A Geometric Problem I Need Help With

[PatrickPowers]

Start with an n dimensional unit cube and an n-ball with unit volume. Center them both at the origin. What proportion of the cube is outside the sphere? I'm too lazy to try for an exact answer : all I want to know is whether the limit is 0, 1, or something in between.

I can't get started until I can compute the area of the base of a right n-pyramid given the height. The area is n-1 dimensional and I can't figure out how to do it. Here's my wrong attempt:

I believe my error is in this line, that something is naive here. If the height is one then the edges are of length sqrt(2). Then the sides of the base are of length two. The area of the base is then 2^(n-1).

After that I know that the volume of the pyramid is hA/n with h the height, A the area of the base, and n the number of dimensions. So the volume would be 2^(n-1)/n. But this is too large. I'm pretty sure it is the area of the base that is too large. So what's the correct way to calculate the area of the base?

[Hugh]

I asked ChatGPT4 and here’s what it said:

Your approach to finding the area of the base of a right n-pyramid given its height is indeed leading you astray, but it's great that you've identified a potential error in your reasoning. The mistake comes from assuming that if the height is one, then the edges of the base are of length \(\sqrt{2}\), and thus the sides of the base would be of length two, leading to an area of \(2^{n-1}\) for the base. This assumption oversimplifies the geometry involved, especially as the number of dimensions increases.

For an n-dimensional pyramid (or, more generally, for any n-dimensional figure), calculating the volume (or the area of its base, which is an \((n-1)\)-dimensional figure) directly from the height using Euclidean distances in this way doesn't scale straightforwardly as it does in 2 or 3 dimensions. The relationship between the height of the pyramid and the side lengths of the base is more complex, particularly because the concept of "side length" becomes less intuitive in higher dimensions.

Let's correct the approach to find the area of the base:

1. **Understanding the Base of the Pyramid**: The base of an n-dimensional pyramid is an \((n-1)\)-dimensional hyperplane figure. For example, in 3 dimensions, the base of a pyramid is a 2-dimensional triangle or square, depending on the pyramid.

2. **Calculating the Area (or Hyper-area) of the Base**: The area of the base in \((n-1)\) dimensions cannot be directly computed by squaring the edge lengths, as we would in 3 dimensions or fewer. Instead, the area (or, more accurately, the "volume" in \((n-1)\)-dimensional space) would depend on the specific shape of the base and its orientation in the n-dimensional space.

3. **Correct Approach to Volume of the Pyramid**: The volume of an n-dimensional pyramid can indeed be expressed as \(\frac{1}{n}\) times the height times the base area (in \((n-1)\)-dimensions). However, calculating the base area requires knowing the specific geometry of the base, which isn't as straightforward as using \(2^{n-1}\).

To address your initial problem of the proportion of an n-dimensional cube that is outside an n-dimensional sphere (n-ball) centered at the same point with both having a unit measure (the cube having side lengths of 2 to make its volume 1 in any number of dimensions, and the sphere having a volume adjusted to also be 1), we approach this with established formulas for volumes of n-balls and relate them to the volumes of n-cubes.

For n-dimensions, the volume of an n-ball of radius \(r\) is given by:
\[ V_{\text{n-ball}} = \frac{\pi^{n/2}}{\Gamma(\frac{n}{2} + 1)} r^n \]

For a unit n-ball (volume = 1), you'd adjust \(r\) accordingly, but since we're comparing proportions, we're more interested in how the volume of the sphere compares to that of the cube as dimensions increase.

As the number of dimensions increases, the volume of the n-ball (sphere) concentrates near its surface, and the corners of the n-cube extend further away from the sphere's surface. This means the proportion of the cube's volume outside the sphere increases with dimension.

To specifically address whether the limit of the proportion of the cube outside the sphere as \(n \rightarrow \infty\) is 0, 1, or something in between, it's known from mathematical studies that this proportion approaches 1. That is, almost all of the volume of the n-dimensional cube lies outside the n-dimensional sphere as the number of dimensions increases. This conclusion is a fascinating aspect of high-dimensional geometry, illustrating how intuitive notions from three-dimensional space do not always extend to higher dimensions.

