Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

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|improve this question
6  
Why don't you use $\varepsilon$-$\delta$ definition? –  no identity Jan 26 '13 at 13:09
3  
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 Jan 26 '13 at 14:15

1 Answer 1

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|improve this answer

Your Answer

 
discard

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.