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 &amp; : x\neq y\\ <br /> 0 &amp; : 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 &amp; : x\neq y\\ <br /> 0 &amp; : 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)&lt;r \} where x\in X and r&gt;0 is the radius.

So we want:

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

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

B((0,0),3)=\{ y\in X^2 : d((0,0),y)&lt;3 \} .