Continuous Functions: Does f(x+δ) = ε?

[Flying_Goat]

A function defined on ℝ is continuous at x if given ε, there is a δ such that |f(x)-f(y)|<ε whenever |x-y|<δ. Does this imply that f(x+δ)-f(x)=ε? The definition only deals with open intervals so i am not sure about this. If this is not true could someone please show me a counter example for it?

Any help would be appreciated. Thanks.

 

[alan2]

No. You have to learn to think differently. Draw a lot of pictures and think about limiting processes, not equalities. A real function is continuous at x if I can draw a rectangular box around the point (x, f(x)), shrink the box arbitrarily small, and the function remains in the box. Consider f(x)=1 for all x. It is continuous everywhere, but for any epsilon>0 there is no delta which satisfies your statement. In fact, f(y)-f(x)=0 for all x,y, yet for any epsilon, no matter how small, I can choose delta arbitrarily large.

 

[Bacle2]

A function defined on ℝ is continuous at x if given ε, there is a δ such that |f(x)-f(y)|<ε whenever |x-y|<δ. Does this imply that f(x+δ)-f(x)=ε? The definition only deals with open intervals so i am not sure about this. If this is not true could someone please show me a counter example for it?

Any help would be appreciated. Thanks.

Another approach is :

The description can also be interpreted as saying that one can find, for any ε>0, a value of δ>0 every point x in the interval:

(y-δ,y+δ) on the x-axis

Is mapped into the interval (f(y)-ε,f(y)+ε )

on the y-axis.

Try playing with relatively-simple functions like x², and see what happens with
the expression f(x+δ)-f(x), for different values of δ, and how you can choose δ to make the difference be within ε.

 

[Flying_Goat]

Thanks for your replies. I understand now.