# Are all metric spaces topological spaces?

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.

-
I think all metric spaces are topological spaces But converse is not true – user82573 Jun 15 '13 at 13:09

## 4 Answers

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, just to double check, would you similarly say that a group is not a semigroup? – alancalvitti Jan 11 '13 at 4:50
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 4:58
Are groups and semigroups equivalent as categories? – alancalvitti Jan 11 '13 at 5:52
no, they are not equivalent as categories. – Ittay Weiss Jan 11 '13 at 8:24
@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 2:50

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.

-
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 4:16
@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 4:42
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 4:45
@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 7:08
I think you are splitting nanotubes. – copper.hat Jan 11 '13 at 9:33

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.

-
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 4:32
More a red herring than a pink elephant. – copper.hat Jan 11 '13 at 7:10

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 $\{ \varnothing,X\}$ – user52188 Jan 22 '13 at 1:42
"why the Ittay answer is not." -- not what? – alancalvitti Jan 22 '13 at 1:49
@alancalvitti sorry. I wanted to say that the answer is no. – user52188 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