If $(X,d)$ is a metric space, then we know the topology generated by the set $$\{ B_d (x,\delta) : \text{$x \in X$ and $\delta > 0$} \}$$ is called the metric topology.

Then how do you call the coarsest topology on $X \times X$ such that the function $d:X \times X \to \mathbb R$ is a continuous function?

Is there any relationship between those two topologies?

  • $\begingroup$ I would call it the topology on $X\times X$ induced or initialized by $d$. $\endgroup$ – drhab Aug 16 '18 at 7:52

I don't think there is a specific name for the coarsest topology on $X \times X$ for which $d : X \times X \to \mathbb{R}$ is continuous. In general if $f$ is a function from a set $Y$ to a topological space $Z$, the coarsest topology on $Y$ for which $f$ is continuous is called the topology on $X$ generated by $f$. Therefore we may call it the topology on $X \times X$ generated by $d$.

For the remainder let's fix the following notations:

  • $\mathcal{O}_d$ is the topology on $X \times X$ generated by $d$;
  • $\mathcal{O}_{\mathrm{m}d}$ is the metric topology on $X$;
  • $\mathcal{O}_{\mathrm{p}d} \sim \mathcal{O}_{\mathrm{m}d} \otimes \mathcal{O}_{\mathrm{m}d}$ is the "metric product topology" on $X \times X$.

$\mathcal{O}_{d}$ is generated by the sets of the form $$d^{-1} [ ( a , b ) ] = \{ ( x , y ) \in X \times X : a < d (x,y) < b \}$$

Note that since $\mathcal{O}_d$ is a topology on $X \times X$ and $\mathcal{O}_{\mathrm{m}d}$ is a topology on $X$, the two topologies cannot be compared. We can, however, compare $\mathcal{O}_d$ and $\mathcal{O}_{\mathrm{p}d}$. Of course, $\mathcal{O}_{d}$ is coarser than $\mathcal{O}_{\mathrm{p}d}$, since $d$ is continuous with respect to $\mathcal{O}_{\mathrm{p}d}$.

These two topologies will differ greatly (except in trivial cases).

For instance, if $X$ has at least two points, then $\mathcal{O}_{d}$ is not Hausdorff (it's not even T0). This is because for all $x , y \in X$ there is no open set containing exactly one of $(x,x)$ or $(y,y)$. On the other hand, $\mathcal{O}_{\mathrm{p}d}$ is always Hausdorff (even perfectly normal, since it is itself a metrizable topology).


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