1. ## Euclidean Spaces

Let's say if we have $\bold{x} \in \mathbb{R}^{n}$ and we want to show that $B_{r}(\bold{x})$(open ball of radius $r >0$) is convex. We know that $d(\bold{x}, \bold{y}) = |\bold{x}- \bold{y}|$ (e.g. the norm).

We can still apply the triangle inequality right? So we have: $|t(\bold{y}-\bold{x})+(1-t)(\bold{z}-\bold{x})| \leq t|\bold{y}-\bold{x}| + (1-t)|\bold{z}-\bold{x}| < r$.

In other words, does $d(\bold{x}, \bold{y}) = |\bold{x}- \bold{y}|$ only apply to $\mathbb{R}$? Or does it apply to general $\mathbb{R}^{n}$ so that we can invoke the triangle inequality? We can use the triangle inequality to prove the general case (which is what I did) right?

2. Originally Posted by particlejohn
In other words, does $d(\bold{x}, \bold{y}) = |\bold{x}- \bold{y}|$ only apply to $\mathbb{R}$? Or does it apply to general $\mathbb{R}^{n}$ so that we can invoke the triangle inequality? We can use the triangle inequality to prove the general case (which is what I did) right?
Perhaps I should not attempt an answer because I may have missed your point.
But one of the requirements for a metric is the triangle inequality holds.
That is, if $d$ is a metric on a space then $d(x,y) \leqslant d(x,z) + d(z,y)$ for all x, y, & z in the space.
Does that get at what you are asking?

3. yes, thanks.

4. how about the euclidean norm? does it hold for that?

5. Originally Posted by particlejohn
how about the euclidean norm? does it hold for that?
Yes it does. Here is how that goes.
$\left\{ {y,z} \right\} \subseteq B_r (x)\; \Rightarrow \;\left\| {y - x} \right\| < r\;\& \;\left\| {z - x} \right\| < r$
$\begin{array}{*{20}c}
{\left\| {\lambda y\_(1 - \lambda )z - x} \right\|} & = & {\left\| {\lambda (y - x) + (1 - \lambda )(z - x)} \right\|} \\
{} & \leqslant & {\lambda \left\| {y - x} \right\| + (1 - \lambda )\left\|{z - x} \right\|} \\ {} & < & {\lambda r + (1 - \lambda )r} \\ {} & = & r \\ \end{array}$

6. so the triangle inequality applies to the following: $|\bold{x}| = \sqrt{x_{1}^{2} + \cdots x_{n}^{2}}$?

The triangle inequality works for all metrics on Euclidean spaces? But not on say $L^{p}$ spaces?

7. I think that you are confusing yourself: the idea of a norm on a linear space with the distance function or metric.
Have a look at these pages:
Norm -- from Wolfram MathWorld
Vector Norm -- from Wolfram MathWorld
In the case of a ‘norm’ as with a metric, the triangle inequality is a requirement.
I think that you need to go to the instructor and clear away this confusion.

8. yeah even Rudin agrees.