Characterization of distance between points in separable space

Question

I want to prove that if $X$ is a separable metric space, then for all $x,y \in X$ and any dense set $\{x_i\}$, we can write:

$$d(x,y) = \sup_{i \in \mathbb{N}} |d(x,x_i) - d(y,x_i) |.$$

Some attempts. One side: fix $x,y$, by denseness there exists a subsequence $x_{n_{k}}$ of $x_n$ such that $x_{n_{k}}$ converges to $x$. This implies $$\sup_{i \in \mathbb{N}} |d(x,x_i) - d(y,x_i) | \geq \sup_{k \in \mathbb{N}} |d(x,x_{n_{k}}) - d(y,x_{n_{k}}) | = d(x,y) $$ even if I am not even sure about how to justify the last passage since it's a supremum and not a limit. Moreover I cannot do the other inqeuality. How can I proceed? Thanks.

Answer 1

Your proof for the inequality $d(x,y) \le \sup_{i \in \mathbb{N}} |d(x,x_i) - d(y,x_i)|$ has the right idea, but as you noticed, it's not rigorous since you didn't justify that $\sup_{k \in \mathbb{N}} |d(x,x_{n_{k}}) - d(y,x_{n_{k}}) | = d(x,y)$.

You can argue as follows instead : since $\{x_n\}$ is dense, you can have find a subfamily $\{x_{n_k}\}$ such that for all $k\ge1$, $d(x,x_{n_k})\le 1/k$. Furthermore, for all $k\ge 1$ $$\sup_{i \in \mathbb{N}} |d(x,x_i) - d(y,x_i) | \geq|d(x,x_{n_{k}}) - d(y,x_{n_{k}}) | \ge d(y,x_{n_k}) -d(x,x_{n_k}).\tag1 $$ And by triangle inequality $$d(x,y)\le d(x,x_{n_k})+ d(x_{n_k},y)\implies d(y,x_{n_k})\ge d(x,y) - d(x,x_{n_k}) \tag2 $$ Combining $(1)$ and $(2)$ together with the fact that $-d(x,x_{n_k})\ge -1/k$, we find $$\sup_{i \in \mathbb{N}} |d(x,x_i) - d(y,x_i) | \ge d(y,x) -2d(x,x_{n_k})\ge d(y,x) -2/k, $$ and since this holds for any $k\ge1$, the desired inequality follows by letting $k\to\infty$.

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To show that $d(x,y) \ge \sup_{i \in \mathbb{N}} |d(x,x_i) - d(y,x_i)|$, apply the following

Reverse triangle inequality : for all $x,y,z\in X$, we have $$d(x,y)\ge |d(x,z)-d(y,z)| $$

to some well chosen $z$ (make sure to prove the above inequality yourself if you didn't already know it).

Answer 2

In any metric space $(X,d)$, you can easily prove that $$\sup_{z\in X}|d(x,z)-d(y,z)|=d(x,y).$$ If moreover $A$ is a dense subset of $X$ then $$\sup_{z\in X}f(z)=\sup_{z\in A}f(z)$$ for $f(z):=|d(x,z)-d(y,z)|$, because $f$ is continuous.