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) \(X, Y\). We make \(X \times Y\) by taking the \sigma algebra generated by sets of the form \(A \times B\). in the case where we have topologies on \(X\) and \(Y\) 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 \(X\) be a measurable space with \(|X| > 2^{\aleph_0}\). Then the product algebra on \(X^2\) does not contain the diagonal.

In particular, if \(X\) 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 \(X\) be a set, \(\mathcal{A} \subseteq P(X)\) and \(U \in \sigma(\mathcal{A})\). There exist \(A_1, \ldots, A_n, \ldots\) with \(U \in \sigma(A_1, \ldots, A_n, \ldots)\)

Proof: The set of \(U\) satisfying the conclusion of the theorem is a \(\sigma\) algebra containing \(\mathcal{A}\).

Lemma: Let \(U \subseteq X^2\) be measurable. Then \(U\) is the union of at most \(2^{\aleph_0}\) sets of the form \(A \times B\).

Proof:

By the preceding lemma we can find \(A_n\) with \(U \in \sigma (\{ A_m \times A_n \})\)

For \(x \in \{0, 1\}^{\mathbb{N}}\), define \(B_x = \bigcap C_n\) where \(C_n = A_n\) if \(x_n = 1\) and \(A_n^c\) otherwise.

Sets of the form \(B_x \times B_y\) then form a partition of of \(X^2\). Thus the sets which can be written as a union of sets of the form \(B_x \times B_y\) form a \(\sigma\) algebra. This contains each of the \(A_m \times A_n\), and so contains \(U\). Thus \(U = \bigcup { B_x \times B_y : B_x \times B_y \subseteq U }\). There are at most \(2^{\aleph_0}\) sets in this union. Hence the desired result.

Finally we have the proof of the theorem:

Let \(D\) be the diagonal. Suppose \(D\) is measurable. Then \(D\) is the union of at most \(2^{\aleph_0}\) sets of the form \(A \times B\). Because \(|D| > 2^{\aleph_0}\) at least one of these sets must contain two points. Say \((u, u)\) and \((v, v)\). But then it also contains \((u, v) \not\in D\). 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 \(\{0, 1\}^{\kappa}\), 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 [tex]I(A)[/tex] 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 [tex]I(A)[/tex] 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 [tex]x prec y[/tex] if [tex]xy = x[/tex].

Thie definition may seem like there should also be a right-multiplication version, but in fact there isn’t. If [tex]x prec y[/tex] then [tex]x = x^* = (xy)^* = y^* x^* = yx[/tex] (one of the many reasons it was important that these things are self adjoint).

So, if [tex]x prec y[/tex] and [tex]y prec x[/tex] then [tex]x = xy = y[/tex]. So the relation is antisymmetric. It’s evidently transitive and reflexive, so we have a partial order.

It has top and bottom elements, [tex]0[/tex] and [tex]1[/tex].

We have an operator [tex]neg x = 1 – x[/tex] which maps [tex]I(A)[/tex] into itself. It has the following properties:

• It’s order reversing.
• [tex]negneg x = x[/tex]
• [tex]x vee neg x = 1[/tex]
• [tex]x wedge neg x = 0[/tex]

Note that it is not determined by these latter two properties. [tex]I(A)[/tex] is not a distributive lattice!

Is [tex]I(A)[/tex] even a lattice? I don’t know. It looks like it.

In the case of [tex]End(H)[/tex] we have [tex]I(A)[/tex] is isomorphic to the lattice of closed subspaces of [tex]H[/tex], with [tex]neg[/tex] being orthogonal complement and [tex]prec[/tex] being [tex]subseteq[/tex]. This is a lattice, with [tex]X wedge Y = X cap Y[/tex] and [tex]X vee Y[/tex] being the closed linear span of [tex]X cup Y[/tex].

Further, fix [tex]x in X setminux Y[/tex]. Then for [tex](pi_X pi_Y)^n x to 0[/tex]. So one might expect that [tex](pi_X pi_Y)^n to pi_{X cap Y}[/tex]. Unfortunately this isn’t true. Usual problem with exchanging limits. It’s true in some sort of weak topology, so if [tex]A[/tex] can be represented as a closed subspace of [tex]End(H)[/tex] in this topology then this limit is in [tex]A[/tex] and we have the desired result. I imagine this only works if [tex]A[/tex] 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 [tex]1[/tex], but projections [tex]x, y[/tex] can have [tex]||x – y||[/tex] taking any value in [tex][0, 1][/tex].

