Continuity of metric function Let $X$ is a vector space and $d$ is a metric function on $X$ and $\|\cdot\|$ is a norm on $X$ and $\langle\cdot,\cdot\rangle$ is an inner product function on $X$
It is to easy to prove $\|\cdot\|$ and $\langle\cdot,\cdot\rangle$ are continuous functions on $X$ and there are many theorems prove that. 
But can I say that the metric function $d$ is continuous on $X$? And how can I prove that? 
Or I can't say that because it has no meaning ( continuity of metric function ) 
 A: Since
$$
d:X\times X\to [0,\infty)
$$
is a metric, it satisfies the triangle inequality:
$$
d(a_1,a_2)\le d(a_1,a_3)+d(a_3,a_2) \quad \forall a_1,a_2,a_3\in X.
$$
Given $a=(a_1,a_2)\in X\times X$, we have for every $x=(x_1,x_2)\in X\times X$:
$$
d(x_1,x_2)\le d(x_1,a_1)+d(a_1,a_2)+d(a_2,x_2),
$$
and therefore
$$\tag{1}
d(x_1,x_2)-d(a_1,a_2)\le d(x_1,a_1)+d(x_2,a_2).
$$
Also, for every $x=(x_1,x_2)\in X\times X$ we have:
$$
d(a_1,a_2)\le d(a_1,x_1)+d(x_1,x_2)+d(x_2,a_2),
$$
i.e.
$$\tag{2}
-\left[d(x_1,x_2)-d(a_1,a_2)\right]\le d(x_1,a_1)+d(x_2,a_2).
$$
Combining (1) and (2) we get:
$$
\left|d(x_1,x_2)-d(a_1,a_2)\right|\le d(x_1,a_1)+d(x_2,a_2)\quad \forall x=(x_1,x_2)\in X\times X,
$$
which shows that $d$ is continuous at $a=(a_1,a_2)\in X\times X$, and therefore $d$ is continuous on $X\times X$ because $a$ is arbitrary.
A: You could prove that the metric $d$ is a continuous function from $A\times A$ to $\mathbb{R}$ by considering the product topology on $A\times A$.
Let $M = \left({A, d}\right)$ be a metric space.
Let $\tau$ be the topology on $A$ induced by $d$.
Let $\left({A \times A, T}\right)$ be the topological product of $(A, \tau)$ and $(A, \tau)$. http://en.wikipedia.org/wiki/Product_topology
Then we can prove that $d: A \times A \to \mathbb{R}$ is continuous (using the topological definition).
Do you want to try proving it now that you've got the basic idea where to start or do you want the proof?
We can also use the $\epsilon$-$\delta$ way, I just prefer this.
EDIT:
OK, since any open set in $\mathbb{R}$ is a union of open intervals, it is sufficient to show that the inverse image under $d$ of any open interval in $\mathbb{R}$ is an open set in $\left({A \times A, T}\right)$. (So the inverse image of any open set is then the union of open sets, which is then open).
So take any $a, b ∈ ℝ$. Let $(x_1,x_2) ∈ d^{-1} ((a,b)) \subseteq A \times A$.  Then $a < d(x_1,x_2) < b$. Now since $(a,b)$ is open in $\mathbb{R}$, there exists $\varepsilon>0$ such that the open ball of radius $2\varepsilon$ around $d(x_1, x_2)$, $B_{2\varepsilon} (d(x,y)) \subseteq  (a,b)$. We want to show that $d^{-1} ((a,b))$ is open in the product topology. 
Note:( $(a,b)$ is an open interval, which is a connected set, not two pairs of points, while $(x_1,x_2)$ and the rest are just pairs of points from $A$.)
So consider $B_\varepsilon (x_1) × B_\varepsilon (x_2) \subseteq A\times A$, the cartesian product of two open balls of radius $\varepsilon$ around $x_1$ and $x_2$. Then since in the product topology, the product of these two open balls is a neighborhood around $(x_1, x_2)$, we just need to show that this product is contained in  $d^{-1} ((a,b))$, to show that  $d^{-1} ((a,b))$ is open. 
For any $(y_1,y_2) \in B_\varepsilon (x_1) \times B_\varepsilon (x_2)$, $d(y_1,y_2) \leq d(y_1,x_1) + d(x_1,x_2) + d(x_2,y_2) < d(x_1,x_2) + 2\varepsilon$.
Similarly $d(x_1,x_2) ≤ d(x_1,y_1) + d(y_1,y_2) + d(y_2,x_1) < d(y_1,y_2) + 2\varepsilon$.
Hence $a < d(x_1,x_2) - 2\varepsilon < d(y_1,y_2) < d(x_1,x_2) + 2\varepsilon < b$, i.e. $d(y_1, y_2) \in (a,b)$ so $(y_1, y_2) \in d^{-1}((a,b))$.
Therefore $B_\varepsilon (x_1) \times B_\varepsilon (x_2) \subseteq d^{-1} ((a,b))$, so $d^{-1} ((a,b))$ is open. So $d$ is continuous.
