Addition of balls in normed vector space I'm having a problem where two balls are given in normed vector space and it is asked to prove that addition of these balls is ball as well.
I know that $B(a,r) = rB(0,1) + \{a\} $ and $rB(0,1) = r\{ x: \|x\| < 1\} + {a}$
But how to show that sum of balls $B_1(a,r_1) + B_2(b,r_2)$ is also ball, for example $C( ... )$ ?
I think the following is not quite correct. I seek understanding.

$(r_1\{x:\|x\| <1 \} + \{a\}) + (r_2\{x: \|x\| < 1 \} + \{b\})$
For addition, it seems right to bring $r_{1,2}$ inside the set: $\{rx: \|x\| < 1 \}$ and form sets with $y$ as follows: $y = rx \Rightarrow x = \frac{y}{r}$,
$\left\{y: \left\|\frac{y}{r}\right\| < 1 \right\}$.
Now adding formed sets 
\begin{align}
\left(\left\{y_1:\left\|\frac{y_1}{r_1}\right\| < 1 \right\} +\{a \}\right) + \left(\left\{y_2: \left\|\frac{y_2}{r_2}\right\| <1 \right\} + \{ b\}\right)\\= \left\{y_1+y_2: \left\|\frac{y_1}{r_1} + \frac{y_2}{r_2}\right\| < 2  \right\} + \{a\} + \{b\}
\end{align}

 A: First verify that $r_1B(0,1)+r_2B(0,1)=(r_1+r_2)B(0,1)$ Any element of the left side is of the type $r_1 x +r_2y$ with $||x||<1$ and $||y||<1$. Since $r_1 x +r_2y=(r_1+r_2)(cx+dy)$ where $c=\frac {r_1} {r_1+r_2}$ and $d=\frac {r_2} {r_1+r_2}$ and since $||cx+dy||<c+d=1$ we see that $r_1B(0,1)+r_2B(0,1) \subset(r_1+r_2)B(0,1)$. To prove the reverse inclusion let $x=(r_1+r_2)y$ where $||y||<1$. Then $x=r_1y+r_2y \in r_1 B(0,1)+r_2 B(0,1)$ We have proved that $r_1B(0,1)+r_2B(0,1)=(r_1+r_2)B(0,1)$. Now you should be able to complete teh proof using that fact that $a+B(x,r)=B(a+x,r)$ for any $a,x,r$. The answer is, of course $B(a,r_1)+B(b,r_2)=B(a+b,r_1+r_2)$. 
A: As suggested by @Kavi Rama Murthy, the answer is:

$$B(a, r_1) + B(b, r_2) = B(a + b, r_1 + r_2)$$

Let $x \in B(a, r_1) + B(a, r_2)$. By definition, $x= u + v$ , where $u \in B(a, r_1)$ and $v \in B(b, r_2)$.
Therefore:
$$\|x - (a + b)\| = \|u - a + v - b\| \le \|u - a\| + \|v - b\| < r_1 + r_2$$
so $x \in B(a + b, r_1 + r_2)$.
Conversely, let $x \in B(a + b, r_1 + r_2)$. Notice that
$$x = \underbrace{\frac1{r_1 + r_2}\Big(r_1(x - b) + r_2a\Big)}_{\in B(a, r_1)} + \underbrace{\frac1{r_1 + r_2}\Big(r_2(x - a) + r_1b\Big)}_{\in B(b, r_2)}$$
Indeed:
$$\left\|\frac1{r_1 + r_2}\Big(r_1(x - b) + r_2a\Big) - a\right\| = \frac{r_1}{r_1 + r_2}\underbrace{\|x - (a + b)\|}_{< r_1 + r_2} < r_1$$
$$\left\|\frac1{r_1 + r_2}\Big(r_2(x - a) + r_1b\Big) - b\right\| = \frac{r_2}{r_1 + r_2}\underbrace{\|x - (a + b)\|}_{< r_1 + r_2} < r_2$$
Therefore $x \in B(a, r_1) + B(b, r_2)$.
