A function f:X->Y is continous iff for every epsilon>0 there is delta>0 such that if dx(x,y)<delta then d_y(f(x),f(y))<epsilon. Since every distance in the uniform topology is at most 1, the theorem will always be true for any function from R^omega to R^omega with the uniform topology, am I wrong?

gasavasiv

gasavasiv

Answered question

2022-10-30

So I was studying topology and I came across the next theorem:
A function f : X Y is continous iff for every ϵ > 0 there is δ > 0 such that if d x ( x , y ) < δ then d y ( f ( x ) , f ( y ) ) < ϵ.
Since every distance in the uniform topology is at most 1, the theorem will always be true for any function from R ω to R ω with the uniform topology, am I wrong?

Answer & Explanation

Miah Scott

Miah Scott

Beginner2022-10-31Added 19 answers

Counter-example: f ( x i ) = ( y i ) where y i = 1 if x i 0 and y i = 0 if x i < 0. This function is not continuous because ( 0 , 0 , . . . , 0 , ( 1 ) n n , ( 1 ) n + 1 n + 1 , . . . ) ( 0 , 0 , . . . ) but d ( ( 0 , 0 , . . . , 0 , ( 1 ) n n , ( 1 ) n + 1 n + 1 , . . . ) , f ( 0 , 0 , . . . ) ) = 1 for all n. [Here ( 0 , 0 , . . . , 0 , ( 1 ) n n , ( 1 ) n + 1 n + 1 , . . . ) starts with n 1 zeros].

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