A little more formal.
Notice two things first, namely and so . Thus,
A little less "limity". Suppose that there existed some such that is continuous at that point. Let be arbitrary. Given any we have that
. Notice that since is continuous at we know that for the previously mentioned there exists
some such that . So, let , then notice that choosing .
The conclusion follows. The point of all of these is to shift the point of continuity to an arbitrary point.
Note that from this it readily follows that the only function which is continuous at a single point and is the function .
First you need to prove that f(0)= 0. That is true because f(a+0)= f(a)+ f(0)= f(a).
If f is continuous at any x= b, then it is continuous at x= 0.
That is true because where y= x- b. Then . Since f is continuous at x= b, so we have and so [tex]\lim_{y\to 0}f(y)= 0= f(0) and f is continuous at 0.
If f is continuous at x= 0 then it is continuous at every point, a.
Almost the same proof: [tex]lim_{x\to a} f(x)= where y= x- a. But f(a+y)= f(a)+ f(y) so and since f is continous at 0, . That is, if f is continous at x= 0, then so f is continuous at x= a.