How many faces does the $n$-dimensional cube $I^n$ have?

Question

I am taking an online Coursera Calculus course, and this question popped up as one of the challenge problems, reproduced below. I had a difficult time understanding the answer, which was $I^n$ has $2n$ faces.

This problem concerns the boundary operator $\partial$ from the bonus material. Denote by I the closed unit interval $[0,1]$. Then, as observed, $\partial I=\{0\}\cup\{1\}$ is the union of two points. Let's get a little creative. Denote by $I^n$ the "$n$-cube", that is, the Cartesian product of $n$ intervals:

$$I^n = I \times I \times \cdots \times I$$

This is a well-defined and perfectly reasonable $n$-dimensional cube. (Just because you can't visualize doesn't mean it can't exist!) Note that $I^1=I$ and $I^0$ is a single point (a zero-dimensional cube!). As a step towards building a "calculus of spaces", let us write $∂I^1=2I^0=2$ as a way of saying that the boundary of an interval consists of two points and that $I^0=1$.

The boundary of an n-dimensional cube consists of a certain number of $(n−1)$-dimensional cubes (called "faces"). For example, a square ($I^2$) has four faces. Using what you know about derivatives, answer this: how many faces does $I^n$ have?

Answer 1

Hint: As you build the $n$ dimensional cube from the $n-1$ dimensional one, you extend it in a new direction. As such, all the $n-2$ dimensional boundaries of the $n-1$ dimensional cube become $n-1$ dimensional boundaries of the $n$ dimensional cube. You should be able to imagine this for $n=1,2,3$. Use this to set up a recurrence relation that you can solve.

Answer 2

There are lots of ways to solve this problem. The best is to find one for yourself. If you don't have a clue, start with the cases which are easy to understand: a line segment has two vertices, a square has four sides, and a cube has six faces. Now try to understand the connection between the dimension, $n$, and the number of faces, $2n$.

But here is one solution: a face is determined by choosing one coordinate $i$, with $1 \leq i \leq n$ and taking either all elements with $i$th coordinate $0$ or all elements with $i$th coordinate $1$. (Thus for instance the top face of the unit cube $[0,1]^3$ is $\{ (x,y,1) \ | \ 0 \leq x,y \leq 1 \}$.) So there are $2n$ faces altogether.

Answer 3

all comes from $\partial (A\times B) = \partial A B \cup A \partial B$

$\partial I^1 = 2$

To calculate the faces for $I^2$:

$\partial (I^2)=\partial (I^1 \times I^1) = \partial(I^1)\times I^1 \cup I^1\times \partial(I^1) = 2I^1 \cup 2 I^1$

therefore $2+2=4$ faces for $I^2$ To calculate the faces for $I^3$:

$\partial(I^1 \times I^2) = \partial I^1 \times I^2 \cup I^1\partial(I^2) = 2 \times I^2 \cup I^1 \times (2I^1 \cup 2 I^1) = 2I^2 \cup 2I^2 \cup 2I^2$

Therefore $I^3$ has $2+2+2$ faces.

following this reasoning, you can derive $I^n$