Number of hypercubes intersected by a hyperplane in a uniform partitioned hypercube Suppose I have a $d$-dimensional unit hypercube $[0,1]^d$. I partition each dimension uniformly into $k$ intervals. Now the unit hypercube is partitioned into $k^d$ small hypercubes. I'm wondering how many of these hypercubes can a hyperplane intersect as $k$ goes to infinity. My conjecture is that this number can be bounded by $C\cdot k^{d-1}$ where $C$ is independent of $k$. But I cannot come up with a proof. Any suggestions on what I should be looking for?
Thanks in advance.
 A: Here is a rough (and unoptimized) idea: let's work in $3D$ because it can still be visualised and I think it makes a plausible case.
Imagine a grid of unit cubes filling out space (i.e. a $3D$ version of squared paper). Now place another unit cube into this grid at an arbitrary point and rotated in an arbitrary fashion. It will clearly intersect at most $A$ cubes, where $A$ is some fairly small number (e.g. it certainly won't intersect more that $100$ grid cubes). This setup obviously also works in exactly the same way if we scale everything by a common factor.
Now take your $3D$ unit cube $[0,1]^3$ that you've partitioned into $k^3$ small cubes. Your plane passing through the cube is oriented in some way. Now take another cube with integer sidelenght $B$, rotated so as to align with the plane (i.e. the plane is parallel to one of the cube's sides) and placed so that it contains $[0,1]^3$ (for this to be possible, $B$ has to be big, e.g. $B = 100$ will certainly always suffice). Furthermore, cut the rotated cube into $(Bk)^3$ small cubes of sidelength $1/k$.
Now it is clear that the plane intersects at most $2(Bk)^2$ of the small rotated cubes since the big rotated cube is aligned so as to be parallel to the plane (the factor of $2$ is there because the plane might slice right in between two layers of the rotated cube). But each small rotated cube intersects at most $A$ of the original small cubes that came from cutting up $[0,1]^3$ by our previous cubic grid argument. (Here the grid doesn't fill out all of space, but that only improves the situation.)
Putting this all together: the intersection of the plane and $[0,1]^3$ is some set that is contained in the intersection of the plane and the big rotated cube. The intersection of the plane and the big rotated cube is contained in the union of at most $2(Bk)^2$ rotated small cubes. Each rotated small cube is contained in the union of at most $A$ original small cubes. Hence the intersection of the plane and $[0,1]^3$ is contained in the union of at most $2AB^2k^2$ small original cubes and we are done.
This is clearly not optimized since I didn't evaluate the constants. Also, we can't really visualise it in higher dimesions. But I do believe the same idea should work in all dimensions, though I don't know how to nicely articulate it.
A: I thought about a proof inspired by the answer of J P and an idea from a friend of mine. Just post here for discussion.
Consider a hyperplane represented by $\sum_{i=1}^{d}a_{i}\xi_{i}=a_{0}$. Let $j=argmax_{i}|a_{i}|$. The hyperplane can be represented as $\xi_{j} = \sum_{i=1, i\neq j}^{d}\frac{-a_{i}}{a_{j}}\xi_{i} + \frac{a_{0}}{a_{j}}$ where $|\frac{-a_{i}}{a_{j}}|\leq 1 \quad \forall i=1,\dots,d, i\neq j$.
Consider a projection of all the hypercubes intersected by the hyperplane to the space of $(\xi_{1},\dots, \xi_{j-1},\xi_{j+1},\dots,\xi_{d})$. After the projection, there are at most $k^{d-1}$ hypercubes with dimension $d-1$.
Now the next question is how many of the $d$ dimensional hypercubes can be projected on each of these $d-1$ dimensional hypercubes. Without loss of generality, consider the $d-1$ dimensional hypercube
\begin{equation*}
  \Xi^{d-1}_{0}=  \big\{(\xi_{1},\dots, \xi_{j-1},\xi_{j+1},\dots,\xi_{d}): 0\leq \xi_{i}\leq \frac{1}{k} \quad \forall i=1,\dots,d, i\neq j \big\}
\end{equation*}
Let  $l_{0}$ and $u_{0}$ be tightest the lower and upper bound of $\xi_{j}$ for the $d$-dimensional hypercubes projected on $\Xi^{d-1}_{0}$.   Since $|\frac{-a_{i}}{a_{j}}|\leq 1 \quad \forall i=1,\dots,d, i\neq j$, we have $u_{0}-l_{0}\leq \frac{1}{k}\sum_{i,i\neq j}|\frac{-a_{i}}{a_{j}}|\leq \frac{d-1}{k}$, i.e., the $j$th coordinate of the $d$ dimensional hypercubes projected on $\Xi^{d-1}_{0}$ can vary at most $\frac{d-1}{k}$. If we arrange the length $\frac{d-1}{k}$ optimally, at most $d$ hypercubes can be intersected by the hyperplane. This argument holds for each $d-1$ dimensional hypercubes. Therefore, at most $d\cdot k^{d-1}$ hypercubes can be intersected by the hyperplane.
