Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

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 $(X,d)$.

share|cite|improve this question
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 $U_{ε} (x) × U_{ε} (y) ⊂ d^{-1} (a..b)$.

share|cite|improve this answer
Sorry, I didn't recognize the notation, but by $(a..b)$ do you mean the set of all real numbers from $a$ to $b$ exclusive (i.e. the open interval)? – Jason Nichols Jul 16 at 21:07
Yeah, it’s a notation introduced by Knuth, I believe. It’s quite handy. It avoids the confusion of regarding $(a,b)$ both as a tuple and as an open interval. See the good notations thread on mathoverflow. – k.stm Jul 17 at 17:53

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.