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?

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 $y_2 \in B_{\Bbb{R}^{m_2}}(x_2,r)$.
