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Willard says, "Whenever $(X,\tau)$ is a topological space whose topology $\tau$ is the metric topology $\tau_{\rho}$ for some metric $\rho$ on $X$, we call $(X,\tau)$ a metrizable topological space."

I think giving a proof is the best way to illustrate how I think of these concepts and illustrate the exact points that I am not understanding.

Theorem: Metric space iff metrizable space.

(->) Let $(X,\rho)$ be a metric space. Consider the topology generated by this metric, $\tau_{\rho}$. Then $(X,\tau_{\rho})$ is a topological space whose topology is the metric topology for some metric, so by definition, metrizable.

(<-) Let $(X,\tau)$ be a metrizable space. There $\exists \rho$ a metric such that $\tau$ is the metric topology given by $\rho$. And so $(X,\rho)$ is a metric space.

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A metric space is not the same thing as a metrizable space. When we say metric space, we specify a particular metric. There are, for any metrizable space with more than $1$ point, infinitely many metrics that induce the topology. – André Nicolas Oct 18 '12 at 0:42
If you've done some linear algebra, think of it like changing basis. We know that picking two different bases for one vector space still gives us isomorphic vector spaces, and in the same way we can pick different metrics to get the same metrizable space. – Kevin Carlson Oct 18 '12 at 0:47
up vote 6 down vote accepted

Really formally: A topological space is a pair $(X, \mathcal{T})$ of a set $X$ and a subset $\mathcal{T}$ of the power set of $X$ satisfying the appropriate axioms for open sets. A metric space is a pair $(X, d)$ of a set $X$ and a map $d: X \times X \to \mathbb{R}^{\ge 0}$ satisfying the appropriate axioms for a distance function. It does not make sense to say a topological space "is" a metric space, or vice-versa, because there is an ontological status problem - how can $(X, \mathcal{T}) = (X, d)$ when $X$ and $\mathcal{T}$ are completely different objects? (See important caveat at bottom.)

If I give you a metric space $(X, d)$ then there is a canonical topological space $(X, \mathcal{T})$ with the same underlying set $X$, called the induced topological space, whose topology is generated by the $\epsilon$-balls.

If I give you a topological space $(X, \mathcal{T})$, there may or may not be some metric space $(X, d)$ whose induced topological space is $(X, \mathcal{T})$. If there is one, then we say that $(X, \mathcal{T})$ is metrizable. If $(X, \mathcal{T})$ is metrizable, then there is some metric $d$ such that $(X, d)$ is a metric space whose induced topological space is $(X, \mathcal{T}$, but note that for instance $(X, \frac{1}{2}d)$ also has this property, as does $(X, 7d)$, so there is absolutely nothing canonical about $d$, i.e. you can't recover $d$ from $\mathcal{T}$.

Caveat: Keeping $d$ and $\mathcal{T}$ around is horribly clunky and pretentious and nobody actually does it, and people say that a metric space "is a topological space" all the time, or vice-versa for metrizable topological spaces, e.g., I might say "the topological $\mathbb{R}^2$ is actually a metric space" when I mean "there exists a metric-space structure which induces the usual topology on $\mathbb{R}^2$." Being super-precise with ordered pairs is only useful when sorting out confusion like this one; it's not how to think about these things in practice.

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A metric space is an ordered pair $\langle X,d\rangle$ such that $X$ is a set and $d$ is a metric on $X$. A metrizable space is an ordered pair $\langle X,\tau\rangle$ such that $X$ is a set, $\tau$ is a topology on $X$, and there exists a metric on $X$ that generates the topology $\tau$. These are clearly not the same thing. A metric space has a specified metric, and the topology, though definable from the metric, is unspecified; a metrizable space has a specific topology, and while it is possible to define metrics that generate that topology, none is specified.

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