# Would a 4D Hypercube have Infinite 3D “Volume”?

The typical representation of a 4D hypercube is an object bounded by 3D cubes on each of its boundaries. The analogous to a 3D cube with 2D faces at each boundary.

The area of a face of a 3D object is xy. When we integrate this to get volume, this is the integral of (xy)dz. That is, a cube contains an infinite amount of areas multiplied by an infinitessimal length in the z direction.

Extending to the 4D hypercube, would the 4D "Volume" be the integral of (xyz)dw? That is, an infinite amount of xyz volumes multiplied by an infinitessimal length in the w direction.

The 4D hypercube is typically drawn showing the boundary 3D cubes without much discussion of the "3D" volume contained within its bounds.

• Does an ordinary cube have infinite 2-D area? I don't think it makes much sense to talk about the 3-D volume of a 4-cube. – saulspatz Mar 31 '18 at 23:27
• No, just if you think about cutting an infinitesimal slice in 4D space and getting a 3-dimensional shape. – Michael James Apr 1 '18 at 15:51

The short answer is yes, although to be rigorous we should use the word "measure" instead of volume. For our purposes, measure is the $n$-dimensional analogue of volume; in one dimension, it is length, in two dimensions it is area, in three dimensions it is volume, and in general $n$-dimensions it is the "generalization" of these concepts.
Measure has the intuitive property of countable additivity: if we take an at most countably infinite collection of pairwise disjoint sets $\{A_i\}$, then the measure of $\bigcup_{i = 1}^\infty A_i$ should be the sum of the measure of each $A_i$ separately. Basically, the volume of two non-overlapping cubes is just the sum of the volume of each cube separately. Then allow for "two" to be replaced with "a countably infinite number of" cubes, and that is countable additivity. It also has the intuitive property that if $A \subset B$, $m(A) \subset m(B)$, where $m(A)$ is the measure of $A$.
The 4D hypercube has the property that at each slice perpendicular to the $w$-axis, the result is a 3D cube. This is the higher-dimensional analog to the fact that if we slice a 3D cube perpendicular to the $z$-axis, we get a 2D square.
Now, let $C$ denote the 4D hypercube. We write this as $$C = [0, 1] \times [0, 1] \times [0, 1] \times [0, 1].$$ Letting the first coordinate be $w$, note that $C$ is the disjoint union $$\bigcup_{w \in [0, 1]} (\{w\} \times [0, 1] \times [0, 1] \times [0, 1]).$$ This is, in fact, an uncountable union of a disjoint collection of sets, each of which is a 3D cube with 3D measure = 1. Let $A = \{a_1, a_2, \ldots\}$ be a countable subset of $[0, 1]$. Then by countable additivity, and the fact that the measure of a subset is smaller than the measure of the set containing it, we have the following: \begin{align*} m(C) &= m\left(\bigcup_{w \in [0, 1]} (\text{3D-cube})\right) \\ &\geq m\left(\bigcup_{w \in A} (\text{3D-cube})\right) \\ &= \sum_{w \in A} m((\text{3D-cube})) \\ &= \sum_{w \in A} 1 = \infty. \end{align*} Thus, the 3D measure $m$ of the 4D cube is infinite.