Question

Let d d be the usual metric on LaTeX: \mathbb{R}^2 R 2 and let \rho ρ be the square metric on LaTeX: \mathbb{R}^2 R 2 , that is, LaTeX: d(x,y)=(|x_1-y_1|^2+|x_2-y_2|^2)^{1/2} d ( x , y ) = ( | x 1 − y 1 | 2 + | x 2 − y 2 | 2 ) 1 / 2 , and LaTeX: \rho(x,y)=\max\{|x_1-y_1|,|x_2-y_2|\} ρ ( x , y ) = max { | x 1 − y 1 | , | x 2 − y 2 | } , where LaTeX: x=(x_1,x_2) x = ( x 1 , x 2 ) , LaTeX: y=(y_1,y_2) y = ( y 1 , y 2 ) . Without directly referring to any theorem from Section 1.5, prove explicitly that (a) LaTeX: \rho(x,y)\leq d(x,y)\leq \sqrt{2}\rho(x,y) ρ ( x , y ) ≤ d ( x , y ) ≤ 2 ρ ( x , y ) (10 points) (b) d d and \rho ρ generate the same topology on LaTeX: \mathbb{R}^2 R 2 . (15 points)