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Let $c$ be a closed geometric shape. Let $P$ and $Q$ be two points in $c$. Let $S$ be the family of all closed squares. Let $s_1, s_2,...$ be a sequence of shapes from family $S$ such that for every $i$:

  • $s_i\subseteq c$
  • $s_i$ does not contain the points $P$ and $Q$.
  • The distances from $s_i$ to $P$ and from $s_i$ to $Q$ is at most $1/i$.

I am trying to prove that there is a shape $s\in S$ whose boundary contains $P$ and $Q$. Intuitively, $s$ is the "limit" of the sequence $s_1, s_2,...$. When $i \to \infty$, the distance of $s_i$ from $P$ and $Q$ goes to $0$. Here are two examples:

enter image description here

At the top, there is a sequence of squares (brown) whose distance from $P$ and $Q$ goes to 0, and this sequence "converges" to a square whose boundary contains $P$ and $Q$. At the bottom, such a sequence does not exist.

My questions are:

  • Is the claim always true? If yes, what is a formal proof for it? If no, what assumptions should be added in order to make it true?
  • What other families $S$ have this convergence property? Intuitively it seems to be true for every family $S$ that is the family of affine transformations of some fixed shapes (i.e. all disks, all parallelograms, etc.). Can it be generalized further or in other ways?

EDIT: Here is my attempt at a proof.

Every square in $S$ can be represented by a quadruple $(x_a,y_a,x_b,y_b)$ representing two opposite corners. The sequence $s_1, s_2,...$ corresponds to a sequence of such quadruples. Since all shapes in the sequence are contained in $c$, the sequence of quadruples is bounded, and hence it has an accumulation point, $(x_a,y_a,x_b,y_b)$. This accumulation point represents a square, $s$. Now this accumulation point satisfies:

  1. $s\subseteq c$. Otherwise, since $c$ is closed, $s$ intersects the exterior of $c$, and hence there must be an infinite number of $s_i$ (for sufficiently large $i$) that also intersect the exterior of $c$ - a contradiction.
  2. The distances from $s$ to $P$ is 0. Otherwise, if the distance is $D>0$, then there must be an infinite number of $s_i$ (for sufficiently large $i$) whose distance from $P$ is at least $\delta>0$, where $0<\delta\leq D$. The same goes for $Q$.
  3. The interior of $s$ does not contain $P$. Otherwise, there must be an infinite number of $s_i$ (for sufficiently large $i$) whose interior contains $P$ - a contradiction. The same goes for $Q$.

From 2 and 3, the boundary of $s$ contains both $P$ and $Q$.

Is this proof correct / sufficiently formal / sufficiently clear?

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    $\begingroup$ Your proof seems fine. To generalize for other families $S$, it will probably be useful to know the Blaschke Selection Theorem (and some easy lemmas about how distance from a point to a set relates to Hausdorff distance from the singleton containing that point to that set, and things like that). $\endgroup$ – user21467 Mar 30 '14 at 17:33

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