Just in case the formulas didn’t come out right, here’s a copy of the text:

[PatrickPowers]

Deleted.

[PatrickPowers]

Let's try the original problem again. There is a really easy solution. We show that for large n, more than half of the volume of the unit n-cube lies outside the n-ball with unit volume.

Start with a unit n-cube and an n-ball with unit volume. Center them both at the origin. What proportion of the cube is outside the sphere? First let's find a crude lower bound.

Get a vector from the origin to a corner of the n-cube. Its length is sqrt(n)/2. Find the midpoint of this vector, which is sqrt(n)/4 away from both the origin and the corner. Consider the n-cube with centered at this point that shares a common vertex with the original unit n-cube. Each 1D edge of this new n-cube is of length 1/2. The volume of this new n-cube is (1/2)^n. We further know from a previous result that, for n large, the half of this n-cube that contains the corner of the unit cube is entirely outside the n-ball.

v = (1/2)^n/2

A lower bound on the area outside the ball is this volume times the number of corners.

V = 2^n*v = 2^n*(1/2)^n/2 = 1/2.

More than half of the volume of the n-cube lies outside the n-ball.

[PatrickPowers]

Can I get a crude upper bound next? Consider an n-cube with diagonal d = sqrt(n)/2 and centered at the origin. We know from a previous result that this n-cube is entirely inside the n-ball. We know that the volume of this n-cube is the length of a 1D edge e to the n'th power.

d = sqrt(n*e^2) = sqrt(n)*e

e = d/sqrt(n)

v = e^n

v = (d/sqrt(n))^n

With d = sqrt(n)/2 we get v =(1/2)^n. This gives us an upper bound of 1-(1/2)^n proportion of the unit n-cube that is not inside the sphere. The limit as n goes to infinity is one. Hmmm, we already had that as an upper bound. This is too crude.

[PatrickPowers]

I wasn't able to do it formally but am convinced that the volume of the sphere that is inside the cube decreases with 1/sqrt(n). So the volume of the intersection of the unit volume sphere and cube goes to zero.

The sphere and cube both grow without bound but in different directions so the proportion in common areas decreases.

[quickfur]

One of the strange things that happen as the number of dimensions increase, is that the vast majority of an n-cube's volume lies close to its vertices rather than its center. Furthermore, the distance of its vertices from its center increases without bound. So as the number of dimensions increase without bound, almost none of the cube's unit volume lies in its center, and almost all of it lies near its corners, which means the intersection between the unit sphere and the unit cube will decrease towards 0.

In the same vein, the volume of an n-sphere inscribed inside a unit n-cube will approach 0 as the number of dimensions increase without bound. This is also an extremely weird phenomenon, because for the first few dimensions past 4 the volume actually increases. However, around 6 or 7 dimensions (I forget exactly which) it starts decreasing, and eventually becomes ridiculously small in spite of its radius being constantly 1. In the limit, it becomes zero, whereas the volume of the n-cube remains constant at 1. Equivalently, if we fix the volume of the n-sphere as n increases, then its radius will increase without bound, and the n-cube that inscribes it will become arbitrarily large, and at the limit becomes infinitely more voluminuous than the n-sphere.

[Keiji]

...in spite of its radius being constantly 1.

You mean diameter, right? Since it's inside a unit n-cube.

I do often get confused between "radius 1 inside a 2x2x... cube" and "diameter 1 inside a 1x1x... (unit) cube"...

Either way, you're bang on otherwise - this is a geometry question that's fascinated me ever since I first found out about it too.

[PatrickPowers]

One of the strange things that happen as the number of dimensions increase, is that the vast majority of an n-cube's volume lies close to its vertices rather than its center.

For any n-cube the bulk of the volume lies about the points that are equidistant between the center and the closest vertex. I think you can start with the n=2 case then show this via induction.

Hmmm, how about this. Partition the unit n-cube into 2^n n-cubes, each with one vertex at the center and sharing the opposite vertex with a vertex of the unit n-cube. For each of these subcubes find the points that are equidistant between the center and the closest vertex. These form an (n-1)-plane. Half of the volume of the subcube will be on one side of this plane and half on the other. Each point a distance from this plane is balanced by an equal but opposite point.

