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I think this is true but i cannot prove it. Any answer or hints are welcome.

I have tried to start with $\mathbb{R}$ with euclidean metric. We may consider $\tau :=\{\emptyset,\mathbb{R}\}$ and obvious it is a topology in $\mathbb{R}$. but in general, there are many different kinds of metric spaces so I am not sure how to prove it.

Martin Sleziak
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user56876
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  • I think all metric spaces are topological spaces But converse is not true –  Jun 15 '13 at 13:09

4 Answers4

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As stated the answer is 'no'. A metric space is not a topological space. However, every metric space gives rise to a topological space in a rather natural way. This is the well known construction that takes a metric space $X$ and constructs the topology on $X$ where a set $U$ is open precisely when for every $x\in U$ there exists some $e>0$ such that the open ball $B_e(x)$ is contained in $U$.

Several comments are due. First, this process loses information. For instance, there exists infinitely many metrics on $\mathbb R$ such that all of them produce the same topology of open balls. So, only knowing the induced topology does not allow you to recover the metric.

The construction mentioned is most clearly understood in the context of the categories of metric spaces and topological spaces. Let $Met$ be the category of all metric spaces and continuous mappings and let $Top$ be the category of topological spaces. The construction above is the object part of a functor $Met\to Top$, it sends any function between metric spaces to itself considered as a function between the associated topological spaces. Now, this functor is trivially faithful but interestingly it is full. This last property says that a function $f:X\to Y$ between metric spaces is continuous via the usual $\epsilon-\delta $ definition if, and only if, the same function $f$ considered now to be between the associated topological spaces is continuous (in the sense that the inverse image of an open is open).

This last remark shows why the topology of open balls is the one most commonly used. It establishes a strong relation between metric spaces and topological space. However, this resulting functor is not an equivalence of categories. It does not even have a left or a right adjoint. This failure of the functor $Met\to Top$ to have any sort of inverse is a way to measure (or see) how different metric spaces are from topological spaces.

Later Addition: As it turns out, metric spaces and topological spaces are equivalent, if metric is interpreted broadly enough. The details can be found here, alg. univ. (to appear).

