Every metric space is associated with a topological space in a canonical way. According to this source, this amounts to a full functor from the category of metric spaces with continuous maps to the category of topological spaces with continuous maps.

Is it possible that there exists another way of obtaining a topological space from a metric space that is equally deserving of the label "canonical"? Perhaps something that no one has thought of yet? To say it in a different way, is there a sense in which the aforementioned functor is unique? Lets assume that the morphisms between metric spaces are precisely the continuous maps, although an answer that considers other morphisms between metric spaces is welcome.

Now obviously this is a soft question, as I have neglected to specify what it means for a map to be deserving of the term "canonical." For this reason, let me motivate the question a little.

At some point in an introductory work on analysis, the author will define the meaning of the expression "the topological space associated with (or induced by) a metric space." I'd like to know if this definition is in some sense "the unique correct definition," or whether it is only "one of many possible."

EDIT: Lets put this another way. Obviously there are many function $\mathsf{Met} \rightarrow \mathsf{Top}$, but most of them are pretty boring. So we can restrict ourselves to functors, where $\mathsf{Met}$ and $\mathsf{Top}$ are viewed as categories (technically we need to make also specify what the morphisms of $\mathsf{Met}$ should be.) Anyway, as Martin points out, we're still going to be left with lots of "boring" functors. So I guess the question is, how do we get rid of all the "boring" ones? And once we do, is the canonical functor the only one that's left? Obviously I haven't defined "boring" so this is a very soft question.

Magma suggests the following refinement of the question: does the canonical functor satisfy suitable a universal mapping property?

Here's another angle. Suppose we run into an alien species, which studies topological spaces (and calls them topotopos, and what we would call "an open set of a toplogical space" they call "an openopen of a topotopo"). They also study metric spaces (and call them metrometros.) We send that species a message asking them about the "the openopens of a metrometro." Will their notion of open set of a metric space coincide with our notion? And if so, why?

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    $\begingroup$ The functor is only full, by definition, when you take continuous maps as morphisms between metric spaces. But this is somewhat unnatural. A better choice of morphisms is isometric / Lipschitz / short maps. Instead, any reasonable choice yields a faithful functor $\mathsf{Met} \to \mathsf{Top}$. Regarding your question: Do you want to keep the underlying set and set maps? That is, are you interested in functors $\mathsf{Met} \to \mathsf{Top}$ which commute with the forgetful functors to $\mathsf{Set}$? $\endgroup$ – Martin Brandenburg Jan 30 '13 at 12:05
  • $\begingroup$ Okay that's a good point. But, should I modify my question? $\endgroup$ – goblin Jan 30 '13 at 12:08
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    $\begingroup$ In any case, there are of course lots of functors $\mathsf{Met} \to \mathsf{Top}$ over $\mathsf{Set}$, especially a lot of boring ones. I am pretty sure that you want to exclude them, so that you should specifiy your question a little bit. Namely, what do you expect from your alternative functors? $\endgroup$ – Martin Brandenburg Jan 30 '13 at 12:10
  • $\begingroup$ Well ideally, I don't want them to exist! How do you do that font that you used for Met and Top and Set? $\endgroup$ – goblin Jan 30 '13 at 12:17
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    $\begingroup$ I think that this , like similar unique or special things in category theory, should be restated (in this case) as follows: does this canonical functor satisfy a suitable UMP (universal mapping property)? If we want to use category theory , we must use the language of category theory. So we must talk in terms of universals (UMP), adjoints, Kan extensions, limits, ect. The base concept of all these constructions is UMP. Thus my restatement of the question. $\endgroup$ – magma Jan 30 '13 at 13:06

Let $(X,d)$ be a metric space. Then the canonical topology $\tau_{can}$ is the coarsest topology on $X$ which makes (with the product topology) $d : X \times X \to \mathbb{R}$ continuous.

Proof: It follows from the triangle inequality that $d$ is continuous in the canonical topology (in fact short when we endow $X \times X$ with the sum metric). Conversely, if $\tau$ is a topology such that $d$ is continuous with respect to $\tau$, then in particular for every $x_0 \in X$ the map $X \cong X \times \{x_0\} \to X \times X \xrightarrow{d} \mathbb{R}$ is continuous. The preimage of $(-\infty,r)$ is the open ball of radius $r$ and center $x_0$. This shows $\tau_{can} \subseteq \tau$.

It follows that the canonical forgetful functor $\mathsf{Met} \to \mathsf{Top}$ is terminal among all functors $F : \mathsf{Met} \to \mathsf{Top}$ over $\mathsf{Set}$ such that $d : F(X) \times F(X) \to \mathbb{R}$ is continuous for all $(X,d) \in \mathsf{Met}$. The initial such functor is given by the discrete topology.

So I am pretty sure that aliens will come up with the same topology.

  • $\begingroup$ Great answer! One question: where does the topology on $\mathbb{R}_+$ come from? I think it's the coarsest topology such that "non-strict" intervals of the form $(a,b) = \{x \in \mathbb{R}_+ | a < x < b\}$ are open, for all $a,b \in \mathbb{R}_+$. Can we prove this? $\endgroup$ – goblin Jan 31 '13 at 6:32
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    $\begingroup$ Yes, the euclidean topology on $\mathbb{R}$ coincides with the order topology. This follows immediately from the definitions. And $\mathbb{R}$ is the unique complete ordered field. $\endgroup$ – Martin Brandenburg Jan 31 '13 at 10:10
  • $\begingroup$ Aha, this last sentence can answer my biggest doubt about alien's metric spaces, is whether they use the same real numbers.. $\endgroup$ – Berci Jan 31 '13 at 11:41
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    $\begingroup$ Well they may also take $\mathbb{R} \cup \{\infty\}$ as the values of the metric. Especially when they travel with the speed of light, then I won't be surprised. $\endgroup$ – Martin Brandenburg Jan 31 '13 at 14:20

Partial answer for your 'alien' question.

If we assume that aliens also think about open sets as the complements of closed sets, and if we also agree with them that being closed means being closed under limits of sequences, and that $\lim(x_n)=x$ in metric space is defined as $\lim d(x_n,x)=0$, and they also use the same real numbers, then we will also agree on what open sets are.


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