So, I've stumbled across a definition of a metric space that has given me some pause. The definition of a Metric Space that I've always used has been the following:

A Metric Space is a set $X$ together with a function $d: X \times X \to \mathbb{R}$ satisfying the standard metric axioms.

However, while reading Munkres' Topology, he presents the following definition of a metric space:

A Metric Space is a metrizable space $X$ together with some specific metric $d$ that gives the topology of $X$.

The reason this definition has given me pause is not only the fact that I've never seen it but that the first definition of a metric space has no assumption that $X$ must be metrizable. In fact, it doesn't even assume $X$ is a topological space. Now, of course, with a metric $d$ on $X$, we may equip $X$ with the metric topology induced by $d$ and make $X$ a topological space, but it seems to me each of the two definitions require us to know quite different things about $X$ before calling it a metric space. The first definition seems more "bare-bones" in a sense while the second requires us to know that $X$ possesses certain kinds of structures/properties. Are these two definitions, in fact, equivalent, or if they are inequivalent, are there other reasons at play for desiring an alternative notion of a metric space?

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    $\begingroup$ Munkre's defintion looks somewhat convoluted, but he means exactly what the other definition says. $\endgroup$ – Kabo Murphy Aug 18 at 23:41
  • $\begingroup$ It would be useful to know exactly where in Munkres you find this quotation. $\endgroup$ – Lee Mosher Aug 19 at 0:01
  • $\begingroup$ I'm confused. Aren't they equivalent for exactly the reason you gave, namely if you have a metric, you can make $X$ into a topological space? $\endgroup$ – mathworker21 Aug 19 at 0:04
  • $\begingroup$ @LeeMosher Munkres' definition is given in section 20 of his book (his introduction to the metric topology). I would say it's page 118 in my copy of the book, but this might not be helpful since I know there are multiple versions floating around! $\endgroup$ – user516079 Aug 19 at 0:09
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    $\begingroup$ Formally, a metric space by your first definition would be an ordered pair $(X,d)$ (and one could subsequently construct the metric topology $T$ that corresponds to $d$), whereas by your second definition it would be an ordered triple $(X,T,d)$ (and $T$ is the metric topology that corresponds to $d$). What I take away from this is that Munkres wishes to put the topology on equal footing with the metric $d$. I don't think this definition has been adopted by the general mathematical public. But I can see its appeal nonetheless. $\endgroup$ – Lee Mosher Aug 19 at 0:16

This community wiki solution is intended to clear the question from the unanswered queue.

Your question has been answered in Lee Mosher's second comment.

The difference is that a metric space in the standard definition is a pair $(X,d)$ with a set $X$, whereas in Munkres' definition it is a pair $(X,d)$ with a topological space $X$. As Lee Mosher remarked, in the second case one should more precisely write it as a triple $(X, \mathfrak T,d)$ with a set $X$ and a topology $\mathfrak T$ on $X$.

There is a $1$-$1$-correspondence between "standard pairs" and "Munkres triples". In fact, the functions $$(X, \mathfrak T,d) \mapsto (X,d), \\(X,d) \mapsto (X, \mathfrak T_d,d) ,$$ where $\mathfrak T_d$ is the metric topology generated by $d$, are inverse to each other. It is therefore a matter of taste which definition you prefer.

Perhaps Munkres intention is to focus on the concept of a metrizable space. This is a space, not a set, with a certain property. This might be also the reason why he explicitly says that a metric space is a metrizable space together with some specific metric $d$ that gives the topology of $X$. The word "metrizable" could of course be omitted.


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