Let $(X,d)$ be a non-empty connected metric space. Then if $p_0 \in X$, $\{d(p_0, q): q \in X\}$ is either equal to $\{0\}$ or is an interval containing 0.
By definition of metric spaces, I know that we must have $d(x,y) \geq 0$ and $d(x,y)=0 \leftrightarrow x=y $. So if $p_0 = q$, then we would have $\{0\}$?
If $p_0 \neq q$, then I'm guessing that the argument behind this is that if both $p_0, q \in X$, then the metric ball they're both in would have the interval $(p_0 + q, p_0 - q)$, which would include (somehow) $0$?
Help would be appreciated!
$\{d(p_0, q): q \in X\}$ is a subset of $\{0\} \cup \mathbb{R^+}$, since $d: X$ x $X$ $\to \{0\} \cup \mathbb{R^+}$.
Now, if $\{p_0\} = X$, then the set we're after is just $\{0\}$, since the distance between $p_0$ and any other point in the set is trivially 0, since $p_0$ is the only point in the set. But if there are other points $q \in X$, then there will be a positive distance between them. By connectedness, we know that it will be an interval. And of course $0$ will be in the set, since $d(p_0, p_0) = 0$ is in the set.