X1 d1 SolutionProof If X1 d1 and X2 d2 are metric spaces the

Solution

Proof:

If (X1, d1) and (X2, d2) are metric spaces then each of the following is a metric on X1 × X2,

(a) d((x1, x2),(y1, y2)) = d1(x1, y1) + d2(x2, y2)

(b) d((x1, x2),(y1, y2)) = p d1(x1, y1) 2 + d2(x2, y2) 2

(c) d((x1, x2),(y1, y2)) = max{d1(x1, y1), d2(x2, y2)}

Now, we have to prove (c) equation – PRODUCT SPACE

Solution:

The non-negitivity of the metric d is easy to see. So we omit that.

Also, If (x1,x2) = x = y = (y1,y2) then

        We trivally have d(x,y) = 0

Further, if d(x,y) = 0 then we have that d_j(x_j,y_j) = 0 for j=1,2

But this means that   x_j = y_j for j=1,2 and x=y

It is also easy to see that d(x,y)=d(y,x)

Suppose, without lose of generality

                max{d1(x1, y1), d2(x2, y2)} = d1(x1, y1)

Then note

        d(x, y)    = max{d1(x1, y1), d2(x2, y2)}

= d1(x1, y1)

d1(x1, z1) + d1(z1, y1)

max{d1(x1, z1), d2(x2, z2)} + max{d1(z1, y1), d2(z2, y2)}

= d(x, z) + d(z, y).

Here we used the triangle inequality for the metric space (X1, d1).

Thus d satisfies the triangle inequality and so we have that (X1 × X2, d) is a metric space.