Vtyjkyuk

Here is Prob. 12, Sec. 5.2, in the book Introduction To Real Analysis by Robert G. Bartle & Donald R. Sherbert, 4th edition:

A function $f colon mathbbR to mathbbR$ is said to be additive if $f(x+y)= f(x) + f(y)$ for all $x, y$ in $mathbbR$. Prove that if $f$ is continuous at some point $x_0$, then it is continuous at every point of $mathbbR$.

My Attempt:

Let $c$ be an arbitrary real number.

First suppose that $x_0 = 0$. As $f$ is continuous at the point $0$, so given any real number $varepsilon > 0$ we can find a real number $delta > 0$ such that
$$lvert f(x) rvert = lvert f(x) - 0 rvert = lvert f(x) - f(0) rvert < varepsilon $$
for all $x in mathbbR$ for which
$$ lvert x rvert = lvert x-0 rvert < delta. $$
Then
$$ lvert f(x) - f(c) rvert = lvert f( x - c ) rvert < varepsilon $$
for all $x in mathbbR$ such that
$$ lvert x-c rvert < delta. $$
Thus it follows that $f$ is continuous at every point $c in mathbbR$.

Am I right?

Now let us suppose that $x_0 neq 0$. As $f$ is continuous at $x_0$, so, for every real number $varepsilon > 0$ we can find a real number $delta > 0$ such that
$$ leftlvert f left(x - x_0 right) rightrvert = leftlvert f(x) - f left( x_0 right) rightrvert <
varepsilon $$
for all real numbers $x$ which satisfy
$$ leftlvert x- x_0 rightrvert < delta. $$

Now if we could show from here that $f$ is continuous at $0$, then as before we will have managed to show that $f$ is continuous at every real number $c$.

What next? How to proceed from here to show that our function $f$ is continuous at $0$? Or, is there any other route?

You are almost there. Let $| t-0 | = |t| < delta$, where $delta$ is the one you have in the second box. Then $t + x_0 = xin mathbbR$, so $|t| = |x - x_0| < delta$.

Thus, $|f(t) - f(0)| = |f(t)| = |f(x-x_0)| = |f(x) - f(x_0)| < epsilon$ for $|t| < delta$.

So $f$ is continuous at $0$.