Also, these posets are very very large. In the simplest example, take [tex]A = End(mathbb{C}^2)[/tex]. Then [tex]I(A)[/tex] is an antichain of size [tex]2^{aleph_0}[/tex] plus a top and bottom element.

Note this is also an example of how badly complementation fails to be unique. For any distinct [tex]x, y[/tex] with [tex]x, y not= 1, 0[/tex] we have [tex]x vee y = 1[/tex] and [tex]x wedge y = 0[/tex].

I’ll post more as I think of it.

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

[tex]S : Boolean^{op} to KHauss[/tex]

[tex]C : KHauss^{op} to C^* [/tex]

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

[tex]CS : Boolean to C^*[/tex]

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 [tex]X[/tex] is zero-dimensional from looking at [tex]C(X)[/tex]?

This turns out to have a nice answer. [tex]X[/tex] is zero-dimensional iff [tex]C(X)[/tex] 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 [tex]X[/tex] be compact Hausdorff. [tex]A subseteq C(X)[/tex] generates a dense subalgebra iff the usual topology on [tex]X[/tex] is the topology generated by [tex]A[/tex] (i.e. the coarsest topology with respect to which each element of [tex]A[/tex] is continuous). Equivalently, if the set [tex] { f^{-1}(U) : f in A, U subseteq mathbb{C} } [/tex], where [tex]U[/tex] 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 [tex]X[/tex] be compact Hausdorff. [tex]X[/tex] is zero-dimensional iff [tex]C(X)[/tex] is generated by its idempotents.

Proof:

If [tex]X[/tex] 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 [tex]A[/tex] be a commutative C* algebra. Define [tex]I(A) = { f in A : f^2 = f }[/tex]

Now, if [tex]f, g in I(A)[/tex] then we have [tex]fg in I(A)[/tex] and [tex]1 – f in I(A)[/tex]. So, define the following:

[tex]f wedge g = fg[/tex]

[tex] neg f = 1 – f [/tex]

[tex] f vee g = neg( neg f wedge neg g)[/tex]

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

It’s easy to see that if [tex]f : A to B[/tex] is a homomorphism then [tex]f(I(A)) subseteq I(B)[/tex]. Further this is a Boolean algebra homomorphism. Hence [tex]I[/tex] is functorial in a natural way.

So, what we want to do is clear. We want to realise an arbitrary Boolean algebra as [tex]I(A)[/tex] of some C* algebra.

We’ll study the structure of [tex]I(A)[/tex] a little more before we set off to do so.

First of all, note that if [tex]f perp g[/tex] (i.e. [tex]f wedge g = fg = 0[/tex]) then [tex]f vee g = f + g[/tex]. This means that we can write any [tex]f in left< I(A) right>[/tex] as [tex]f = sum z_k f_k[/tex] with [tex]f_k in I(A)[/tex] non-zero and [tex]f_k perp f_l[/tex] for [tex]k not= l[/tex].

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

We have [tex]||f||^2 = ||f f^*|| = || sum |z_k|^2 f_k ||[/tex].

So [tex]||f|| = || sum |z_k|^{2^n} f_k ||^{2^{-n}} to max |z_k|[/tex]

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 [tex]f, g in I(A)[/tex] be distinct. Then [tex]||f – g|| = 1[/tex].

Proof: [tex]f – g = f(neg g) – g(neg f)[/tex]. These are disjoint, hence [tex]||f – g|| = max{ 1 } = 1[/tex].
Theorem: Let [tex]B subseteq I(A)[/tex] be a Boolean subalgebra which generates a dense subalgebra of [tex]A[/tex] (remember, that’s C* subalgebra, not Boolean). Then [tex]B = I(A)[/tex].