Ittay Weiss
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  • Ittay, just to double check, would you similarly say that a group is not a semigroup? – alancalvitti Jan 11 '13 at 04:50
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    Of course a group is a semigroup. Every group satisfies (with the same operation) the axioms of a semigroup. Also, the category of semigroups is fully embedded in the category of groups. In contrast, the question whether a metric space satisfies the axioms of a topology doesn't even make sense. And, the category of metric spaces does not embed in the category of topological spaces. – Ittay Weiss Jan 11 '13 at 04:58
  • Are groups and semigroups equivalent as categories? – alancalvitti Jan 11 '13 at 05:52
  • no, they are not equivalent as categories. – Ittay Weiss Jan 11 '13 at 08:24
  • You are entitled to take a category theoretic point of view, of course - but it is not the only view to be taken. There is a historical point of view - metrics were important in mathematics and science long before topologies were conceived. Topologies grew out of attempts to formalise continuity in a metric context and beyond. Then there is a pedagogical point of view - for most young mathematicians the open ball topology in a metric space is one of the paradigmatic examples, and rightly so, because of its utility and range of applications. – Mark Bennet Jan 11 '13 at 08:41
  • I completely agree with you Mark. It is important to understand how metric spaces and topological spaces are related. These are not identical concepts nor is one an extension of the other. These are different notions that are related (regardless of whether one takes a categorical point of view or not). – Ittay Weiss Jan 11 '13 at 09:08
  • @IttayWeiss, (Arkhangelskii said topology is the study of nearness without distance) I believe topological spaces are extensions of metric spaces: a metric space determines the topology. However a metrizable topology might not be uniquely metrizable. What do you mean by extension? I mean it in the sense of semigroups are extensions of group. Categorical equivalence is not necessary (why did you mention cat equivalence?) – alancalvitti Jan 11 '13 at 15:14
  • where did I mention categorical equivalence? in what sense then are topological spaces an extension of metric spaces? if you can be precise about what 'extension' means to you in that context. – Ittay Weiss Jan 11 '13 at 18:19
  • @IttayWeiss, in your last paragraph: "However, this resulting functor is not an equivalence of categories" – alancalvitti Jan 19 '13 at 00:42
  • @IttayWeiss, in the same way that semigroups are extensions of groups. There is a unique interpretation: groups are a subcategory of semigroups (I believe the standard definition has them as semigroups with a single idempotent), so metric spaces are a subcategory of topological spaces. – alancalvitti Jan 19 '13 at 00:44
  • @alancalvitti I mentioned equivalence of categories (and the lack of any adjoints) since it is a nice way to see the differences between the two categories. – Ittay Weiss Jan 19 '13 at 02:48
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    @alancalvitti To consider a category $D$ and extension of $C$ it should be the case that $C$ embeds in $D$, not just be a subcategory thereof. The category of groups embeds in the category of semigroups. But the category of metric spaces does not embed in the category of topological spaces. – Ittay Weiss Jan 19 '13 at 02:50
  • @IttayWeiss, thank you, I wasn't aware of this distinction. Just to be sure we're talking about the same sense of 'embed' do you mean a faithful functor? (Even if so there's at least 2 ways of defining faithful). – alancalvitti Jan 19 '13 at 03:03
  • @alancalvitti you are right that there are several ways to give a precise meaning, and faithful and full will certainly have to be imposed. Some reasons not to consider Met --> Top an embedding is that it identifies non-isomorphic objects. Another problem is the choice of morphisms to take when forming Met. The 'correct' notion as far as metric space theory is concerned are the non-expanding maps and not the continuous ones (according to Gromov, not me). With these morphisms Met-->Top is not full. These issues do not occur for Grp-->SemiGrp. – Ittay Weiss Jan 19 '13 at 03:16
  • @IttayWeiss, these considerations may be worth asking as a new Q here, given the fundamental importance of these structures (I for one work in data analysis with finite metric spaces where neighborhood graph preserving maps are more important arrows than non-expansive ones). But I don't get the non-isomorophic identification part: iso's in $\bf Met$ are isometries. So of course you will expect that some non-isometric spaces may be isomorphic in $\bf Top$ - correct? What am I missing? – alancalvitti Jan 19 '13 at 18:24
  • @alancalvitti yes, some non-isometric metric spaces become isomorphic in Top showing that the induced topology does not capture all aspects of the metric. It looses a lot of information. The passage from groups to semigroups does not loose any information. You would expect an extension to not loose any information. – Ittay Weiss Jan 19 '13 at 21:51
  • @IttayWeiss, I asked a separate Q about $\bf Group \to \bf Semigroup$. Also, Edgar Mathias' answer is interesting: there is not a unique topology associated with a metric space. I did not consider that possibility. – alancalvitti Jan 21 '13 at 19:52
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Yes. In a metric space $(X,d)$, the metric defines the open sets as follows: A set $U$ is considered open iff for all $x \in U$, there exists some $\epsilon>0$ such that $B(x,\epsilon) \subset U$. ($B(x,\epsilon) = \{y \mid d(x,y) < \epsilon \}$). The topology induced by the metric consists of these open sets.

You need to prove that this is indeed a topology.

