Is every manifold a metric space?

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?

...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 $q$.

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.