Let $\mathbb{R}^d$ be a the usual Euclidean space and let $Y$ be a fixed non-empty closed subset of $Ball(0,1)$ (the unit ball in $\mathbb{R}^d$ about $0$ of radius $1$).

Let $f$ be the map taking $x \in \mathbb{R}^d$ to the hyperplane passing through $0$ and $x$.
Is the map $g$ defined by $$ g(x)\triangleq f(x)\cap Y, $$ continuous from $\mathbb{R}^d$ to the Hausdorff space $H(Ball(0,1))$?

If not, can we at-least conclude that it is measurable with respect to an atomless measure on $H(Ball(0,1))$?

  • $\begingroup$ I don't understand the definition of $f$, there will be many hyperplanes passing through $0$ and $x$ if $d>2$. $\endgroup$ – Michael Greinecker May 9 '19 at 9:40
  • $\begingroup$ Say we make a selection for each $x$. $\endgroup$ – N00ber May 9 '19 at 10:13
  • $\begingroup$ If $f$ is not continuous, why should $g$ be continuous? $\endgroup$ – Jan-Christoph Schlage-Puchta May 12 '19 at 15:52

Here is what is true: Let $h(p)$ be the hyperplane through $0$ with normal vector $p$ for each $p\neq0$. Then the function $p\mapsto H(p)=h(p)\cap B_1(0)$ is continuous.

Indeed, $H_p=\{x\in\mathbb{R}^d\mid px=0,\|x\|\leq 1\}$ and the Hausdorff distance of $H_p$ and $H_q$ is bounded by $\max_{x\in B_1(0)}|(p-q)x|$. Since the latter is a function continuous in $(p-q)$ and vanishes at $0$, the result follows.

Every sensible continuous selection that makes the original problem well-defined should lead to the same outcome.

| cite | improve this answer | |
  • $\begingroup$ If you consider the segment $[-1,1]\times \{0\}$ in $\mathbb{R}^2$ isn't the function discontinuous at all points of the vertical axis? $\endgroup$ – Del May 11 '19 at 9:07
  • $\begingroup$ @Del The question was about the intersection with the unit ball and not just any compact convex set. $\endgroup$ – Michael Greinecker May 11 '19 at 9:38
  • $\begingroup$ Buy $Y$ is any closed subset of the ball and we are intersecting a hyperplane with it, right? And then we compute the Hausdorff distance between two such sets. Maybe I'm misunderstanding the question $\endgroup$ – Del May 11 '19 at 11:55
  • $\begingroup$ @Del We intersect appropriately parametrized hyperplanes with the unit ball and look at the Hausdorff distances of these intersections. $\endgroup$ – Michael Greinecker May 11 '19 at 13:51

The answer to the question as asked is negative for certain choices of $f$.

For example, let $d=3$ so that any two distinct nonzero points define a unique hyperplane passing through the origin. Let $B$ be the closed ball of radius $10^{-10}$ around the point $(1,0,0)$. We claim there exist a function $f$ as in the question defined on $B$ which is not measurable. Since the restriction of a measurable function to a measurable set is measurable, it follows that any extension of $f$ to $\mathbb{R}^3$ will be nonmeasurable.

Let $A$ be a nonmeasurable subset of $B$. If $x \in A$, let $f(x)$ be the plane passing through $x$, the origin and the vector $(0,1,0)$. If $x \in B \setminus A$ let $f(x)$ be the plane passing through $x$, the origin and the vector $(0,0,1)$.

Let $P$ be the plane spanned by $(1,0,0)$ and $(0,1,0)$. Let $U$ be the open ball of radius $\frac{1}{10}$ in the Hausdorff metric around intersection of $P$ with the unit ball around the origin. Then we have $f^{-1}(U) = A$, so the claim is established.

| cite | improve this answer | |
  • $\begingroup$ Interesting, but this does not hold for any selection $\endgroup$ – N00ber May 9 '19 at 19:30
  • $\begingroup$ The previous answer shows that there exist selections such that the map in question is continuous. The answer I gave shows there exist selections such that the map in question is not measurable. Thus the answer to the question depends nontrivially on the selection. $\endgroup$ – burtonpeterj May 9 '19 at 21:30

Not the answer you're looking for? Browse other questions tagged or ask your own question.