Geometry question and Diameter

Question

Let $F$ be a bounded subset of $\mathbb{R}^2$. Impose a $\delta-grid$ on $F$. (Squares of length and height $\delta$). Prove that any subset $U$ of diameter at most $\delta$ is contained in $3^2$ cubes.

How do I prove this? I drew a picture. I note that each cube has diameter at most $\delta\sqrt(2)$. So a set of diameter $\leq \delta$. How do I formally prove this?

General observations: (1) A circle of diameter $\leq \delta$ can be placed within exactly 1 $\delta-$ cube.

(2) When I draw intervals in $\mathbb{R}^1$, it seems reasonably clear that the worst case scenario is that the set is in 3 intervals. However I am not sure how to prove it. It seems to me that it should it be at most $2^n$ (which is still fine).

(3) If I take each cube that contains the set of diameter less than or equal to $\delta$, then geometrically, it is clear that it can be in at most $3^n$ cubes.

My proof sketch consists of breaking it into cases, if the set is already in a cube then we are done. If the set is contained more than $3^n$ cubes then since each cube has diameter $\sqrt{n}\delta$, then the diameter of the set would would be greater than or equal to $\sqrt{n}\delta3^n$, a contradiction.

Answer 1

You may assume $\delta=1$. Let $U\subset{\mathbb R}^2$ have diameter $\leq1$. Define $$a:=\inf_{(x,y)\in U}x,\qquad b:=\sup_{(x,y)\in U}x,\qquad c:=\inf_{(x,y)\in U}y,\qquad d:=\sup_{(x,y)\in U}y\ .$$ Then $U\subset[a,b]\times[c,d]$. Now $$0\leq b-a\leq1,\qquad 0\leq d-c\leq1\ .$$ Under these circumstances it is clear that $[a,b]\times[c,d]$ can hit at most $3\cdot3$ unit squares.

I think that the claimed inequality can be improved from $3^2=9$ to a smaller number of grid squares.

Answer 2

Given a subset $E \subset \mathbb{R}^2$, the set of all $x$-coordinates of $E$, denoted by $\pi_x E$ is a subset of $\mathbb{R}$ with diameter less than that of $E$, hence less than $\delta$. Thus, there are 2 or 1 successive "horizontal edges" in your grid that contain $\pi_x E$: fix any $x$-value, then all other $x$-values are less than $\delta$ away. Call these intervals $I_1$ and $I_2$ (if applicable.)

Similarly, there are two vertical subintervals, $J_1$ and $J_2$, adjacent, such that $ \pi_y E \subset J_1 \cup J_2$.

Now, $$ E \subset \pi_x E \times \pi_y E \subset \left(I_1 \cup I_2\right) \times \left(J_1 \times J_2\right) \, , $$ which is made up of 4 squares. $4 < 3^2$ :))