Codomain of metric in Munkres

Question

In Munkres' Topology, a metric on set $X$ is a function $d: X \times X \to \mathbb{R}$ such that:

1. $d(x, y) \geq 0$ for all $x, y \in X$ and equality holds if and only if $x = y$.
2. $d(x, y) = d(y, x)$ for all $x, y \in X$.
3. $d(x, y) + d(y, z) \geq d(x, z)$ for all $x, y, z \in X$.

Why is it that the metric $d$ is defined to have codomain $\mathbb{R}$ instead of codomain $[0, +\infty) \subset \mathbb{R}$?

There's two reasons why I'm curious:

1. The condition of being a metric doesn't allow negative values for the metric.
2. If we change the codomain in the definition to be $d : X \times X \to [0, +\infty)$ and we show $d$ is a continuous function, then $[0, c)$ for any $c > 0$ is open in the subspace topology on $[0, +\infty)$ so the preimage $d^{-1}([0, +\infty))$ is an open set in $X \times X$ which contains points of the form $(x, x) \in X \times X$ for some $x$. If we keep the codomain as $\mathbb{R}$, then it seems awkward that to get points of the form $(x, x) \in X \times X$ in the preimage of an open set in $\mathbb{R}$ we have to include the preimage of negative values, e.g. $d^{-1}((-1, 1))$ for the open interval $(-1, 1) \subset \mathbb{R}$, but these negative values have an empty preimage.

Edit per comments: changed from range to codomain per comment!

Answer 1

It's just a matter of preference. Some authors choose to define a metric $d$ as a function $X \times X \to \mathbb{R}$ that is non-negative while others define $d$ as a function $X \times X \to [0,\infty)$. In either case, you can view $d$ as a function whose codomain is $[0,\infty)$. Indeed, let $X,Y$ and $Z$ be topological spaces and let $Y \supseteq f(X)$ be a subspace of $Z$. If $f :X \to Z$ is continuous, then $f$ is also continuous when viewed as a function $X \to Y$.