$f(x+y) = f(x)+f(y)$ continuous at $x_0=0 \implies f(x)$ is continuous over R? Let $x,y \in R$ 
$f(x+y) = f(x)+f(y)$
is it true that if $f$ is continuous at $x_0=0$, than $f$  is continuous in $R$? 
 A: At any arbitrary $x_1\in\mathbb{R}$ and any $\Delta\neq 0$, we have
$$
f(x_1+\Delta)-f(x_1)=f(\Delta)=f(\Delta+x_0)-f(x_0).
$$
As $\Delta\to 0$, the rightmost expression above goes to $0$ due to continuity at $x_0$, so the leftmost expression also goes to $0$. This implies continuity at $x_1$ and therefore in $\mathbb{R}$. Note we don't need the fact that $x_0=0$.
A: Make $x=a_n$ such that $a_n \rightarrow a$ and $y=-a$
$$f\left(a_n-a\right)=f(a_n)-f\left(a\right)$$
doing $n \rightarrow \infty$ and by the continuity at $0$ 
$$f\left(a_n-a\right) \rightarrow 0 \quad \text{(because $f(0)=0$)}$$
$$a_n \rightarrow a$$
and finally
$$f\left(a_n\right) \rightarrow f(a)$$
So we get continuity at $a$.
A: $f$ is continuous at $0$ if and only if 
$$\forall \epsilon>0 \exists \delta>0 : |x|<\delta \implies |f(x)-f(0)|<\epsilon.$$
Since $f(0)=f(0+0)=f(0)+f(0)=2f(0)$ we have $f(0)=0.$ Thus
$f$ is continuous at $0$ if and only if 
$$\forall \epsilon>0 \exists \delta>0 : |x|<\delta \implies |f(x)|<\epsilon.$$
Now, $f$ is continuous at $x_0$ if and only if 
$$\forall \epsilon>0 \exists \delta>0 : |h|<\delta \implies |f(x_0+h)-f(x_0)|<\epsilon.$$ This obviously holds since $$f(x_0+h)-f(x_0)=f(h)$$ and $f$ is continuous at $0.$