[quickfur]

Keep in mind that as the number of dimensions increase, the distance of a unit n-cube's vertices from its center increases without bound. Eventually this shifts the majority of its bulk away from its center and towards its vertices.

[PatrickPowers]

Keep in mind that as the number of dimensions increase, the distance of a unit n-cube's vertices from its center increases without bound. Eventually this shifts the majority of its bulk away from its center and towards its vertices.

The bulk of the matter is equidistant between the center and the nearest vertex. The average distance to the center is the same as the average distance to the nearest vertex. These two distances increase equally as N grows.

It is true that most of the material is closer to the surface as N increases. But the volume near a vertex is so skinny that even though numerous they in total don't hold much. The volume near the center is the same as the total volume that is equally near to the closest vertex. This is evident by symmetry if you partition the n-cube into 2^n cubes, each of which shares a vertex with the original cube and has the opposite vertex at the center. A cross section through the center of one of these subcubes that is perpendicular to the diagonal gives you two halves with identical shape.

An n-cube can be a very spikey shape, which is the source of all these phenomena comparing them with spheres and so forth. Nevertheless n-cubes fit together nicely which makes them the most convenient as a unit of volume.

[quickfur]

[...]The bulk of the matter is equidistant between the center and the nearest vertex. The average distance to the center is the same as the average distance to the nearest vertex. These two distances increase equally as N grows.

It is true that most of the material is closer to the surface as N increases. But the volume near a vertex is so skinny that even though numerous they in total don't hold much. [...]

Yes, the volume around an individual vertex is very skinny and doesn't hold much. But there are so many vertices -- the number of vertices grow exponentially with increasing dimension -- that the total volume close to the vertices overwhelms the volume around the center.

This also leads to another weirdness of higher dimensions: at the limit as n→∞, the number of vertices in an n-cube become uncountable whereas the number of facets remains countable. This fundamental divergence in cardinality leads to difficulties in formulating a consistent ∞-dimensional geometry without further restrictions (such as excluding vectors of non-finite norm in Hilbert space). But it shows you how the volume of the n-cube around the center practically vanishes compared to the volume in the direction of the unbounded number of vertices.

[steelpillow]

the number of vertices grow exponentially with increasing dimension -- that the total volume close to the vertices overwhelms the volume around the center.
...
This also leads to another weirdness of higher dimensions: at the limit as n→∞, the number of vertices in an n-cube become uncountable whereas the number of facets remains countable.

For a measure polytope (n-cube), the number of vertices doubles with each dimension. This is a geometric series, not exponential. The vertices would appear to be countable in an obvious way.

[quickfur]

the number of vertices grow exponentially with increasing dimension -- that the total volume close to the vertices overwhelms the volume around the center.
...
This also leads to another weirdness of higher dimensions: at the limit as n→∞, the number of vertices in an n-cube become uncountable whereas the number of facets remains countable.

For a measure polytope (n-cube), the number of vertices doubles with each dimension. This is a geometric series, not exponential. The vertices would appear to be countable in an obvious way.

The number of vertices doubles with each dimension, meaning that in n dimensions there are 2^n vertices. When n diverges to infinity, you can put each vertex in 1-to-1 correspondence with the binary expansion of a real number between 0 and 1. I.e., the vertex coordinates are <±1, ±1, ...>; map 1 to itself and -1 to 0, and you have an infinite sequence of binary digits, unique per vertex. Since all permutations are taken, that means the binary expansion of every real number between 0 (inclusive) and 1 (exclusive) maps to exactly one vertex of the infinite dimensional cube. Since there are uncountably many real numbers between 0 and 1, there must also be an uncountable number of vertices.

