Intersection of two balls in n-dimension I've been thinking about how two prove this statement (mathematically) but cant seem to figure it out
the definition of a ball two dimensions is $B(x,r)=:\|x-y\|\leq r ~~\forall y\in\mathbb{R} $ where $x$ is the center of the ball and r is the radius.
Now define
$B_\mathbb{R^{m_1+m_2}}((x_1x_2),r)$
a ball in $\mathbb{R^{m_1+m_2}}~~~$ where $x_1 =(x1,x2,x3,\dots,xm_1)$$x_1 =(xm_1+1,xm_1+2,xm_1+3,\dots,xm_2)$
and
$B_\mathbb{R^{m_1}}(x_1,r)\times B_\mathbb{R^{m_2}}(x_2,r)$
the cartesian product of two balls in $\mathbb{R^{m_1}}$ and $\mathbb{R^{m_2}}$
prove that $B_\mathbb{R^{m_1+m_2}}((x_1,x_2),r)\subset B_\mathbb{R^{m_1}}(x_1,r) \times B_\mathbb{R^{m_2}}(x_2,r)$
in the one dimensional case a ball is just an interval so the cartesian product of two intervals is a square and its easy to see geometrically that a circle centered at $(x_1,x_2)$ and radius r is less than a square with sides of length r. One idea I have is to choose an arbitrary point $z \in B_\mathbb{R^{m_1+m_2}}((x_1,x_2),r)$ and show that it is bounded by the cartesian product of the two balls.
the first step I think would be to center the balls at the origin to make calculation simpler.
defining  $B_\mathbb{R^{m_1}}(0,r)\times B_\mathbb{R^{m_2}}(0,r)$ in n dimensions.
would it be $B_\mathbb{R^{m_1}}(0,r)\times B_\mathbb{R^{m_2}}(0,r)=:max\{\|x_1\|,\|x_2\|\}$ (i think) where do i go next?
 A: First I would rephrase the problem so there are not so many subscripts.
Let's work with $\mathbb R^{m+n} = \mathbb R^m \times \mathbb R^n.$
And we usually use the letter $x$ to describe the coordinates of an arbitrary point in space (or in the ball), not the center of the ball.
As you suggest, the problem is simpler if we suppose that the coordinates at the center of the ball are all zero; later we might consider non-zero coordinates at the center.
Also, we know the balls we're dealing with are all in real Cartesian coordinate spaces, so it's enough to specify the dimension of the space, that is,
$B_k$ rather than $B_{\mathbb R^k}.$
So the $(m+n)$-dimensional ball of radius $r$ with center $0$ in $\mathbb R^{m+n}$ is
$$ B_{m+n}(0,r)
 = \{ (x_1,x_2,\ldots,x_{m+n})\mid x_1^2 + x_2^2 + \cdots + x_{m+n}^2 \leq r^2 \}. $$
The Cartesian product of $B_m(0,r)$ and $B_n(0,r)$ is
\begin{eqnarray}
 B_m(0,r) \times B_n(0,r)
&=& \{ (x_1,x_2,\ldots,x_m)\mid x_1^2 + x_2^2 + \cdots + x_m^2 \leq r^2 \}\\
&& {}\quad \times \{ (x_{m+1},x_{m+2},\ldots,x_{m+n})\mid x_{m+1}^2 + x_{m+2}^2
   + \cdots + x_{m+n}^2 \leq r^2 \} \\
&=& \{ (x_1,x_2,\ldots,x_{m+n})\mid x_1^2 + \cdots + x_m^2 \leq r^2 , \,
       x_{m+1}^2 + \cdots + x_{m+n}^2 \leq r^2 \} \\
\end{eqnarray}
To show that $B_{m+n}(0,r) \subseteq B_m(0,r) \times B_n(0,r),$
you merely need to show that if $(x_1, x_2, \ldots, x_{m+n})$ is a point in
$\mathbb R^{m+n}$ such that
$$ x_1^2 + x_2^2 + \cdots + x_{m+n}^2 \leq r^2 $$
then both of the following inequalities also are true:
$$
\begin{cases}
x_1^2 + x_2^2 + \cdots + x_m^2 \leq r^2,\\[0.5ex]
x_{m+1}^2 + x_{m+2}^2 + \cdots + x_{m+n}^2 \leq r^2.
\end{cases}
$$
To show this, observe that
$$
x_1^2 + x_2^2 + \cdots + x_{m+n}^2 = 
 (x_1^2 + x_2^2 + \cdots + x_m^2) + (x_{m+1}^2 + x_{m+2}^2 + \cdots + x_{m+n}^2)
$$
and that
\begin{gather}
x_1^2 + x_2^2 + \cdots + x_m^2 \geq 0,\\[0.5ex]
x_{m+1}^2 + x_{m+2}^2 + \cdots + x_{m+n}^2 \geq 0.
\end{gather}
Therefore
$$ x_1^2 + x_2^2 + \cdots + x_m^2 \leq x_1^2 + x_2^2 + \cdots + x_{m+n}^2 $$
and
$$ x_{m+1}^2 + x_{m+2}^2 + \cdots + x_{m+n}^2
\leq x_1^2 + x_2^2 + \cdots + x_{m+n}^2. $$

For a ball with center somewhere other than all zero coordinates,
we can suppose we have a ball
$B_{m+n}(\mathbf a, r)$ centered at
$$ \mathbf a = (a_1, a_2, \ldots, a_{m+n}). $$
Then instead of $x_1^2 + x_2^2 + \cdots + x_{m+n}^2 \leq r^2$ we must write
$$ (x_1 - a_1)^2 + (x_2 - a_2)^2 + \cdots + (x_{m+n} - a_{m+n})^2 \leq r^2 $$
and so forth.
A: Consider $\|\cdot\|$ the maximum norm.
If $x \in B_{\Bbb{R}^{m_1 + m_2}}((x_1,x_2);r)$, then
$$\|(x_1,x_2) - x\| \leq r.$$
Making the identification $(v_1,...,v_{m_1},\underbrace{0,...,0}_{m_2\text{ times}})$ with $(v_1,...,v_{m_1}) \in \Bbb{R}^{m_1}$, you can write $x = (y_1,y_2)$ where $y_1 \in \Bbb{R}^{m_1}$ and $y_2 \in \Bbb{R}^{m_2}$. Thus,
$$\|(x_1-y_1,x_2-y_2)\| \leq r.$$
Note that
$$\|x_1 - y_1\| \leq \max\{|x_1 - y_1|,|x_2 - y_2|\} = \|(x_1-y_1,x_2-y_2)\| \leq r$$
in the same way, $\|x_2 - y_2\| \leq r$. Therefore, $y_1 \in B_{\Bbb{R}^{m_1}}(x_1,r)$ and $y_2 \in B_{\Bbb{R}^{m_2}}(x_2,r)$.
A: Assuming you are referring to the standard Euclidean norm, an answer to your question "where to go next"? is to simply observe that
$$\max\{\|x_1\|,\|x_2\|\}\leq (\|x_1\|^2+\|x_2\|^2)^{1/2}$$
and the r.h.s is the norm of $B_{\mathbb{R}^{m_1+m_2}}$.
