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Given a metric space $(X,d)$, how to prove that the function $d \colon X \times X \to \mathbf{R}$ is continuous?

If we take any two arbitrary real numbers $a$ and $b$ such that $a < b$, then we need to show that the set $d^{-1} (a,b)$ given by

$$ d^{-1} (a,b) := \{ (x, y) \in X \times X | a < d(x,y) < b \} $$

is open in the product topology on $X \times X$.

A basis for this product topology is the collection of all cartesian products of open balls in $(X,d)$.

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Why don't you use $\varepsilon$-$\delta$ definition? – Norbert Jan 26 '13 at 13:09
the triangle inequality should prove useful – robjohn Jan 26 '13 at 13:18
wwhat is the metric on $X\times X$? – vesszabo Jan 26 '13 at 13:31
I can figure out a $\delta$ - $\epsilon$ $\,$ proof by taking the following metric on the Cartesian product $X \times X$: $ d_{X \times X} ( (x,y) , (x_0, y_0) ) := d(x, x_0) + d(y, y_0)$, where $d$ denotes the metric on $X$. However, we have to give a proof that is independent of a particular choice of a metric on the Cartesian product. – Saaqib Mahmuud Jan 26 '13 at 13:36
I would be interested to read such a proof! Good luck :D – Mercy King Jan 26 '13 at 14:15

For $a, b ∈ ℝ$, let $(x,y) ∈ d^{-1} (a..b)$, i.e. $a < d(x,y) < b$. Now choose $ε$ such that $U_{2ε} (d(x,y)) ⊂ (a..b)$ and look at $U_ε (x) × U_ε (y)$.

For any tuple of points $(x',y') ∈ U_ε (x) × U_ε (y)$ you have $$d(x',y') ≤ d(x',x) + d(x,y) + d(y,y') < d(x,y) + 2ε$$ as well as $$d(x,y) ≤ d(x,x') + d(x',y') + d(y',y) < d(x',y') + 2ε$$ This means $a < d(x,y) - 2ε < d(x',y') < d(x,y) + 2ε < b$.

Therefore $U_{ε} (x) × U_{ε} (y) ⊂ d^{-1} (a..b)$.

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