Proof:

Suppose not, then we can find some idempotent [tex]f notin left< B right>[/tex]. Let [tex]epsilon > 0[/tex]. We can find [tex]delta > 0[/tex] such that for [tex]||f – g|| < delta|| we have ||g^2 - g|| < epsilon[/tex]. [tex]g in left< B right>[/tex] with [tex]||f – g|| < delta[/tex]. Then [tex]g = sum z_k g_k[/tex] for [tex]g_k in B[/tex] perpendicular. So [tex]g^2 - g = sum (z_k^2 - z_k)[/tex]. In particular [tex]||g^2 - g|| = max ||z_k^2 - z_k||[/tex]. So, for every [tex]k[/tex] we have [tex]||z_k^2 - z_k|| < epsilon[/tex]. Now, choose [tex]epsilon[/tex] small enough so that this requires that [tex]z_k < frac{1}{4}[/tex] or [tex]z_k > frac{3}{4}[/tex]. Let [tex]h = sumlimits_{z_k > frac{3}{4}} g_k[/tex]. Then [tex]h[/tex] is an idempotent and [tex]||h – g|| < frac{1}{4}[/tex. Hence [tex]||h - f|| < frac{1}{4} + delta[/tex]. We may choose [tex]delta[/tex] to be as small as we like, so let [tex]delta < frac{1}{4}[/tex]. Then [tex]||f - h|| < frac{1}{2}[/tex]. But [tex]f[/tex] and [tex]h[/tex] are both idempotents, and by our previous corollary if they were distinct then we'd have [tex]||f - h|| = 1[/tex]. Hence [tex]f = h[/tex], and so [tex] f in left< B right>[/tex].

Hence [tex]B = I(A)[/tex]
QED

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

Start with some Boolean algebra [tex]B[/tex]. Generate the free algebra with [tex]B[/tex] as its generators. For [tex]x in B[/tex] we’ll denote the corresponding element of the algebra by [tex]x^C[/tex]. Quotient out by the requirement that [tex]1^C = 1[/tex] and [tex]0^C = 0[/tex], that [tex](x wedge y)^C = x^C y^C[/tex] and that [tex](x vee y)^C = x^C + y^C – xy[/tex].

Now, for [tex]f = sum z_k x_k^C[/tex] define [tex]f^* = sum z_k^* x_k^C[/tex].

Again, we can write [tex]f[/tex] as a disjoint sum of the [tex]x_k^C[/tex]. Having done so define [tex]||f|| = max |z_k|[/tex].

Easy checks show that this gives an algebra norm which satisfies the C* condition. Thus its completion is a C* algebra. Call this algebra [tex]C(B)[/tex]
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 [tex]B = I(C(B))[/tex], as [tex]B[/tex] is a Boolean subalgebra of [tex]I(C(B))[/tex] which generates a dense C* subalgebra of [tex]C(B)[/tex].

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 [tex]C[/tex] is isomorphic to [tex]CS[/tex] (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 [tex]f : C(B) to Bbb{C}[/tex]. [tex]Bbb{C}[/tex] is [tex]C(mathbb{2})[/tex] where [tex]mathbb{2}[/tex] is the 2-element Boolean algebra [tex]{0, 1}[/tex]. So these homomorphisms naturally biject with the Boolean algebra homomorphisms [tex]B to mathbb{2}[/tex]. 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 [tex]B[/tex] is the Boolean algebra of finite and cofinite subsets of [tex]mathbb{N}[/tex] then [tex]C(B)[/tex] is the C* algebra of convergent complex valued sequences.
• [tex]C(P(mathbb{N})) = l^{infty}[/tex]
• If Meas is the Boolean algebra of measurable subsets of [tex][0, 1][/tex] modulo the ideal of null sets, then [tex]C(Meas) = L^{infty}[0, 1][/tex]

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 [tex]neg[/tex]. Unfortunately I can’t seem to prove that [tex]I(A)[/tex] 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 [tex]wedge[/tex] 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 [tex]X[/tex]). 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. [tex]beta mathbb{N}[/tex] and [tex]{ 0, 1 }^{2^{mathbb{N}} }[/tex]. The first is not sequentially compact for any one of a number of reasons, the second is not sequentially compact because the sequence [tex]x_n : a to a_n[/tex] has no convergent subsequence.

