I'm trying to learn some topology as a hobby, and my understanding is that all manifolds are examples of topological spaces. Similarly, all metric spaces are also examples of topological spaces. I want to explore the relationship between metric spaces and manifolds, could it be that all manifolds are examples of metric spaces?

  • $\begingroup$ I doubt it because I think you can make manifolds where points would have to be infinitely far apart from each other. $\endgroup$ Commented Nov 15, 2015 at 11:42
  • $\begingroup$ @GregoryGrant my knowledge is small, but isn't a manifold defined as a metric space in which every local neighborhood of elements in the metric space is homeomorphic to $\mathbb{R}^n$ ? I should add, I'm coming from the idea of a metric space as taught/used in differential geometry, and I'm quite unlearned in topology of any kind. $\endgroup$
    – galois
    Commented Nov 18, 2015 at 9:34
  • $\begingroup$ @jaska Actualy a manifold is a topological space where each point has a neighborhood that's homeomorphic to $\Bbb R^n$ for some fixed $n$. Not all topological spaces are metric spaces, but all metric spaces are topological spaces. $\endgroup$ Commented Nov 18, 2015 at 11:44
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    – user170039
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4 Answers 4


...all manifolds are examples of topological spaces.$\newcommand{\Reals}{\mathbf{R}}$

This is true by definition; there are numerous formal definitions of a manifold, but all explicitly refer to a topology.

Similarly, all metric spaces are also examples of topological spaces.

In a hair-splitting sense this is arguably not quite correct. Instead, a metric space $(X, d)$, i.e., a non-empty set $X$ together with a function $d:X \times X \to \Reals$ satisfying the axioms of a metric, is naturally associated to a topology: Take $T$ to be the topology generated by the family of open balls in $(X, d)$.

...could it be that all manifolds are examples of metric spaces?

Now we're in less hair-splitting territory: A manifold does not come equipped with a "natural" metric (even if we assume the metric induces the manifold topology), so the literal answer is "no".

That said, you can ask whether or not every manifold $(M, T)$ admits a topological metric $d$. The answer is "it depends on your definition".

  • If your definition of "manifold" doesn't stipulate second countable (the topology is generated by countably many open sets), there exist manifolds such as the long line whose topology is not metrizable. Similarly, a non-Hausdorff manifold is not metrizable.

  • If your definition stipulates $(M, T)$ is Hausdorff, second-countable, and paracompact, then "yes, in the sense that every manifold admits a topological metric inducing the topology $T$". If $(M, T)$ is connected and smooth, a topological metric can be "constructed" by using paracompactness and a partition of unity to induce a Riemannian metric on $M$, and defining the distance $d(p, q)$ to be the infimum of the lengths of piecewise geodesics joining $p$ to $q$.

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    $\begingroup$ A Hausdorff, second countable and locally compact space (which is implied by locally euclidean) is automatically paracompact, so you don't need to ask paracompactness in addition to manifolds (or you may, but without asking it to be second-countable). $\endgroup$
    – Pedro
    Commented Feb 13, 2016 at 18:52
  • $\begingroup$ How are you defining "topological metric"? A metric on a topological space whose open balls generate the topology? $\endgroup$
    – tparker
    Commented Jun 2, 2018 at 3:55
  • $\begingroup$ @Pedro I was wondering about paracompactness in Andrew D. Hwang's answer. Indeed Hausdorff, second countable and locally Euclidean (with or without the same dimension for each point) is indeed paracompact right? $\endgroup$
    – user636532
    Commented Aug 5, 2019 at 3:21
  • $\begingroup$ Super helpful answer. $\endgroup$ Commented Feb 27, 2021 at 16:31

It depends on whom you ask. As a general topologist I would say no, there are manifolds (locally Euclidean (of a fixed dimension) Hausdorff spaces) that are not metrisable, like the long line. These can be quite interesting.

But in many cases, manifolds are assumed to be second countable (or some equivalent definition) and often also connected as well. And in such a context all manifolds are special metric spaces.

So it depends on the context. If you want differentiable manifolds (or smooth manifolds etc.) we almost always go into a setting where everything is supposed to be metrisable. And this is where they most often arise.


Assuming the usual definition of a topological manifold as a locally Euclidean space which is both Hausdorff and second-countable, it turns out that every manifold $M$ is a metrizable space. That is, you can put a metric on $M$ which induces the topology of $M$. This follows from example from Urysohn's metrization theorem. The metric is highly non-unique and a manifold doesn't come with a preferred metric which turns it into a metric space.


This is too long for a comment. You will probably think it is pedantic, but I promise it isn't!

In mathematics, we make a distinction between structures and properties. A metric space is a structure, and "being a manifold" is a property that a topological space can have. Therefore, the question doesn't make sense. (In practice, some working mathematicians are less careful with this distinction than others!)

More detail:

A topological space is a structure. If someone comes up to you and say

Look at this set. Is it a topological space?

you should reject this question as meaningless. They haven't told you what subsets are supposed to be open.


Is this topological space a metric space?

is a meaningless question. However, the following

(*) Is there a metric on the underlying set of this topological space whose associated topological space is this one?

is a perfectly reasonable question.

A less wordy thing to say is

Is this topological space metrizable?

when we mean (*). Here, metrizable is a property.

A manifold is a certain type of topological space, which is to say, it is a topological space with the property that each That is, if someone comes up to you and says

Look at this topological space. Is it a manifold?

then in fact they have asked you a reasonable question. The answer is yes if for all points of that space, there exists an open neighborhood homeomorphic to an open subset of $\mathbb{R}^n$ and no otherwise.

So the right question is

Is every manifold metrizable?

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    $\begingroup$ You can define “manifold-ness” just as well as a structure, instead of a property: a manifold is a topological space $M$ plus a mapping that associates every point in $M$ with some open subset $\Omega$ of $M$, an open subset $R$ of some $\mathbb{R}^n$, and a homeomorphism between $\Omega$ and $R$. $\endgroup$ Commented Nov 15, 2015 at 16:08
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    $\begingroup$ @leftaroundabout: Is the unit sphere in $\mathbb{R}^n$ a manifold, according to your definition? It seems to me that it's not, because it doesn't come equipped with a mapping of the kind you demand. $\endgroup$
    – Vectornaut
    Commented Nov 15, 2015 at 22:51
  • $\begingroup$ @Vectornaut In fact $\Bbb R^n$ itself is not equipped with such a mapping. The structure vs. property distinction pendulum swings more to structure, however, as soon as we speak e.g. of smooth manifolds, which means that we have a topological space together with an atlas such that ... With this in mind, I think it is reasonable to view manifolds as structures rather than properties. $\endgroup$ Commented Nov 15, 2015 at 22:58
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    $\begingroup$ @HagenvonEitzen: Personally, I find it very worthwhile to keep in mind that being a topological manifold is a property, while being a smooth manifold is a structure. Perhaps the distinction is not worthwhile to everyone, though. $\endgroup$
    – Vectornaut
    Commented Nov 15, 2015 at 23:03
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    $\begingroup$ @leftaroundabout, if I want to think of the unit sphere as a metric space, I absolutely do have to tell you what metric I'm using. Indeed, I can think of at least two obvious and quite different choices of a metric on the unit sphere. $\endgroup$
    – Vectornaut
    Commented Nov 15, 2015 at 23:24

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