Background Suppose $(x_i)$ and $(y_j)$ are two sequences in a metric space $(X,d)$ that converge respectively to $x$ and $y$. I am trying to show that the sequence $(d( x_i, y_j))$ converges to $d(x,y)$
Thoughts I think probably the first thing to do is to show that the metric is itself a continuous function. I believe this is so because if $V$ is an open subset of $\mathbb{R}$ in the range of $d$, say $V := (a, b)$ then the preimage of $V$ is the set $ d^{-1}(V) = \{(x,y) \in X \times X: a < d(x,y) < b \} $ which is an open ball of radius $|b-a|$. Assuming this part is OK, since $d$ is continuous, it commutes with the limit operation but in this case, there are actually two limits: one as $i \rightarrow \infty$ and one as $j \rightarrow \infty$ I think maybe there is a couple of ways around this; one, by fixing $x := x_0$ and taking the limit as $j \rightarrow \infty$ and then by similarly fixing $y := y_0$ and take the limit as $i \rightarrow \infty$. Perhaps another way is to simply use the same sequence index and consider each point $(x_i, y_i)$ as a term in the sequence.
Question I think I understand intuitively what's going on, but I'm not sure of how to make the argument precise. Is the approach I outlined promising or is there a cleaner approach that I'm overlooking?