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What is the boundary of $ \partial ([a,b] \times [a,b])$?

I know that $ \partial ([a,b] \times [a,b])=\partial [a,b]^2=\overline{[a,b]^2}\setminus ([a,b]^2)^\circ=[a,b]^2\setminus (a,b)^2$

Visually the boundary $ \partial [a,b]$ is the edge of a rectangle with side lengths $a$ and $b$, but how can I write this down mathematically rigorously?

Idea: $ \partial ([a,b] \times [a,b])=\partial [a,b]^2=\overline{[a,b]^2}\setminus ([a,b]^2)^\circ=[a,b]^2\setminus (a,b)^2=\{[a,a+x]\cup [a+x,a]\cup [a+x,b]\cup [b,a+x]\, |\, x\in [0, b-a]\}$

Asaf Karagila
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CherryBlossom1878
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2 Answers2

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For nice enough spaces, there is actually a product rule for boundaries. We can write $\partial (X \times Y) = \partial(X) \times Y \cup X \times \partial (Y)$. This happens for example with your box, with $X = Y = [a,b]$. The boundary of a single interval is the two endpoints, so both products are of the form 'endpoints times perpendicular edge,' which gives the two edge segments in that direction. The union says to do this both ways, and you see the boundary box.

This is actually one way of thinking of the original product rule from calculus. When Leibniz figured that out, he was actually thinking of the product geometrically, as the area of a box. When you increment the functions, the box is perturbed in each direction, and he discarded the corner as it was a product of infinitesimals and then regarded as $0$. Then the change in the area of the box corresponds to the boundary, and in that case we have the familiar $d(fg) = (df)g + f(dg)$.

For the relationship between these two, I'll link this great thread, which uses some measure theory. Here you go.

A. Thomas Yerger
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In general, $\partial(A \times B) = \partial A \times B \cup A \times \partial B$. Of course this is assuming that you consider the boundary inside let‘s say $\mathbb{R}^2$, since it is a notion relative to an ambient space.

In this situation, we get $\partial([a, b]^2) = \{a, b\} \times [a, b] \cup [a, b] \times \{a, b\}$.

HDB
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