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Let $c>0$ be given. I look for $n\ge 1$ and a collection of closed subsets $(F_i: 1\le i\le n)$ such that

$$\bigcup_{1\le i\le n} {\rm int}(F_i)= \mathbb R^d,$$

and for every $x\in \mathbb R^d$, there exists $k\in P(x)$ satisfying $x\in {\rm int}(F_k)$ and

$$\inf \big\{|x-y|: y\in \partial F_k\big\}\ge c,$$

where $P(x):=\{1\le i\le n: x\in {\rm int}(F_i)\}$. Here ${\rm int}(F_k)$ denotes the interior of $F_k$ and $\partial F_k$ denotes the boundary of $F_k$.

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    $\begingroup$ I think your condition on $k$ is stated twice, no? In any case I suppose the two half-spaces $F_1=\{x:x_1\geq-c\}$ and $F_2=\{x:x_1\leq c\}$ should work. $\endgroup$
    – Pierre PC
    Commented Feb 12 at 14:02
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    $\begingroup$ (Or, as stated, $F_1=\mathbb R^d$...) $\endgroup$
    – Pierre PC
    Commented Feb 12 at 14:05
  • $\begingroup$ @PierrePC Many thanks for this quick example. Indeed, I look for the closed subsets $F_k$ of the form $F_k: = \{x : a\le |x|\le b\}$, while your construction still seems to work right? $\endgroup$
    – GJC20
    Commented Feb 12 at 14:11
  • $\begingroup$ The condition can be stated as "... and for every $x\in\mathbf{R}^d$, there exists $k$ such that $B(x,c)\subset F_k$". $\endgroup$
    – YCor
    Commented Feb 12 at 14:59
  • $\begingroup$ But anyway the question is trivial as stated, so I assume that some requirements are missing. [Please make edits accordingly, rather than adding them in comments.] $\endgroup$
    – YCor
    Commented Feb 12 at 14:59

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