[steelpillow]

The number of vertices doubles with each dimension, meaning that in n dimensions there are 2^n vertices. When n diverges to infinity, you can put each vertex in 1-to-1 correspondence with the binary expansion of a real number between 0 and 1. I.e., the vertex coordinates are <±1, ±1, ...>; map 1 to itself and -1 to 0, and you have an infinite sequence of binary digits, unique per vertex. Since all permutations are taken, that means the binary expansion of every real number between 0 (inclusive) and 1 (exclusive) maps to exactly one vertex of the infinite dimensional cube. Since there are uncountably many real numbers between 0 and 1, there must also be an uncountable number of vertices.

Yes, in a sense your binary mapping maps to real numbers, but it maps to a particular subset of the rational numbers, itself a subset of the reals. Specifically, to multiples of something like (2^-n). So it is more accurate to say that you are mapping to a subset of the rationals. The rationals are countable, and your binary mapping very obviously so. For example we may count them from the lower dimensions up as, say (leaving out the trailing zeroes here), 0,1,00,01,10,11,000,001... and so on. Countability does not mean we can count to the end of an infinite series, it means that we have a counting algorithm which, if we go on long enough, includes every member of the series. There is no such algorithm for the reals, but there is for the rationals. Your statement that "Since all permutations are taken, that means the binary expansion of every real number between 0 (inclusive) and 1 (exclusive) maps to exactly one vertex of the infinite dimensional cube. " is false because you are mapping a countable set onto an uncountable set, and then assuming that you can map the uncountable set back again. You cannot. Consequently, your conclusion also fails.

It is perhaps also worth adding that the way such counting works is to "approach" or "tend to" infinity but never get there: infinity is not a member of the series as such. We then perform a somewhat debatable sleight of hand to generalise to the endpoint of the series. In my day this was referred to "the limit as n tends to". Although the sleight of hand has been somewhat refined since then, its outcome remains the same.

[quickfur]

No you misunderstood the mapping I described. You cannot map them from lower dimensions and up because all lower dimensional coordinates do not span all coordinate axes. Every vertex has a non-terminating list of coordinates, therefore each vertex maps to a non-terminating binary digit sequence. Since the rationals only cover a subset of this (the sequence eventually has a repeating segment), they only cover a subset of the vertices. It is the rationals that are a subset of the vertices, not the other way round.

[steelpillow]

Nobody is mapping them "from lower dimensions and up". I have no idea what you are reading into my remarks.

[quickfur]

You said:

For example we may count them from the lower dimensions up as, say (leaving out the trailing zeroes here), 0,1,00,01,10,11,000,001... and so on.

This will give you finite subsets of the full coordinates of the vertices (the mappings that end in trailing zeros, aka the coordinates that end with minus signs that we map to 0), but not the vertices with non-terminating sequences of +'s and -'s. Any countable list of such will only give you a strict subset of the full set of vertices in the ∞-cube.

What I'm saying is that the vertices of the infinite-dimensional cube have coordinates <±1, ±1, ±1, ...>, with all changes of sign taken, as is the case with all finite-dimensional cubes. Each combination of signs corresponds with a single vertex. So we can map + to 1 and - to 0, and thus each vertex corresponds with a unique infinite binary digit string. As such, there is a 1-to-1 mapping between each vertex and a unique real number 0≤r<1. From which we conclude that there are an uncountable number of vertices thus represented, since the real numbers between 0 and 1 are uncountable.

You can't span the full set of vertices only by taking the union of all finite initial segments of coordinates like you seem to propose above, because any finite initial segment must end in implied trailing zeros, so the union of all such finite initial segments only give you the vertices whose coordinates eventually end with trailing -1's. Clearly, there are many other vertices that do not end this way (the vertex <1, -1, 1, -1, 1, -1, ...> is one example, among many), so the list thus constructed can only be a subset of the full set of vertices. It can only be a countable subset of the full set of vertices, which by the argument in the previous paragraph must be uncountable.

[PatrickPowers]

Aha now I understand it. That's a clever construction.

[steelpillow]

Any countable list of such will only give you a strict subset

You appear not to have grasped what a countable infinity is yet. By your arguments, the rationals are not countable (or, worse, binary numbers are not rational). Those are very elementary mistakes.
There is little point in repeating Cantor's proof here. May I suggest you read up on it.

