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.

couldbe 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 '10 at 19:58