# Why does $S = ([0,1] \times [0,1]) \cap \mathbb{Q}^2$ have no area?

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.

• Perhaps it could be considered an "informalism"? Mar 29, 2021 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. Mar 30, 2021 at 0:00
• 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. ptvtp.files.wordpress.com/2011/10/…
– Anon
Mar 30, 2021 at 0:09
• 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, 2021 at 0:14
• This seems like Jordan measure (the name despite it's not really a measure). Mar 30, 2021 at 0:41

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 $$S_n := [\sqrt{2} + \frac{1}{n}, 1] \times [0,1]$$.