[mr_e_man]

Quickfur knows what he's talking about.

I will repeat Cantor's proof here.

Suppose we have a countably infinite list of vertices. Denote the n'th vertex as v_n, with coordinates v_n,k, for natural numbers n and k.
Since these are vertices of the cube, their coordinates are v_n,k ∈ {+1, -1}.
Consider the point x with coordinates x_k = - v_k,k.
Since -(+1) = -1 and -(-1) = +1, we have x_k ∈ {+1, -1} for all natural numbers k, so this point is also a vertex of the cube.
But since -(+1) ≠ +1 and -(-1) ≠ -1, we have x_k ≠ v_k,k. Thus, for any natural number n, the vertex v_n is not equal to x, since they differ in the n'th coordinate.
So there is a vertex x not in the list. This proves that the cube's vertices are uncountably infinite.

[steelpillow]

Quickfur knows what he's talking about.

I will repeat Cantor's proof here.

Suppose we have a countably infinite list of vertices. Denote the n'th vertex as v_n, with coordinates v_n,k, for natural numbers n and k.
Since these are vertices of the cube, their coordinates are v_n,k ∈ {+1, -1}.
Consider the point x with coordinates x_k = - v_k,k.
Since -(+1) = -1 and -(-1) = +1, we have x_k ∈ {+1, -1} for all natural numbers k, so this point is also a vertex of the cube.
But since -(+1) ≠ +1 and -(-1) ≠ -1, we have x_k ≠ v_k,k. Thus, for any natural number n, the vertex v_n is not equal to x, since they differ in the n'th coordinate.
So there is a vertex x not in the list. This proves that the cube's vertices are uncountably infinite.

In "Denote the n'th vertex as v_n, with coordinates v_n,k, for natural numbers n and k," what does k signify? Is it the number of dimensions we have got up to so far, or something? Without proper definitions, you are just producing impenetrable nonsense.

[Klitzing]

"... with coordinates v_n,k", i.e. k is obviously just the coordinate index.
E.g. let v_1 be the vertex with coordinates 1 throughout, then you'd have v_1,k = 1 for all k.

--- rk

[quickfur]

We're sort of veering off topic here, but I'd like to approach this from a different angle.

First, the set of rationals.

We know that this is the set of fractions of integers. For simplicity let's consider the positive rationals, which have the form a/b for some natural numbers a, b≠0. Since the natural numbers can be factorized into powers of primes, we can write this as (2^a₁·3^a₂·5^a₃...) / (2^b₁·3^b₂·5^b₃...) where 2, 3, 5, ... are the prime numbers, and a₁, a₂, ... and b₁, b₂, ... are the corresponding (non-negative) exponents in the prime factorization of a and b.

Dividing out the common prime factors, we arrive at this representation of the rationals: a/b = 2^c₁·3^c₂·5^c₃... where c₁, c₂, ... are integer exponents which may be negative (i.e., if they appear in the denominator in the previous representation).

Now, since a and b are finite (by definition), the sequence of exponents a₁, a₂, ... and b₁, b₂, ... must eventually terminate. Therefore, the sequence of integer exponents c₁, c₂, ... must also eventually terminate, or equivalently, there exists some natural number k for which c_i = 0 for all i > k.

Thus, we can uniquely represent a positive rational number as a finite sequence (c₁, c₂, ... c_n) for some finite n. Equivalently, if we allow c_i to continue indefinitely for all i, then there exists some index k for which c_i = 0 for all i > k. IOW, the sequence of exponents eventually ends in trailing zeroes.

Now consider the set of all positive rational numbers. Obviously, this is the same set as the one we obtain by collecting all of the (c₁, c₂, ...) sequences we constructed above for each rational number. I.e., the set Q of rational numbers has a 1-to-1 correspondence with the set of finite sequences of integers. Now here's the important point: no matter how large a rational number we might pick, it has a representation as a finite sequence of integer exponents. Or, equivalently, it is represented by an infinite sequence of integer exponents that eventually terminate in trailing zeroes.