One thing I noticed recently: This is actually only one example. We can identify [tex]beta mathbb{N}[/tex] with the set of ultrafilters on [tex]mathbb{N}[/tex]. i.e. a subset of [tex]P(P(mathbb{N}))[/tex], which is in turn identified in a natural way with [tex]{ 0, 1 }^{2^{mathbb{N}} }[/tex]. Under this identification, the subset topology and the normal topology on [tex]beta mathbb{N}[/tex] agree. Further, the [tex]x_n[/tex] defined above are the principal ultrafilters at [tex]n[/tex]. So their closure is precisely a copy of [tex]beta mathbb{N}[/tex].

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 [tex]beta mathbb{N}[/tex]?

Well, no. At least, not quite. So far the only counterexampe I have depends on the value of the reaping number, [tex]tau[/tex]. It’s fairly standard (I’ll write a post on it at some point) that [tex]{0, 1}^{tau}[/tex] is not sequentially compact. If [tex]tau < 2^{aleph_0}[/tex] then this is an example of a compact non-sequentially compact space of weight [tex]< 2^{aleph_0}[/tex], so can't contain a copy of [tex]beta mathbb{N}[/tex], which has weight [tex]2^{aleph_0}[/tex]. 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 [tex]2^{aleph_0} = aleph_1[/tex], or possibly if diamond holds, then the answer will be yes. You can at least pare down any example into a compactification of [tex]mathbb{N}[/tex] for which the sequence [tex](n)[/tex] does not have any convergent subsequences. Suppose [tex]x_n[/tex] 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 [tex]k to x_{n_k}[/tex] is our desired compactification of [tex]mathbb{N}[/tex]. Allow me to phrase the problem in what may be a slightly more suggestive way. Let [tex]X, Y[/tex] be topological spaces. Say [tex]Y[/tex] is a minor of [tex]X[/tex] if it is the continuous image of a closed subset of [tex]X[/tex], and write [tex]Y prec X[/tex]. Also say [tex]X[/tex] is a major of [tex]Y[/tex]. Note that [tex]prec[/tex] forms a preorder on topological spaces. This is not a partial order on isomorphism classes. For example, [tex][0, 1][/tex] and [tex]S^1[/tex] are both minors of eachother and are not isomorphic. Also, [tex]beta mathbb{N}[/tex] and [tex]{0, 1}^{2^{mathbb{N}}}[/tex] are minors of eachother which are not isomorphic.

A standard result is that every compact hausdorff space of weight [tex]leq kappa[/tex] is a minor of [tex]{0, 1}^{kappa}[/tex]. So given a minor closed class in order to show that it contains all spaces of weight [tex]leq kappa[/tex] it suffices to show that it includes [tex]{0, 1}^{kappa}[/tex]. e.g. in order to show all compact spaces of weight [tex]< tau[/tex] are sequentially compact, it suffices to show that [tex]{0, 1}^{kappa}[/tex] is for every [tex]kappa < tau[/tex]. 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 [tex]leq kappa[/tex] (note: The compactness assumption is neccesary here. In general weight is not neccesarily non-decreasing on continuous images).
• For any space [tex]X[/tex], the class of minors of [tex]X[/tex].

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

[tex]beta mathbb{N}[/tex] has the following interesting property. If [tex]X[/tex] is compact (Hausdorff) and [tex]f : X to beta mathbb{N}[/tex] is a continuous surjection then there is an embedding [tex]g : beta mathbb{N} to X[/tex] such that [tex]fg = id[/tex]. The proof is fairly straightforward, but the important part to take home is that if [tex]beta mathbb{N}[/tex] is a minor of [tex]X[/tex] then it is a subspace of [tex]X[/tex]. So the class of spaces which contain [tex]beta mathbb{N}[/tex] is major closed.

Final point: Every compact space which is not sequentially compact contains a separable subspace which is not (indeed a compactification of [tex]mathbb{N}[/tex]. There are only [tex]2^{2^{aleph_0}}[/tex] separable Hausdorff spaces, so there is a set of representatives for the isomorphism classes of such subspaces. Thus we can find some set [tex]R[/tex] such that [tex]X[/tex] is not sequentially compact iff it has a minor in [tex]R[/tex]. So far we’ve shown we can choose [tex]|R| leq 2^{2^{aleph_0}}[/tex], and we want to show we can choose [tex]|R| = 1[/tex]. 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.