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My reason for asking is that the theory of metric spaces is so clean and so many significant theorems can be proved for an arbitrary metric space (which makes it plausible to me that metric spaces are the right way to study the general theory of distance), but I get the impression that topology is much more confused and that you can't prove anything significant for a arbitrary topological space. If that is correct, how can a person be confident that topological spaces express continuity in its purest form?

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    $\begingroup$ Topological spaces are easier to study than metric spaces, and there are important examples of spaces with no metric or no natural metric, for instance Zariski topologies in algebraic geometry and weak topologies on function spaces. As for to "express continuity in its purest form"; is that any more than hot air? $\endgroup$ May 31, 2010 at 13:10
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    $\begingroup$ Who says that topological spaces are the "purest form" of continuity? There are plenty of generalizations of concepts like convergence and continuity (pertopological spaces, semiuniform convergence spaces etc.). Some important types of convergence can't be expressed as a topology, e.g. the continouos convergence of functions is not topologizable in general. $\endgroup$ May 31, 2010 at 13:10
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    $\begingroup$ Closed. It is not at all clear what you are asking, since you don't seem to have included a definition of "continuity" or "purity of form". $\endgroup$
    – S. Carnahan
    May 31, 2010 at 13:32
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    $\begingroup$ I disagree with the decision to close this question. But a better formulation, such as "what alternatives are there to the general notion topological space for the theory of continuity?" might be appropriate. $\endgroup$ May 31, 2010 at 13:33
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    $\begingroup$ @Omar, please come over to tea.mathoverflow.net, where you'll find a very strong consensus that "potentially leading to an interesting discussion" is not something particularly desirable about a mathoverflow question. We aim for precise questions, with identifiably correct answers. Admittedly "not a real question" is a bit misleading, but we're constrained somewhat by the software. In this case, see Scott C's comment for an explanation of why people didn't like this question. $\endgroup$ May 31, 2010 at 15:47

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Geometry and "theory of distance" are not the only motivation for general topology. Another motivation comes from information processing and computation.

Briefly, let us consider the notion of "observable subset" where by "observable" we mean "can be confirmed by a (finite-time) computation". More precisely, a subset $S \subseteq X$ of a datatype $X$ is computationally observable if there exists a program $p$ such that

for all $x \in X$, $p(x)$ terminates if, and only if $x \in S$.

Note the asymmetry between an observable subset and its complement. You can think of the fact that $p(x)$ terminates as giving you half a bit of information.

We make the following basic observations:

  1. The empty subset is observable by a program which never terminates.
  2. The whole set is observable by a program which always terminates.
  3. If $S$ and $T$ are observable by $p$ and $q$, respectively, then $S \cap T$ is observable by the program $p; q$ (execute $p$, wait for it to finish then execute $q$).
  4. If $(S_n)$ is a sequence of subsets observable by a (computable) sequence $p_n$ of programs, then $\bigcup_n S_n$ is observable by the program which runs $p_0, p_1, p_2, \ldots$ in parallel by dovetailing and finishes as soon as one of them does.

If we replace "observable" by "open" and generalize to arbitrary unions instead of just countable computable ones, we get the usual definition of topology.

You should convince yourself that it is unreasonable to expect that a countable intersection of observable subsets is observable: if it were, we could solve the Halting problem.

The topological spaces which arise in theoretical computer science from the point of view that "open = computationally observable" are typically not Hausdorff, let alone metrizable. They are however very nice spaces with lots of good properties.

You asked about continuity. Under the view I described we can show that computable maps are continuous in a very natural way: if $f : X \to Y$ is computable and $S \subseteq Y$ is observable by $p$ then $f^{-1}(S)$ is observable by $q(x) = p(f(x))$.

To summarize

"Observable sets are open, computable maps are continuous."

So yes, the usual definition of continuous maps as one whose inverse image preserves open sets is right. The definition of topology, however, could be discussed a bit further. There is something fishy about going from computable countable unions to general ones just like that.

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There is evidence that metric spaces are not sufficiently general nor sufficiently "canonical" to serve as a basis for our understanding of continuity.

Zariski topology, of common use in algebraic geometry, is a topology that is not separated. Hence it cannot be described with a distance.

In functional analysis, the weak topology on the dual of some function space is metrizable only if the function space itself is separable. This hypothesis may be undesirable (e.g. when dealing with the general theory of Hilbert spaces). Another example in functional analysis, is given by the space of Schwartz distributions $D'(R^n)$, which has a natural topology as the dual of $C^\infty_c(R^n)$. This topology is not metrizable.

Moreover, there is also an uniqueness problem. There may be different interesting non-equivalent distances on a space, that give the same topology. Take $R^2$ and consider the euclidean and non-euclidean distances. These distances generate the same topology, but they contain a lot of information that do not relate to topology alone. A distance carries in general more information than just facts related to continuity. Completeness, curvature (...) are properties that can be given a meaning for metric spaces, but do not depend on the topology alone. These properties can be used to distinguish between metric spaces generating the same topology.

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