## Is it true that the only interesting topologies are metric topologies and weak topologies?

In "Infinite dimensional analysis, A hitchhikers guide" by Aliprantis and Border, they write that these 2 classes of topologies "by and large include everything of interest".

@Pete Clarke: I was asking if it holds for mathematics in general, and not just for functional analysis. From the answers I get the impression that their statement is a fairly good first approximation.

@Gerald Edgar: I thought I read in Mathoverflow that the Zariski topology can be regarded as a weak topology (at least in some cases).

-
Probably MO is the wrong board for this question... posters here are fond of Zariski topology and such things. But probably Aliprantis & Border mean these are everything of interest for infinite-dimensional analysis. But even there they are wrong if you want to do Schwartz ditributions. – Gerald Edgar Apr 16 2010 at 13:20
I'd just like to note that I voted for Gerald's comment because of the middle sentence, with the implication that the question would be better if refined a little to be clear what context it is in (for those of us who sadly lack any knowledge of said book). Having read the comment again, it could be read as "This looks like a functional analysis question and MO is full of algebraic geometers so you're probably in the wrong place.". I'm sure that that's not what Gerald meant and I'd like to be sure that everyone knows that MO is a great place for functional analysis! – Andrew Stacey Apr 16 2010 at 19:58
I think it is worth mentioning that the book by Aliprantis and Border is a book about the functional analysis and measure theory used in mathematical economics. – Michael Greinecker Jan 28 2011 at 7:16

Picking up on Gerald's interpretation of the question (namely, that it really focusses on infinite dimensional vector spaces) then I say: absolutely not!

For example, piecewise-smooth paths in some Euclidean space has a topology that is neither of these (it's an uncountable inductive limit of Frechet spaces). (Not that I particularly recommend this space!)

Close to Gerald's comment, "dual-Frechet" spaces (that is, the dual of a Frechet space with the strong topology) have very nice properties, almost as nice as Frechet spaces themselves. This class includes distributions with the "right" topology.

And that's the point, really. If you're only interested in, say, distributions for what they can say about compactly supported functions, then the weak topology is probably fine. However, if you are interested in distributions in their own right then the weak topology is very unlikely to be okay.

Here's an example from my research: I like infinite dimensional manifolds, and I quite like loop spaces. To construct the Dirac operator on a loop space, I needed to put an inner product on the cotangent bundle. So I needed, in effect, a continuous injective map $(L\mathbb{R}^n)^* \to H$ ($H$ being some standard Hilbert space). I can't do this with the weak topology any continuous map from $(L\mathbb{R}^n)^*$ with the weak topology to a normed vector space has to factor through a finite dimensional space. With the strong topology, though, it was no problem.

So, deal with metric and weak topologies if you like; but real analysts use the strong topology[1].

[1] Not sure what complex analysts use.

-
+1 for the so-bad-its-good pun :D – Johannes Hahn Apr 16 2010 at 17:08
Thanks! Incidentally, if anyone can think of a better footnote, I'd be happy to edit it (with credits, but I don't think that I can split the reputation). – Andrew Stacey Apr 16 2010 at 20:00

Every topological space X has the initial topology (or weak topology) with respect to the family of continuous functions from X to the Sierpiński space. (see http://en.wikipedia.org/wiki/Initial_topology.)

This is the two point space {a,b} with open sets: emptyset, {a} and {a,b} only. If U is any subset of X, then the function fU, mapping points in U to a and the rest to b, will be continuous if and only if U is open.

