# The set of boundary vectors of compact convex body has empty interior

Let $$K$$ be a compact convex body in the Euclidean space $$\mathbb R^n$$ and $$\partial K$$ be its topological boundary in $$\mathbb R^n$$.

Definition. A vector $$\mathbf v\in\mathbb R^n$$ is called $$K$$-boundary if there exists a point $$\mathbf a\in\partial K$$ such that $$[\mathbf a,\mathbf a+\varepsilon\mathbf v]\subseteq\partial K$$ for some $$\varepsilon>0$$.

Conjecture: The set of $$K$$-boundary vectors has empty interior in $$\mathbb R^n$$.
More precisely, this set has Hausdorff dimension $$\le n-1$$.

Question. Is this conjecture true? And if yes, where is it proved?
It sounds like a basic fact in Convexity Theory.

Remark 2. For $$n=2$$, Conjecture is true: the set of $$K$$-boundary vectors coincides with a countable union of linear subspaces of dimension $$\le 1$$ and hence it has empty interior in $$\mathbb R^2$$. Moreover, it is meager and Lebesgue null in $$\mathbb R^2$$.

Remark 3. For $$n=3$$, Conjecture also is true: the set of $$K$$-boundary vectors has Hausdorff dimension 2 and hence is Lebesgue null and meager in $$\mathbb R^3$$, see my partial answer below.

Remark $$n$$ (Added in $$n$$-th Edit): It seems that the 3-dimensional agrument used in my answer below easily generalizes to higher dimensions and proves

Theorem. For any compact body $$K$$ in a Euclidean space $$\mathbb R^n$$ and any $$k\le n$$ the subspace $$\big\{L\in G(k,\mathbb R^n):\exists x\in\partial K\;\dim(\partial K\cap(x+L))=k\big\}$$of the Grassmannian $$Gr(k,\mathbb R^n)$$ has Hausdorff dimension $$n-k-1$$.

So, now the problem is to find this theorem somewhere in literature (I tend to believe that such a result had to appear somewhere).

• @WlodAA My conjecture can be someone's theorem. Everything is relative (as Einstein said :) – Taras Banakh Mar 27 at 7:33
• @WlodAA I also this it is true, but somehow the proof escapes. The set of K-boundary vectors is in fact $F_\sigma$, so the conjecture actually says that this set is meager. But maybe for non-empty interior it will be easier to prove. – Taras Banakh Mar 27 at 7:35
• Aiming at measure zero could be another approach. – Wlod AA Mar 27 at 7:45
• @erz I do not know if this will help: for any convex polyhedron the set of boundary vectors is the union of hyperplanes parallel to faces so a nowhere dense set, but the convex hull of all segments in the boundary is the initial convex polyhedron. And what kind of information can be deduced from this fact? But locally, indeed, the union of segments on the boundary that contain a given point in its interior is an interesting object, which can be used to classify points of the boundary: vertex points, edge points, flat points etc. But what next? – Taras Banakh Mar 27 at 8:46
• Another idea (sorry, if it is again unhelpful): let us fix a hyperplane $H$, take away the (at most two) pieces of $\partial K$ that contain open sets of a hyperplane parallel to $H$, and try to collect the $K$ boundaries on what is left, but parallel to $H$. Will the obtained set have empty interior? – erz Mar 27 at 9:06

Here is a proof of Conjecture in dimension $$n=3$$.

Let $$K$$ be a compact convex body in $$\mathbb R^3$$. For $$k\le 3$$, let $$Gr(k,3)$$ be the space of $$k$$-dimensional linear subspaces of $$\mathbb R^3$$. So, $$Gr(1,3)$$ is the space of lines and $$Gr(2,3)$$ is the space of planes in $$\mathbb R^3$$.

A point $$x\in\partial K$$ will be called

$$\bullet$$ flat if there is a plane $$P\in Gr(2,3)$$ such that $$x$$ is an interior point of the intersection $$\partial K\cap(x+P)$$ in $$x+P$$;

$$\bullet$$ edge if $$x$$ is not flat and there is a line $$L\in Gr(1,3)$$ such that $$x$$ is an interior point of the intersection $$\partial K\cap(x+L)$$ in $$x+L$$.

