Exercise 1.12

Suppose d 1 ( x , y ) = | x y | , d 2 ( x , y ) = | ϕ ( x ) ϕ ( y ) | , where ϕ ( x ) = x ( 1 + | x | ) . Prove that d 1 and d 2 are metrics on which induce the same topology, although d 1 is complete and d 2 is not.

Answers

Proof. First, each d i ( i = 1 , 2 ) induces a topology τ i whose open balls are all

B i ( a , r ) { x R : d i ( a , x ) < r } ( a R , r > 0 ) . (1)

Next, remark that the monotonically increasing mapping ϕ : R ] 1 , 1 [ is odd and that

ϕ ( x ) x 1 . (2)

ϕ is therefore a τ 1 -homeomorphsim of R onto ] 1 , 1 [ . A first consequence is that, at fixed a R , given any positive scalar 𝜀 , the τ 1 -continuousness of ϕ supplies an open ball B 1 ( a , η ) on which | ϕ ( a ) ϕ | < 𝜀 . In terms of balls B i , this reads as follows,

B 1 ( a , η ) B 2 ( a , 𝜀 ) . (3)

The second consequence is that the τ 1 -continuousness of ϕ 1 yields similar inclusions

B 2 ( a , 𝜀 ) B 1 ( a , η ) (4)

provided η > 0 . At arbitrary 𝜀 , the special case η = η is the concatenation

B 2 ( a , 𝜀 ) B 1 ( a , η ) B 2 ( a , 𝜀 ) ; (5)

which proves that τ 1 = τ 2 . Finally, all inequalities n < i < j over N together yield

d 2 ( i , j ) = | ϕ ( i ) ϕ ( j ) | n 0 . (6)

The sequence n = 0 , 1 , 2 , is therefore τ 2 -Cauchy. We will nevertheless establish that it τ 2 -diverges. To do so, we start by offering the τ 2 -converge to some λ : The triangle inequality immediately dismiss that assumption, as follows,

d 2 ( 0 , λ ) d 2 ( 0 , n ) d 2 ( λ , n ) = ϕ ( n ) d 2 ( λ , n ) n 1 . (7)

We then conclude that d 2 fails to be complete. □

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2023-09-06 20:37
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