We consider the function
given by
This function is not continuous in 0 {\displaystyle {}0} . For ϵ = 1 2 {\displaystyle {}\epsilon ={\frac {1}{2}}} and every positive δ {\displaystyle {}\delta } , there exists a negative number x ′ {\displaystyle {}x'} such that d ( 0 , x ′ ) = | x ′ | ≤ δ {\displaystyle {}d(0,x')=\vert {x'}\vert \leq \delta } . But for such a number we have d ( f ( 0 ) , f ( x ′ ) ) = d ( 1 , 0 ) = 1 ≰ 1 2 {\displaystyle {}d(f(0),f(x'))=d(1,0)=1\not \leq {\frac {1}{2}}} .