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$D_r(x)$ denotes a closed ball of radius $r$ centered at $x$.

Definition. Let $M \subset \mathbb{R}^n$.

  • $D_r (M): = \bigcup\limits_{x \in M} D_r (x)$
  • $Int_r (M): = \{x ~|~ D_r(x) \subset M\}$

Question: Is it true that $M \subset \mathbb{R}^n$ the following conditions are equivalent

  • There is $r> 0$ such that $Int_r (D_r (M)) = M$ and $D_r (Int_r (M)) = M$
  • M is a n-dimensional smooth submanifold

Related: https://en.wikipedia.org/wiki/Parallel_curve

Update: This question was previously called question 2 and there was also a trivial question 1 (see edit history), which was refuted in the first comment.

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    $\begingroup$ What about a filled rectangle in $\mathbb{R}^2$ with $r > 0$ small enough? (Question 1 is about shapes invariant under the closing operation en.wikipedia.org/wiki/Closing_(morphology), Question 2 about invariance under both closing and opening). $\endgroup$
    – Luc Guyot
    Commented Nov 25, 2021 at 11:41
  • $\begingroup$ For Question 2, wouldn't the union of a filled rectangle with two half-disks attached along their diameters to two opposing sides (pill-like or capsule-like shape) give a negative answer? $\endgroup$
    – Luc Guyot
    Commented Nov 25, 2021 at 12:32
  • $\begingroup$ > filled rectangle Oh, yes, indeed. $\endgroup$
    – Arshak Aivazian
    Commented Nov 25, 2021 at 17:13
  • $\begingroup$ It seems to me that for $n$-dimensional smooth submanifolds the operation $D_r (M)$ for small r simply puts a segment in the direction of the outward normal. Then no point $D_r (M) \setminus M$ belongs to $Int_r D_r (M)$, because its $r$-neighborhood does not belong to $D_r (M)$ (namely, its $r$-neighborhood contains some next points in the direction of the same normal). . So $Int_r D_r (M) = M$. Is that correct? $\endgroup$
    – Arshak Aivazian
    Commented Nov 25, 2021 at 17:25
  • $\begingroup$ It seems to me that, and vice versa, for points close to the boundary, the distance from the inner point to the boundary is achieved in the direction of (some) inner normal vector. Then, for small r, the $Int_r (M)$ operator simply removes half-intervals of length r from the boundary in the direction of the inner normal vectors. Then $D_r (\text{end of the remote half-interval})$ contains the entire remote half-interval. So $D_r Int_r (M) = M$. $\endgroup$
    – Arshak Aivazian
    Commented Nov 25, 2021 at 17:49

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