The inclusions between unit balls in $\ell^p$ spaces I need to show that $d_∞(x, y) ≤ d_2(x, y) ≤ d_1(x, y) ≤ nd_∞(x, y)$
where $d_1=|x_1-0|+|y_1-0|$ and I'm setting $|x_1-0|+|y_1-0|<0$. Illustrating the $B(0,1)$ balls (centered at 0 with radius 1) gives me a square with 4 points of intersections $(1,0),(0,1),(0,-1),(-1,0)$.
I then add the metrics $d_2$=$\sqrt{|x_1-0|^2+|y_1-0|^2}$ and $d_∞=\max_
{1≤i≤n}|x_i − y_i|$
Here is what they look like:
Green=$d_∞$
Red=$d_2$
Blue=$d_1$.
Now this is completely counter intuitive since it's nested in the opposite order than I expected. Did I go wrong anywhere?


The balls have been drawn correctly. E,g, note that being in the smallest ball $d_1((x,y),(0,0)) \le 1$ implies (if the inequalities are correct, which they are) that both $d_\infty((x,y),(0,0)) \le 1$ and $d_2((x,y),(0,0)) \le 1$ as well. So indeed the $d_1$ ball is contained in the other two.
For $n=2$:
$d_{1}((x,y),(u,v)) = |x-u| + |y-v| \le 2\max(|x-u|,|y-v|)$, as we estimate both terms with the max. For $n$ coordinates we get a factor $n$. 
Also $\max(|x-u|,|y-v|)^2 \le |x-u|^2 + |y-v|^2$ as the max squared is one of the terms in the sum, and all terms are positive. Now take two square roots to see that $d_{\infty} \le d_2$.
