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Is there a difference between a metrizable space and a metric space ? For me if $X$ is a metrizable space, then there is a metric $d$ s.t. $(X,d)$ is a metric space. And obviously a metric space is metrizable.

The reason I'm asking this question is in the book Topology second edition of J. Munkres, page 179 they prove (theorem 28.2) that if the $X$ is metrizable, then a set is compact $\iff$ it's sequentially compact. Now page 276 (Theorem 45.1) they prove that a metric space is compact $\iff$ it's complete and totally bounded.

My question : In the proof of the theorem 28.2 in the part $(3)\implies (1)$ they in fact proved that the space is totally bounded, and then proved that it's moreover compact. But they introduced the totally boundness to prove theorem 45.1 only. I don't really understand why they introduce the concept of totally boundness that late (i.e. page 275) whereas it's exactly the technique used to prove theorem 28.2. So I guess that there is a subtlety that I didn't get between metric space and metrizable space (because for me it almost look to be the same thing, and the method used in theorem 28.2 should also work to prove theorem 45.1).

user349449
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  • Possible duplicate: https://math.stackexchange.com/questions/215986/there-is-no-difference-between-a-metrizable-space-and-a-metric-space-proof-incl?rq=1 – uniquesolution Aug 14 '18 at 12:46

4 Answers4

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A metric space is a metric space, so a pair $(X, d)$ where $X$ is a set of points and $d$ is a metric on $X$.

A metrizable space is a topological space, so a pair $(X, \tau)$ where the topology $\tau$ is nice enough that it is possible to find a metric $d$ (and thus infinitely many different metrics) on $X$ such that the induced topology on the metric space $(X, d)$ is exactly $\tau$.

Arthur
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  • I can see that my question is not that clear. But anyway : why in the Munkres, the concept of totally boundness has been introduced that late ? Because in theorem 28.2 they proved that sequencial compact $\implies $ totally bounded $\implies $ compact. For a metrizable space are there sets that are totally bounded for some metric and not totally bounded to other metrics ? I guess no because different metric in a space induce the same topology... – user349449 Aug 14 '18 at 12:56
  • @MathBeginner The real number line in the standard metric isn't totally bounded, but it is in the metric given by $d(x, y) = |\arctan(x)-\arctan(y)|$. These two metrics give rise to the same topology. – Arthur Aug 14 '18 at 13:04
  • @MathBeginner total boundedness and also completeness depends on the metric, so it's only defined for metric spaces, not for metrisable spaces. A metrisable space can have two compatible metrics, one of which is complete and the other not. – Henno Brandsma Aug 14 '18 at 18:30
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When speaking about a "metric space" it is usually assumed some concrete metric is given, whereas a "metrizable space" is a topological space whose topology can be shown to be equivalent to some topology induced by a metric. In some cases the metrizability of a space follows indirectly, without pointing out a specific metric.

uniquesolution
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If $(X,d)$ is a metric space, it is also a topological space. Just consider the topology $\mathcal{T}_d$ generated by the open balls for the distance $d$.

However, when you have a topological space $(X,\mathcal{T})$, it is not obvious that there exists some metric $d$ such that $\mathcal{T}_d = \mathcal{T}$. In fact, sometimes this is not the case. When it is possible to find such a metric, we say that the topological space $(X,\mathcal{T})$ is metrizable. Note that the notion of metrizability is always relative to some topology $\mathcal{T}$.

Augustin
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There is an unanswered question regarding the order of presentation. Why is total boundedness introduced so late? If "in metric spaces, compact iff sequentially compact" was done in p179, why wait till p276 to define total boundedness?

In Kolmogorov & Fomin,

  • the notion of total boundedness was introdued in p97,
  • "a metric space is compact iff it is totally bounded and complete" is proved on p100,
  • "the subset of a complete metric space is relatively compact iff it is totally bounded" is proved on p101.

I believe that this natural order of presentation is what you are looking for.

Sasha Queequeg
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