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I am currently in an Advanced Calculus 2 class and am using the C. H. Edwards "Advanced Calculus of Several Variables" text. In chapter 4 when discussing area in $\mathbb{R}^2$, the text says that the set $S = ([0,1] \times [0,1]) \cap \mathbb{Q}^2$ has no area. The reasoning the book gives is that given any set of non-overlapping rectangles $\{R_i'\}$ for $i=1,\dots,n$ where $R_i' \subseteq S$ for all $i$, $\sum_{i=1}^na(R_i') = 0$, and for any set of rectangles $\{R_i''\}$ for $i = 1,\dots, m$ where $S \subseteq \bigcup_{i = 1}^nR_i''$, $\sum_{i = 1}^ma(R_i'') \geq 1$. I understand if these statements are true, there can be no area for the set $S$, but I am having trouble understanding why for any $i$, $a(R_i') = 0$, and I also don't understand how $\sum_{i = 1}^ma(R_i'') \geq 1$. Any help would be appreciated!

Clarification: This is the definition of the area of a bounded set $S \subseteq \mathbb{R}^2$ given in our book:

Given a bounded set $S \subseteq \mathbb{R}^2$, we say that its area is $\alpha$ if and only if given $\epsilon > 0$, there exists both

  1. A finite collection $R_1',\dots,R_k'$ of nonoverlapping rectangles, each contained in $S$, with $\sum_{i=1}^ka(R_i') > \alpha - \epsilon$
  2. A finite collection $R_1'',\dots,R_l''$ of rectangles whose union contain $S$, with $\sum_{i=1}^la(R_i'') < \alpha + \epsilon$

If there exists no such number $\alpha$, we say that the set $S$ does not have area, or that its area is not defined.

realribs
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  • Perhaps it could be considered an "informalism"? – morrowmh Mar 29 '21 at 23:57
  • The way we well-define area (aside from boxes) in analysis is through measure. A set is a zero set if you can cover it with a collection of boxes of total volume at most $\epsilon$, for any $\epsilon > 0$, and it can be shown that a countable set has measure zero. $S$ is countable, so it is measure zero. – While I Am Mar 30 '21 at 00:00
  • Could you please post your book's definition of area (if it is defined)? I looked up you book online, and it makes no mention of the text you've said in Chapter 4. https://ptvtp.files.wordpress.com/2011/10/c-h-edwards-auth-advanced-calculus-of-several-variables-elsevier-inc-academic-press-inc-1973.pdf – Anon Mar 30 '21 at 00:09
  • Basically, you are asked to show that $a(R'i) = 0$ and $[0,1] \times [0,1] \subseteq \cup{i = 1}^n R''_i $. – Anon Mar 30 '21 at 00:14
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    This seems like Jordan measure (the name despite it's not really a measure). – Divide1918 Mar 30 '21 at 00:41
  • @peter.petrov You can't knock off arbitrary points of a rectangle and it still remain a rectangle. – Anon Mar 30 '21 at 01:22
  • Countable sets have measure zero. – stoic-santiago Mar 30 '21 at 03:03

1 Answers1

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Claim $1$: $R_i \subseteq S \Rightarrow ar(R_i) = 0 $

Let $(x,y), (x',y') \in R_i \Rightarrow (x', y) \in R_i$ as $R_i$ is a rectangle. Suppose $x \neq x'$. WLOG $x < x'$. As irrationals are dense in $\mathbb{R}$, $\exists x_0 \in \mathbb{R} \setminus \mathbb{Q}$ s.t. $x < x_0 < x'$. Thus $(x_0, y) \in R_i$ which is a contradiction. Thus $x = x'$. Similarly, we can show $y = y'$. Hence $(x,y) = (x',y') \Rightarrow |R_i| = 1 \Rightarrow a(R_i) = 0$.

Claim $2$: $S \subseteq \bigcup_{i = 1}^n R_i \Rightarrow [0,1] \times [0,1] \subseteq \bigcup_{i = 1}^n R_i$.

As $R_i$ is closed for each $i$, and finite union of closed sets is closed, $\bigcup_{i = 1}^n R_i$ is closed. Then, $ \begin{align*} S \subseteq \bigcup_{i = 1}^n R_i &\Rightarrow \bar{S} \subseteq \overline{\bigcup_{i = 1}^n R_i} = \bigcup_{i = 1}^n R_i \Rightarrow [0,1]^2 \subseteq \bigcup_{i = 1}^n R_i \end{align*} $

Note: $[0,1]^2$ need not necessarily be a subset of an infinite collection of rectangles which contain $S$. An example would be the sequence of rectangles given by $(R_n) \cup (S_n)$ where $R_n := [0, \sqrt{2} - \frac{1}{n}] \times [0,1]$ and $S_n := [\sqrt{2} + \frac{1}{n}, 1] \times [0,1]$.

Anon
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