Calculating the area of a set in $\mathbb{R}^2$.

Question

Suppose $E$ is a bounded subset of $\mathbb{R}^2$ and $\{P_i\}_{i=1}^{n}$ is a finite family of rectangles covering $E$, i.e. $E \subset \cup_{i=1}^{n} P_i$. Define the area $S(E)$ of $E$ to be the infimum $$ S(E) = \inf \sum_{i=1}^{n} |P_i|, $$ where infimum is taken over all possible finite coverings and $|P_i| = (b_i-a_i)(B_i-A_i)$, $P_i = [a_i,b_i]\times[A_i,B_i]$. How do we calculate the $S(E)$ if, say $E = [0,1]\times \{\frac1n \mid n \in \mathbb{N}\}$ or $E = ([0,1]\cap A)\times ([0,1]\cap A)$ where $A=\{\frac{n}{2^k} \mid n,k \in \mathbb{Z}, k \geq 0, n > 0\}$?

Answer 1

For the first problem let $\epsilon>0$ and let $P_1=[0,1]\times[0,\epsilon].$ Then there are only finitely many uncovered line segments, say for $n=1,...,k$. Then let $P_i=[0,1]\times[1/(i-1)-\epsilon/2^{i},1/(i-1)+\epsilon/2^{i}]$ for $2\leq i \leq k+1$. This covers and $$ \sum_{i=1}^{k+1}|P_i|<2\epsilon. $$ Since this is true for all $\epsilon>0$, we have that the area is zero.

For the second one, I believe if you require finitely many rectangles in your cover, the area is 1. Consider any finite cover $\{P_i\}$. Then $\bigcup P_i$ is closed since each $P_i$ is closed. Thus $\mathbb{R}^2\setminus \bigcup P_i$ is open. Assume there is a point $a\in [0,1]\times[0,1]$ not covered by $P_i$. Then let $\epsilon$ be such that $B(a,\epsilon)\subset \mathbb{R}^2\setminus \bigcup P_i$. It isn't hard to find a point in $E$ that is in this ball, giving us a contradiction.(Indeed $E$ is dense in $[0,1]\times[0,1]$!) Thus it must be that any finite cover of $E$ covers all of $[0,1]\times[0,1]$. This gives us $S(E)\ge 1$ and as you observed, $[0,1]\times[0,1]$ covers, so $S(E)=1$.