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?

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 $B_{\mathbb{R}^{m_1+m_2}}$.
