TITLE: How is the metric topology the coarsest to make the metric function continuous?
QUESTION [11 upvotes]: Let $X$ be a metric space with metric $d$. If $\mathcal{T}$ is a topology on $X$ such that the function $d\colon X \times X \to \mathbb{R}$ is continuous, then how to show that $\mathcal{T}$ is finer than the topology induced by the metric $d$?
In other words, how to prove that if $X$ has a metric $d$, then the topology induced by $d$ is the coarsest topology relative to which the function $d$ is continuous?

REPLY [2 votes]: It suffices to show that if $\{x_i\}_{i\in I}$ is a net converging to $x$ in $\mathcal{T}$ then $\{x_i\}_{i\in I}$ converges to $x$ in $d$ (because then any set closed in $\mathcal{T}$ will be closed in the metric topology). But if $x_i \rightarrow x$ in $\mathcal{T}$ then $(x_i,x) \rightarrow (x, x)$ in $\mathcal{T}\times \mathcal{T}$ and so, since $d$ is continuous w.r.t. $\mathcal{T}\times \mathcal{T}$, we have $d(x_i,x)\rightarrow d(x,x)=0$, i.e. $x_i \rightarrow x$ in $d$. I want to mention that it's often simpler to deal with nets than open sets in topology and I think this problem is a good example of that.