In $\mathbb R^3$ define a brick to be of the form $[a,b] \times [c,d] \times [e,f] $
Define a body to be a finite union of bricks.
For any object $X$, let $|X|$ denote its "volume". Further, denote it's projection onto the plane spanned by the $i$th and $j$th standard basis vectors by $X_{ij}$
I am asked to show why it is false to say that if we set $T = S \times S \in \mathbb R^6$, then:
$|S|^2 = |T| \leq |T_{12}||T_{34}||T_{56}| = |S_{12}||S_{31}||S_{23}| $
I suspect the problem lies in the second equality symbol and so I try to demonstrate this with the example of $S = [0,1] \times [0,2] \times [0,1] \; \cup [1,2] \times [0,1] \times [0,1]$ and so $|S| = 3$
This gives $S_{12} = [0,1] \times [0,2] \; \cup [1,2] \times [0,1], \; S_{23} = [0,2] \times [0,1], \; S_{13} = [0,1] \times [ 0,1]$
Now then:
$$T = [0,1] \times [0,2] \times [0,1] \times [0,1] \times [0,2] \times [0,1] \\ \cup [0,1] \times [0,2] \times [0,1] \times [1,2] \times [0,1] \times [0,1] \\ \cup [1,2] \times [0,1] \times [0,1] \times [0,1] \times [0,2] \times [0,1] \\ \cup [1,2] \times [0,1] \times [0,1] \times [1,2] \times [0,1] \times [0,1]$$
Then we see that:
$T_{12} = [0,2] \times [0,2], \; T_{34} = [0,1] \times [0,2], \; T_{56} = [0,2] \times [0,1]$
So we see that: $|S|^2 = 9 = 4 + 2 + 2 + 1 = |T|$ so the first equality holds.
$|T_{12}||T_{34}||T_{56}| = 4*2*2 = 16$
But: $|S_{12}||S_{23}||S_{13}| = 3*2 * 2 = 12$
This then shows that the second equality is violated.
I wanted to ask though if what I did here was correct, because I'm not so sure that I calculated everything correctly.
I would really appreciate any help, thank you!