# Intersection with a fixed set in Hausdorff metric space [closed]

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 $$H(Ball(0,1))$$?

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

• 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? – Del May 11 '19 at 9:07
• @Del The question was about the intersection with the unit ball and not just any compact convex set. – Michael Greinecker May 11 '19 at 9:38
• 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 – Del May 11 '19 at 11:55
• @Del We intersect appropriately parametrized hyperplanes with the unit ball and look at the Hausdorff distances of these intersections. – 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.

• Interesting, but this does not hold for any selection – N00ber May 9 '19 at 19:30
• 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. – burtonpeterj May 9 '19 at 21:30