A real number, however, cannot in general be represented by any finite sequence of numbers; most of the real numbers require an infinite sequence of numbers to be represented (e.g., as an infinite non-repeating digit sequence).

//

Coming back to my ∞-dimensional cube's vertices: this is the set of all changes of sign of <1, 1, 1, ...>. Suppose we try to enumerate all of them by collecting the vertices of all finite-dimensional cubes, which obviously should be a subset of the ∞-dimensional cube's vertices. So we can start off like this:

V₁ = { <-1>, <1> } // the 1-dimensional cube's (i.e., the line segment's) vertices.
V₂ = { <-1, -1>, <-1, 1>, <1, -1>, <1, 1> } // the 2-dimensional cube (i.e. square)'s vertices
V₃ = { <-1, -1, -1>, <-1, -1, 1>, <-1, 1, -1>, <-1, 1, 1>, <1, -1, -1>, <1, -1, 1>, <1, 1, -1>, <1, 1, 1> } // the 3-cube's vertices
...

At each step, we extend the coordinates with <-1> so that the resulting vertices are part of the ∞-dimensional cube's vertices. I.e., <-1> becomes <-1, -1, -1, ...>, <1> becomes <1, -1, -1, -1, ...>, and <1, 1> becomes <1, 1, -1, -1, -1, ...>, and so on. Basically just tack an endless trail of -1's to the coordinates above.

So let's collect all of the V's together:

V^* = V₁ ∪ V₂ ∪ V₃ ∪ ...

Now the question is, does V^* include all of the ∞-dimensional cube's vertices?

By construction, any vertex you choose from V^* must have come from V_i for some i. And we know that every vertex in V_i consists of i coordinates that could be +1 or -1, but after that all the remaining coordinates will be -1. This is because V_i consists of the vertices of the i-dimensional cube, which obviously has only i coordinates. No matter which element you pick from V^*, it must have come from one of these V_i sets. Therefore, all elements in V^* must eventually terminate in trailling -1's.

How many elements does V^* have? It's essentially a collection of all finite sequences of integers. (Well, a suitable subset of that.) Therefore, its cardinality cannot be greater than the cardinality of the set of rational numbers. I.e., it's countable.

//

Now consider the point P = <1, -1, 1, -1, 1, -1, ...>, i.e,. the point whose coordinates alternate between 1 and -1. Obviously this is a valid combination of signs of the <1, 1, 1, ...>, so it must also be one of the vertices of our ∞-dimensional cube. Unfortunately, P is not a member of V^*, because it does not eventually terminate in trailing -1's; it alternates between 1 and -1 indefinitely. In fact, there are many other points like this. For example, <1, 1, -1, 1, 1, -1, 1, 1, -1, ...> where every third coordinate is -1, and the rest are 1. This also cannot be in V^* because it also doesn't eventually terminate in -1's. There's also <-1, 1, -1, 1, 1, -1, 1, 1, 1, -1, ...> where the gap between -1's gradually increments. Actually, we don't even need to look for these exotic vertices; even plain old <1, 1, 1, 1, ....> where all coordinates are 1, isn't in V^*.

Or, for that matter, take any element E of V^*, then -E isn't in V^*. Because E must eventually terminate in trailing -1's, so -E must eventually terminate in trailing 1's, so -E cannot be in V^*. I.e., the set of vertices in V^* do not contain their antipodal points. Furthermore, there exist points not in V^*, whose antipodes are also not in V^* (so it's not enough to just add all antipodes to V^* to complete it). For example, the alternating point <1, -1, 1, -1, ...> that we gave above: its antipode is simply <-1, 1, -1, 1, ...>, which also doesn't terminate in -1's.

This shows that V^* does not contain all of the vertices of the ∞-dimensional cube. Why? The fundamental reason is that V^* was constructed from finite sequences of coordinates. Yes, we collected all of them across all finite dimensions together into V^*, but all of them are finite sequences. None of them include vertices like <1, -1, 1, -1, ...> that do not have a finite initial segment followed by some constant trailing coordinate.

