topologies. They are of course iff they’re discrete. (Compact examples of?) these are supposed to be the natural infinite generalisation of finite topological spaces.

One thing we do in point-set topology is look at cardinal invariants of topological spaces. These don’t behave all that well when they’re finite, so we often redefine them to take as their minimum value. That way we can avoid inconvenient nonsenses like numbers for which we have or . Silly things like those. On the other hand, the basic definitions more or less work and I was curious as to how they look in the finite example.

Two important ones are the density , the smallest cardinality of a dense subset, and the cellularity , the sup of the cardinalities of a pairwise disjoint collection of open sets (In the finite case I’m going to require the sets to be non-empty. In the infinite case this is irrelevant because ).

For any topological space we have - every non-empty open set contains an element of any given dense set, and if two sets are disjoint they can’t contain the same element. For metric spaces they are always equal, but in general they can be arbitrarily far apart.

Curiously, it turns out for finite topological spaces they’re also always equal. The reason is as follows.

In a finite topological space, every element is covered by some maximal element of the space - an such that . Thus the set of maximal is dense in . Now, any dense set must contain this set, because for each maximal there is with , and so . Thus the collection of maximal elements of is a dense set of minimal cardinality.
Now, for maximal we have for . Thus the complement of is closed, and is open. Thus the set of singletons of maximal elements of is a pairwise disjoint collection of open subsets of . Hence we have and so .

Another important cardinal invariant is the weight. The weight of is the smallest cardinality of a basis of . I originally thought there wasn’t much interesting to say about this in the finite case, but on the tube home today I noticed I was wrong.

For finite spaces the weight satisfies .

The proof of this is a fairly trivial induction. Suppose and the result holds for every space of cardinality . Let be a basis for . By our previous observation, contains an open point . Because the singleton is open it must contain a basis element, and so must in fact be a basis element. Let . Then is a basis for , which is a space of cardinality . Hence by our inductive hypothesis we have and so .

A couple of notes on this: Obviously the hypothesis is neccesary, as the indiscrete topology on any set shows. This inequality is the best possible, as the discrete topology shows (there are other examples. e.g. the sierpinski space).
This latter inequality is very much a special property of finite topologies. In general there are no good inequalities between and . For compact Hausdorff spaces we have .

Finally, I am of course probably reinventing the wheel with all of this. I’ve no illusions that any of this is remotely new, but equally I’ve got no relevant sources except a couple random posts on the web about finite topological spaces.