Is the distance function continuous?

[Design]

Homework Statement

Prove that the distance function d : Rⁿ x Rⁿ -> R, defined as d(x,y) = |x-y| is continous.

The Attempt at a Solution

|x-y| >= | |x| - |y| |
|x+y| <= | |x| + |y| |

Not sure what to do from here on

thank you

 

[ystael]

How would you attempt to prove any other function $$f: \mathbb{R}^n \times \mathbb{R}^n \to \mathbb{R}$$ is continuous? There's nothing special about the fact that it's the Euclidean distance function.

 

[Design]

So For all Epsilon>0 there exist a delta > 0
so |x-a| < delta and |y-b|<delta and | |x-y| - |a-b|| < epsilon

Where do i go from there?

 

[ystael]

So For all Epsilon>0 there exist a delta > 0
so |x-a| < delta and |y-b|<delta and | |x-y| - |a-b|| < epsilon

Where do i go from there?

This would be a correct statement of what you need to prove, except for one problem: You have the quantifiers wrong. The definition of continuity for a function $$f: \mathbb{R}^n \times \mathbb{R}^n \to \mathbb{R}$$ is: $$f$$ continuous at $$(a, b)$$ if, for every $$\epsilon > 0$$ there exists $$\delta > 0$$ such that, for every $$(x, y)$$ satisfying $$|(x, y) - (a, b)| < \delta$$, we have $$|f(x, y) - f(a, b)| < \epsilon$$. Look carefully at the italicized part there and see how it differs from what you wrote.

You may have a different definition of continuity for functions of two variables which says: ... there exists $$\delta > 0$$ such that, for every $$(x, y)$$ satisfying $$|x - a| < \delta$$ and $$|y - b| < \delta$$, we have ... This is equivalent to the standard definition of continuity, but that statement is a theorem which must be proved.