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Determine the intersection of a plane $P$ and cube $U$ in $\mathbb R^3$ using topology.

Note: This can be solved analytically (using brute force); this question asks for a topological argument (ideally, one that might generalize to higher dimensions.)

My partial work is below, but there are some gaps. Can you help me fill them?

Solution: The intersection of a plane $P$ and cube $U$ in $\mathbb R^3$ is either the empty set, a single point, a segment, a triangle, a quadrilateral, a pentagon, or a hexagon, but nothing else.

Each of the six faces $F_n$ of $U$ defines a plane $R_n$; call their union $R_*$. Assume that $P$ intersects the interior of $U$ and that $P$ is not parallel to any $R_n$ (the other cases are easy and are left out here.) Then $P \cap R_*$ consists of six lines $\ell_n$; call their union $L_*$. Then the boundary of $I$ is a subset of $L_*$ [Q1] .

Since $U$ is a convex compact and $P$ a convex closed set of their intersection $I$ is a convex compact set. Furthermore, since their interiors intersect, $I$ has dimension 2 [Q2] .

Finally, as above, $I$ bounded by a subset of 6 lines. The only figures which satisfy this are triangles, quadrilaterals, pentagons, and hexagons [Q3].

Q1: It seems clear that $\partial I \subset L_*$, but how do I prove it?
Q2: Here too, how do I prove that it's dim 2? Is this a transversal intersection of codim 0 and codim 1? How do I show it's transversal?
Q3: I've shown $I$ must be one of these; but how do I know each is possible?


Update

I believe I've solved Q1, to which I request verification.

Observe that $\partial U \subset of R_*$ (where "$\partial$" denotes the boundary) and that $\partial P = \emptyset$, and recall that the boundary of an intersection of sets is a subset of the union of their boundaries. Therefore $\partial I \subset (\partial U \cap \emptyset) \subset R_*$, and since $\partial I \subset P$, then $\partial I \subset P \cup R_* = L_*$.


Background

To clarify the motivation: Vladimir Arnold poses the question (#23):

What polygons may be obtained as sections of a cube by a plane? Can we get a pentagon? A heptagon? A regular hexagon?

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I've seen this approach analytically, where you use 12 equations to find the points of intersection, eliminate those outside the range of the cube, and manipulate the coefficients to show the various shapes they form. But this is cumbersome, repetitive, and produces no insight.

What intrigued me is: What can we show about the intersection without committing to specific metrics? For example, we can show it is convex, compact, two dimensional, and bounded by at most six intersections of two planes. This alone provides almost the complete solution.

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    $\begingroup$ Just a wild guess, as this is not similar to anything I've done or thought about before, but perhaps you are to regard the cube as a sphere (topologically the same as a cube), which simplifies the types of intersections possible? This is assuming that "using topology" means you only want the possible topological type for the intersection (i.e. give all possible intersections, where two intersections are considered the same if they are homeomorphic). $\endgroup$ Jan 24 at 19:58
  • $\begingroup$ Without additional context, it is hard to say what you should be doing here. The concepts of "plane, cube, convex, triangle, quadrilateral, pentagon, hexagon" are all geometric, not topological, as is most of your argument. Apparently, the problem came to you in a context were what was intended by "determine" and "using topology" was clear, but we do not know what that context is. What course are you taking? $\endgroup$ Jan 25 at 19:27
  • $\begingroup$ @PaulSinclair I've added more context - please let me know if that clarifies. By topological I meant only irrespective of specific metric values but perhaps that's the wrong usage of the word (perhaps qualitative would be correct.). $\endgroup$ Jan 25 at 21:44
  • $\begingroup$ So what you want to find is what types of polygons can result from such an intersection? Please say so clearly! "Deternine the intersection" does not indicate in any way "types of polygon" is what you were after. Even when you start taking about polygons in your follow-up, I had no idea that this was your endpoint, not just some step on the way. $\endgroup$ Jan 26 at 4:36
  • $\begingroup$ Personally, it twould not even occur to me to approach this by equations. I would examine it by cases, starting with the case of the plane containing an edge (rectangles only), for planes not containing edges, various possible combinations of picking three points on distint sides or vertices, and what the planes through those points provides. Precise calculation of numbers does not add anything to this. $\endgroup$ Jan 26 at 4:41

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To determine what polygons can arise from intersecting a cube with a plane, note that the plane will intersect the planes of the faces in lines (unless it is the plane of the face, which case the polygon formed is a square). Because the faces are convex, the intersection of that line with the face itself will be a line segment. These line segments are what forms the polygon. Since there is at most one segment per face, and a cube has six faces, the polygon can have at most six sides.

This means the only polygons possible are triangles, quadrilaterals, pentagons and hexagons. it is not hard to show examples of each: Cutting off one corner of the cube results in a triangle. Rectangle cuts abound. By cutting off a full edge by a plane not parallel to it, you can get non-rectangular quadrilaterals. To get a hexagon, choose a full diagonal of the cube and take plane perpendicular to it which passes through the center. To get a pentagon, wobble that plane just enough to make it pass through a single vertex.

Regular triangles, squares and hexagons are easily possible. It seems doubtful to me that you could get a regular pentagon, though. But I have not examined it that closely enough to be sure.

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