Solution:
Consider two metric spaces on R2 ;
1.Euclidean metric de
d e ( x ‾ , y ‾ ) = ( x 1 − y 1 ) 2 + ( x 2 − y 2 ) 2 d_e(\underline{x},\underline{y})=\sqrt{(x_1-y_1)^2+(x_2-y_2)^2} d e ( x , y ) = ( x 1 − y 1 ) 2 + ( x 2 − y 2 ) 2
2. The max-metric dm
d m ( x ‾ , y ‾ ) = d_m(\underline{x},\underline{y})= d m ( x , y ) = max{ |x1 -y1 |,|x2 -y2 |}
We have;
d m ( x ‾ , y ‾ ) d_m(\underline{x},\underline{y}) d m ( x , y ) =max{|x1 -y1 |,|x2 -y2 |}≤ \leq ≤ ( x 1 − y 1 ) 2 + ( x 2 − y 2 ) 2 \sqrt{(x_1-y_1)^2+(x_2-y_2)^2} ( x 1 − y 1 ) 2 + ( x 2 − y 2 ) 2 =d e ( x ‾ , y ‾ ) d_e(\underline{x},\underline{y}) d e ( x , y )
Also;
d e ( x ‾ , y ‾ ) = ( x 1 − y 1 ) 2 + ( x 2 − y 2 ) 2 = ( ∣ x 1 − y 1 ∣ ) 2 + ( ∣ x 2 − y 2 ∣ ) 2 d_e(\underline{x},\underline{y})=\sqrt{(x_1-y_1)^2+(x_2-y_2)^2}=\sqrt{(|x_1-y_1|)^2+(|x_2-y_2|)^2} d e ( x , y ) = ( x 1 − y 1 ) 2 + ( x 2 − y 2 ) 2 = ( ∣ x 1 − y 1 ∣ ) 2 + ( ∣ x 2 − y 2 ∣ ) 2
d e ( x ‾ , y ‾ ) ≤ d m 2 ( x ‾ , y ‾ ) + d m 2 ( x ‾ , y ‾ ) d_e(\underline{x},\underline{y})\leq\sqrt{d_m^2(\underline{x},\underline{y})+d_m^2(\underline{x},\underline{y})} d e ( x , y ) ≤ d m 2 ( x , y ) + d m 2 ( x , y ) =2 d m ( x ‾ , y ‾ ) \sqrt{2}d_m(\underline{x},\underline{y}) 2 d m ( x , y )
Hence,
d m ( x ‾ , y ‾ ) ≤ d e ( x ‾ , y ‾ ) ≤ 2 d m ( x ‾ , y ‾ ) d_m(\underline{x},\underline{y})\leq d_e(\underline{x},\underline{y})\leq \sqrt{2}d_m(\underline{x},\underline{y}) d m ( x , y ) ≤ d e ( x , y ) ≤ 2 d m ( x , y )
for all x,yϵ \epsilon ϵ X.
Hence, metrics dm and de are equivalent.
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