-
 It's better than that. The initial topology so generated is in fact isomorphic (as a lattice) to the set of continuous maps from $X$ to the Sierpinski space $\Sigma$, i.e., $O(X) \cong C(X,\Sigma)$. This is something every student of topology should be aware of. – Andrej Bauer Apr 18 2010 at 11:45 Andrej, yes, and thanks for the remark. Every function from X into Sigma is f_U for some U, and as I explained, U is open if and only if f_U is continuous. So the isomorphism is the association of U with f_U. But one thing I find incongruous in your suggestion is that although we may regard C(X,Sigma) as a lattice by imposing an order on Sigma (say, with a less than b), we have not put the order topology on Sigma. – Joel David Hamkins Apr 18 2010 at 20:13

Frechet spaces, limits of Frechet spaces were mentioned before. I'd like to emphasize a particularly important example (which was also mentioned before, but I want to extend it a bit): The space of test functions $\mathcal{D}(\Omega)$ is a strong inductive limit of Frechet spaces, neither metrizable nor do they carry the weak topology. These spaces are the foundation of distribution theory and therefore most important.

on the other hand the other two usual spaces of test function $\mathcal{E}(\Omega)$ and the schwartz space $\mathcal{S}(\mathbb{R}^n)$ are metrizable because they can be topologized by a countable family of (semi)norms.

The distribution space $\mathcal{D}'(\Omega), \mathcal{E}'(\Omega), \mathcal{S}'(\mathbb{R}^n)$ can be endowed with the weak topology (that is the pointwise convergence). But the strong topology is also common and this is again neither weak nor metrizable.

-
 Spaces $\mathcal{D}'(\Omega), \mathcal{E}'(\Omega), \mathcal{S}'(\mathbb{R}^n)$ the strong topologies are neigher weak nor metrizable. And yet textbooks commonly check things only with sequences in these spaces (rather than nets or filters). And commonly get it right. Interesting... – Gerald Edgar Apr 18 2010 at 11:52 Maybe the strong duals are sequential spaces? – Johannes Hahn Apr 18 2010 at 16:34 @ Gerald Edgar. The fact that using sequences works for the latter two spaces is not happenstance. It is because they are inductive limits of sequences of Banach spaces linked by compact mappings. This class of spaces was investigated by J. Sebastião e Silva (they are now called Silva spaces). A similar remark applies to the first space but this is more delicate---they are projective limits of sequences of Silva spaces, but of a special kind, namely with partitions of unity (in the functional analytic sense which is related, of course, to the classical version). – jbc Sep 18 at 5:28

So the meaning of "weak topology" is that which Joel defines. This was actually a new idea to me, but it seems very general (more general that what I would have called the "weak topology" in Functional Analysis). As Joel points out, interpreted in maximal generality, any topological space carries the weak topology in this sense.

The book actually very quickly specialises to weak topologies generated by continuous, real-valued functions. Then your space has to be completely regular. I think it would be fair to say that this does rule out some interesting examples.

-

Of course, any question of the form "Is it the case that the things considered interesting by such-and-such small set of people are all the interesting things?" is going to have a negative answer. Worse, taking statements such as yours too literally and outside of appropriate context will just cause you to be closed-minded and unable to understand other people's ideas.

Even if you are interested only in functional analysis you will quickly hit against topologies which are not included among your topologies. For example, given an infinite-dimensional Hilbert space $H$, what is a good topology to put on its lattice of closed subspaces? (This topology won't even be Hausdorff.)

For a second example, consider how we reconstruct a compact Hausdorff space $X$ from its $C^{*}$-algebra $A = C(X,\mathbb{C})$ of continuous complex maps. As the points of the reconstructed space we take the maximal ideals in $A$, and declare that the closed sets are the ideals of $A$ (please correct me if I am getting closed/open and ideal/filter wrong here, I always do). Even though at the end of the day this topology turns out to be metrizable, that is entirely orthogonal to how the topology is defined and thought of.

-
 I certainly don't want to be closed minded. In contrast to metric spaces I find topological spaces extremely hard to understand, I can see the trees but I can't see the forest; I am seeing the definitions and the theorems but I am not seeing the ideas that are running underneath them. Perhaps the reason for this is that I don't have enough experience with non-metrizable topologies. I was looking for a heuristic that was more precise than "topological spaces generalize metric spaces". – teil Apr 18 2010 at 12:39