Michael Hardy
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copper.hat
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    I don't think it is correct to say that the metric define the open sets. The information giving a metric space does not mention any open sets. You can use the metric to define a topology, granted with nice and important properties, but a-priori there is no topology on a metric space. – Ittay Weiss Jan 11 '13 at 04:16
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    @IttayWeiss: I think you're being too harsh. As you mentioned in your own post, the open ball topology is functorial, and it's (as far as I'm aware) the only topology one ever considers when one wants to make topological-style arguments about metric spaces. Historically, I believe that topological spaces were defined precisely to generalize metric spaces with their open-ball topologies. So it's not like this topology is coming out of thin air. – Paul VanKoughnett Jan 11 '13 at 04:42
  • Of course it's an important topology but still you can't speak of the open sets in a metric space. And certainly, the existence of a natural choice of a topology that nicely captures continuity does not make a metric space the same thing as a topological space. – Ittay Weiss Jan 11 '13 at 04:45
  • If a metric space (X,d) is the same as the associated open ball topology then every question about the original metric supposedly can be answered just by looking at the topological space. This is of course impossible. – Ittay Weiss Jan 11 '13 at 05:06
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    @IttayWeiss: You are nit picking. It is fair to say, and not misleading in the slightest, that metric spaces are topological spaces. In "Real & complex analysis", Rudin writes "The most familiar topological spaces are the metric spaces". Kantorovich & Akilov write "One important class of topological spaces is the class of metric spaces". In "Introductory Real Analysis", Kolmogorov & Fomin write "Metric spaces are topological spaces of a rather special (although very important) kind". These are quotes from the first three analysis books I picked from my library. – copper.hat Jan 11 '13 at 07:08
  • In the answer I gave I tried to explain precisely why metric spaces are not the same as topological spaces and to explain how one can think of a metric space topologically and give a precise way of seeing the difference between the two concepts. I have no doubt that the great minds whom you quote understood these issues very well. I do think that it is important to make the distinction precise and today with the language of CT it's quite easily done without any nit picking. Just clear definitions and constructions. – Ittay Weiss Jan 11 '13 at 08:29
  • I am afraid the distinction in categorical terms is lost on me. Clearly a metric space has more structure than a topological space, just as a Lipschitz function has more 'structure' than a continuous function. They are clearly not equivalent concepts. But to state that a metric space is not a topological space is, at best, misleading. – copper.hat Jan 11 '13 at 08:50
  • A Lipschitz mappings is continuous because you can prove that the condition of being Lipschitz implies the condition of continuity holds. However, for a metric space you can't even start showing that the axioms of a topological space hold, not before you define an associated structure with that metric space. The fact that it is in some sense a canonical association does not mean the metric space is the same as the associated topological space. It just means you can canonically associate a topological space to every metric space. I.e., a functor Met-->Top. – Ittay Weiss Jan 11 '13 at 09:11
  • moreover, the sentence "the metric defines the open set" is, I believe, wrong and misleading. There are no open sets before you define the ones defined by the metric via open balls. And then saying that the metric defines the topology you just defined using the metric is circular. – Ittay Weiss Jan 11 '13 at 09:19
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    I think you are splitting nanotubes. – copper.hat Jan 11 '13 at 09:33
  • I came here after leaving an answer to a closely related question: http://math.stackexchange.com/questions/355214/. To copper.hat: Ittay Weiss is quite correct to say that a metric space is not a topological space. Similarly I would say that a topological space is not a set: in each case there is an additional structure. But @Ittay Weiss: it is too strong to say that "you can't speak of the open sets in a metric space". Proof: people speak this way all the time, including some who don't know what a topological space is. – Pete L. Clark Apr 08 '13 at 21:15
  • From a categorical perspective, the point is that everyone who knows about metric spaces and topological spaces knows about the functor $(X,d) \mapsto (X,\tau_d)$ from metric spaces to topological spaces. Isn't this the necessary and sufficient condition for speaking about anything: a commonly understood context? – Pete L. Clark Apr 08 '13 at 21:18
  • Is the point here that usual definition of a metric space defines a metric, but doesn't explicitly define the naturally induced topology? – copper.hat Apr 08 '13 at 21:49
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For every metric space $(X,d)$ there is a natural induced topological structure. This natural choice of topology $\tau$ is given by the topology generated by open balls. That is, $\{ B(x,r) : x \in X, r > 0\}$ forms a subbase for $\tau$.

The topology is called induced because the open sets determined by $d$ and the open sets determined by $\tau$ agree.

nullUser
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    It is not enough to show that you can associate a topological space with a metric space to show that a metric space is a topological space. After all, I can also associate a pink elephant with a metric space. That doesn't show that a metric space is a pink elephant. – Ittay Weiss Jan 11 '13 at 04:32
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    More a red herring than a pink elephant. – copper.hat Jan 11 '13 at 07:10
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You can equip a space metric with various topologies. The most common is generated by the balls. Take the the lower topology containing the balls. Also you can take a trivial topology like you said in $\mathbb R$. Is the same.

  • That's interesting Edgar. I did not consider the possibility of multiple topologies from one metric space. That complicates the issue - as undergrads we're all taught a metric space is a topological space, not multiple nontrivial ones. Also, what do you mean by trivial topology? Both the discrete and indiscrete topologies are considered trivial. – alancalvitti Jan 21 '13 at 19:54
  • I think you need to understand why the Ittay answer is not. He is right. With trivial topology I mean ${ \varnothing,X}$ –  Jan 22 '13 at 01:42
  • "why the Ittay answer is not." -- not what? – alancalvitti Jan 22 '13 at 01:49
  • @alancalvitti sorry. I wanted to say that the answer is no. –  Jan 22 '13 at 10:34
  • Ittay provided part of the answer (lack of embedding $\bf Met \to \bf Top$, and you provided the other part. Also, I meant that in addition to the indiscrete topology, one could also endow the discrete topology (typically also considered trivial). But you're right there's many not one. – alancalvitti Jan 22 '13 at 16:13