Solution:
Consider two metric spaces on R2;
1.Euclidean metric de
de(x,y)=(x1−y1)2+(x2−y2)2
2. The max-metric dm
dm(x,y)=max{ |x1-y1|,|x2-y2|}
We have;
dm(x,y)=max{|x1-y1|,|x2-y2|}≤ (x1−y1)2+(x2−y2)2 =de(x,y)
Also;
de(x,y)=(x1−y1)2+(x2−y2)2=(∣x1−y1∣)2+(∣x2−y2∣)2
de(x,y)≤dm2(x,y)+dm2(x,y) =2dm(x,y)
Hence,
dm(x,y)≤de(x,y)≤2dm(x,y)
for all x,yϵ X.
Hence, metrics dm and de are equivalent.
Comments