Does it make sense to define a "metric topological space" $(M, d, \tau)$

Question

When doing things related to compactness, sometimes you have to switch definition from sequential compactness which is defined on a metric space $(M, d)$, to things related to covering compactness which is defined on $(M, \tau)$ (or $(M,d)$?)

I find it troublesome to switch between "Let $A \subset (M, \tau)$" and "let $A \subset (M, d)$ ". Why don't we simulatenously work in both spaces. Then I remembered that a metric $d$ is not a topology, and $\tau$ is not a metric. So is it possible to define this triple $(M, d, \tau)$? If so it usual to do so?

Answer 1

Both sequential and covering compactness are purely topological notions. Moreover, if the metric $d$ induces the topology $\tau$ then there is no confusion.

Answer 2

You can define $(M,d)$ to be a metric space and let $\tau$ be the induced topology on $M$, but $(M,d)$ suffices for common purposes.

Answer 3

Unless specified otherwise, one usually associates the topology which is induced by the metric. That means if we have a metric space $(M,d)$, one implicitely views this as the topological space $(M,\tau)$ (if needed), where $\tau$ is the topology induced by the metric $d$, i.e. $\tau = \{U(x,r),:x\in M, r>0\}$ where $U(x,r)=\{y\in M: d(x,y)<r\}$.

The same is true for stronger structures than the metric, i.e. one views the normed space $(M,\|\cdot\|)$ as the metric space where $d(x,y)=\|x-y\|$ and the inner procuct space $(M, \langle \cdot,\cdot\rangle)$ as the normed space where $\|x\|=\sqrt{\langle x ,x \rangle}$.