prove continuity Let $ f:\Bbb R \to \Bbb R $ satisfy the property $ f(x+y)=f(x)+f(y)$ for all $x,y$ in $ \Bbb R $
I have to show that 
1)$f(0)=0 , f(-x)=-f(x),$ for all $x$ in $\Bbb R$, and $f(x-y)=f(x)-f(y)$ $y$ in $\Bbb R$
2) If $f$ is continuous at $x=a$ then $f$ is continuous at every point in of $\Bbb R$
I can prove the first part, but I can't understand how prove second part.
 A: Hint: Note that
$$
\lim_{x \to a} f(x) = \lim_{y \to 0} f(a + y) = \lim_{y \to 0}[f(a) + f(y)]
$$
A: If $f$ is continuous at a point $x$, then given $\epsilon>0$, there exists a number $\delta>0$, such that
$$|f(x+h)-f(x)|<\epsilon \tag 1$$
whenever $0<h<\delta$.
Note, by assumption $f(x+y)=f(x)+f(y)$.  Thus, we have 
$$|f(x+h)-f(x)|=|f(x)+f(h)-f(x)|=|f(h)| \tag 2$$
And, we know that since $f$ is continuous at $a$, then given $\epsilon>0$, there exists a $0<\delta$, such that
$$|f(a+h)-f(a)|=|f(h)|<\epsilon \tag 3$$
whenever $0<h<\delta$
Finally, we see that $(2)$ and $(3)$ together yields $(1)$ and that completes the proof.
A: If we take $x_{0}\in\mathbb{R}
 $ we have $$\lim_{x\rightarrow x_{0}}\left(f\left(x\right)-f\left(x_{0}\right)\right)=\lim_{x\rightarrow x_{0}}f\left(x-x_{0}\right)
 $$ and now if we put $$ x-x_{0}=y-a
 $$ the limit is equal to $$=\lim_{y\rightarrow a}f\left(y-a\right)=\lim_{y\rightarrow a}\left(f\left(y\right)-f\left(a\right)\right)=0
 $$ because $f
 $ is continuous at $a
 $.