Claim 1. For each flat point $$x\in\partial K$$ there exists a unique plane $$p(x)\in Gr(2,3)$$ witnessing that $$x$$ is flat.

Let $$F$$ be the set of all flat points in $$\partial K$$.

Claim 1 allows us to define the map $$p:F\to Gr(2,3)$$ assigning to each flat point $$x$$ the unique plane $$p(x)$$ witnessing that $$x$$ is flat.

Claim 2. The image $$p(F)\subseteq Gr(2,3)$$ of $$F$$ is at most countable.

Claim 3. For every edge point $$x\in\partial K$$ there exists a unique line $$L\in Gr(1,3)$$ witnessing thar $$x$$ is an edge point.

Let $$E\subseteq\partial K$$ be the set of edge points in $$\partial K$$. Claim 3 allows us to define a function $$\ell:E\to Gr(1,3)$$ assigning to each edge point $$x$$ the unique line $$\ell(x)$$ witnessing that $$x$$ is an edge point.

Endow the projective plane $$Gr(1,3)$$ with the metric assigning to any lines $$\ell_1,\ell_2\in Gr(1,3)$$ the Hausdorff distance between the doubletons $$\ell_1\cap S^2$$ and $$\ell_2\cap S^2$$.

For every $$n\in\mathbb N$$ let $$E_n$$ be the set of edge points $$x\in E$$ for which there exist points $$a,b\in\partial K$$ such that $$\|a-b\|=\frac1n$$, $$x=\frac12(a+b)$$ and $$[a,b]\subseteq \partial K\cap\ell(x)$$. Here $$\|\cdot\|$$ denotes the standard Euclidean norm of the space $$\mathbb R^3$$.

Claim 4. For every $$n\in\mathbb N$$ the restriction $$\ell{\restriction}_{E_n}:E_n\to Gr(1,3)$$ is a Lipschitz map.

For every edge point $$x\in E$$ let $$x+\ell(x)^\perp$$ be the plane in $$\mathbb R^3$$ that contains $$x$$ and is orthogonal to the line $$\ell(x)$$.

Claim 5. For every $$n\in\mathbb N$$ and every point $$x\in E_n$$ there exists a neighborhood $$O_x$$ of $$x$$ in $$E$$ such that for every $$y\in O_x$$ the line $$y+\ell(y)$$ has non-empty intersection with the set $$E_{2n}\cap (x+\ell(x)^\perp)$$.

Claims 4 and 5 imply the following

Claim 6. For every $$n\in\mathbb N$$ and every $$x\in E_n$$ there exists a neighborhood $$O_x$$ of $$x$$ in $$E_n$$ such that $$\ell(O_x)\subseteq \ell\big(E_{2n}\cap (x+\ell(x)^\perp)\big)$$.

The set $$E_{2n}\cap (x+\ell(x)^\perp)$$ is contained in the intersection $$\partial K\cap (x+\ell(x)^\perp)$$ which is a topological circle of Hausdorff dimension 1. Since the map $$\ell{\restriction}_{E_{2n}}$$ is Lipschitz, the image $$\ell(O_x)\subseteq \ell(E_{2n}\cap(x+\ell(x)^\perp))\subseteq \ell(\partial K\cap(x+\ell(x)^\perp))$$ has Hausdorff dimension $$\le 1$$.

Then the set $$\ell(E)$$ has Hausdorff dimension $$\le 1$$ being the union of countably many sets of Hausdorff dimension $$\le 1$$. Consequently, the union $$\bigcup_{x\in E}\ell(E)$$ has Hausdorff dimension $$\le 2$$ and hence is Lebesgue null in $$\mathbb R^3$$.

Since the set $$p(F)$$ is countable, the union $$\bigcup_{x\in F}p(x)$$ of countably many planes has Hausdorff dimension $$2$$ and hence is Lebesgue null in $$\mathbb R^3$$.

Observing that the set $$B$$ of $$K$$-boundary vectors is a subset of $$\bigcup_{x\in F} p(x)\cup \bigcup_{x\in E}\ell(x)$$, we see that $$B$$ is Lebesgue null and hence has empty interior in $$\mathbb R^3$$.