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Let $R \in \mathbb{R}^2 $ be a closed rectangle (i.e, a set of the form $[a, b]\times [c, d]$). Its area is defined by $S(R)=(b-a)(d-c)$.

Now, suppose $R=\bigcup_{i=1}^{\infty}R_i$ when $R_i$ are also rectangles. I need to prove that $S(R)\leq \sum_{i=1}^{\infty}S(R_i)$. Note that no measure theory is allowed, only "basic" tools.

It's quite harder than I expected. Any suggestions?

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    Very nice question! – 2017-01-13
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    It's easy to solve using compacts, or measure theory , but it's actually more pretty to solve using only basic tools! – 2017-01-13
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    What level of math is this? Do you know anything about compact sets? Anything about measure theory? – 2017-01-13
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    I know a lot about compact sets and measure theory. Using compactness is allowed, measure theory isn't. Edit: just noticed Heine-Borel is a direction. I'll think about it – 2017-01-13
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    First of all. Let's prove this for $[0,1] \times [0,1]$. Because of I could get bijection between this rectangular and others. Actually Heine-Borell lemme give us very important thing, because of $R$ compact there is finite sequence of rectangular from **any set of rectangulars which cover R**. So let's assume that $ 1 = S(R) > \bigcup_{i=1}^{n} S(R_{i})$, this should be true always. So we should find appropriate sequence of rectangulars which give us contradiction. – 2017-01-13
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    But using norms this is obvious to show using triangle-inequality – 2017-01-13

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