So I was studying topology and I came across the next theorem:A function f : X → Y is continous iff for every ϵ &gt; 0 there is δ &gt; 0 such that if d x ( x , y ) &lt; δ then d y ( f ( x ) , f ( y ) ) &lt; 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?

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So I was studying topology and I came across the next theorem:A function   f  :  X  →  Y is continous iff for every   ϵ  &amp;gt;  0 there is   δ  &amp;gt;  0 such that if       d    x    (  x  ,  y  )  &amp;lt;  δ then       d    y    (  f  (  x  )  ,  f  (  y  )  )  &amp;lt;  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?

Miah Scott · Accepted Answer

Counter-example:   f  (      x    i    )  =  (      y    i    ) where       y    i    =  1 if       x    i    ≥  0 and       y    i    =  0 if       x    i    &amp;lt;  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].

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?

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