Proving d_N is a Metric with Discrete Metric d_X

[Ted123]

Homework Statement

[PLAIN]http://img833.imageshack.us/img833/6932/metric2.jpg

The Attempt at a Solution

I've shown $d_{X\times Y}$ is a metric by using the fact that $d_X$ and $d_Y$ are metrics.

What is a simpler description of $d_N$ with $d_X$ the discrete metric?

Is it just: $$d_N(x,y) = \left\{ \begin{array}{lr} <br /> N & : x\neq y\\ <br /> 0 & : x=y <br /> \end{array} <br /> \right.$$

 

[micromass]

What is a simpler description of $d_N$ with $d_X$ the discrete metric?

Is it just: $$d_N(x,y) = \left\{ \begin{array}{lr} <br /> N & : x\neq y\\ <br /> 0 & : x=y <br /> \end{array} <br /> \right.$$

No. How did you get there??

Just calculate

$$d_N((x_1,x_2),(y_1,y_2))$$

and see what the possible outcomes are.

 

[Ted123]

No. How did you get there??

Just calculate

$$d_N((x_1,x_2),(y_1,y_2))$$

and see what the possible outcomes are.

Yeah I see what I assumed wrong.

$d_N(x,y)$ is the number of coordidates in which x and y differ.

How do I describe these open balls?

The definition is: $B(x,r)=\{ y\in X : d(x,y)<r \}$ where $x\in X$ and $r>0$ is the radius.

So we want:

$B((0,0),1)=\{ y\in X^2 : d((0,0),y)<1 \}$ ;

$B((0,0),2)=\{ y\in X^2 : d((0,0),y)<2 \}$ ;

$B((0,0),3)=\{ y\in X^2 : d((0,0),y)<3 \}$ .