In order to include all the vertices of the ∞-dimensional cube, we must include infinite sequences of coordinates. You cannot approach these "from below"; collecting all finite initial segments of such sequences is not enough. For example, we may try to complete V^* by adding points of the form <1>, <1, -1>, <1, -1, 1>, <1, -1, 1, -1>, ..., but all of these initial segments actually are already in V^*. The problem is that all of them eventually terminate (we eventually pad them with trailing -1's), whereas the full sequence <1, -1, 1, -1, ...> does not terminate (and does not end with trailing -1's).

The only way to include such points is to admit infinite sequences of coordinates in our construction of V^*. This is equivalent of collecting all infinite digit sequences into a set -- the resulting set is the set of real numbers, of uncountable cardinality. In the same way, the set of vertices of the ∞-dimensional cube is also uncountable. (In my previous post I've already shown a 1-to-1 correspondence with the real numbers between 0 and 1.)

//

The crux of the issue here is that collecting all finite initial segments of sequences is fundamentally weaker than collecting the infinite sequences themselves. The former can only approximate the full set, but will always fall short. It can only have a countable cardinality, whereas the latter can have uncountable cardinality. Just like the rationals can only approximate the reals, but contain infinite "gaps" (irrationals like √2, √3, pi, etc.). The only way you can get the full set of real numbers is to start with sequences that are already infinite in length (e.g., Dedekind cuts, etc.). The set of reals cannot be reached by building up from finite initial segments.

Likewise, the set of vertices of the ∞-dimensional cube cannot be reached by collecting all the vertices of finite-dimensional cubes; you must start with vectors of infinite length to begin with, otherwise there will always be missing vectors. The union of vertices of all finite-dimensional cubes is fundamentally incomplete; they cannot span all of the vertices of the ∞-dimensional cube, but are missing many elements -- their own antipodes in ∞-dimensional space, for example, as well as many other points where both the point and its antipode aren't included. In fact, such a collection of points from finite-dimensional cubes can only be a meager subset of the full set of vertices of the ∞-dimensional cube.

[steelpillow]

"... with coordinates v_n,k", i.e. k is obviously just the coordinate index.
E.g. let v_1 be the vertex with coordinates 1 throughout, then you'd have v_1,k = 1 for all k.

--- rk

That's what I thought. This is just the diagonalisation argument for real numbers.

OK, I think I see what is happening. The problem with using it here is that Cantor set out to represent real numbers as infinite binary strings, and to prove something about real numbers. However, here we are taking countable vertices in k dimensions, as k tends to infinity. It is not safe to assume that the limit of a countable series is an uncountable series, that was not Cantor's starting point. When he came up with a number not in the set, he had defined it as countable reals. But in our case we cannot do more than count to some arbitrary vertex and discover that we have not counted them all yet. Same diagonal construct, but a different understanding of exactly what it is modelling. Stay grounded!

[quickfur]

Now it's my turn to veer off-topic, but there's an interesting borderline approach that perhaps may appeal more to the likes of steelpillow, who IIRC does not buy Cantor's diagonalization argument.

The thing about uncountable sets, if one were to accept Cantor's ideas, is that there are so many members in the set that you cannot possibly name them all. In fact, you can never name more than a countable subset of them (i.e., precisely those members which admit a finite description), which has Lebesgue measure zero. The vast majority of these members are anonymous, amorphous entitites that forever lie beyond any finite description. Thus, the overwhelming majority of points that constitute the continuum of the real numbers, for example, are unnameable, anonymous numbers that make up the bulk of the continuum; those scant few numbers that we can actually name and identify are so few that they are mere specks dotted across the continuum. In fact, if we only regard the numbers that are nameable (definable / describable / etc.), then we do not have a continuum at all; only a disconnected point cloud. This point cloud includes more than the rationals, of course; they include numbers like √2, e and pi and various concrete irrational and transcendental numbers that we can name and define. However, all of them still only form a disconnected point cloud. The elusive numbers that fill up the gaps in between are actually unnameable and indescribable, because they are numbers that require an infinite amount of space to define.

This means that, for all intents and purposes, we can collapse these anonymous indescribable numbers into a single exceptional entity, thereby creating a countable model of mathematics that nevertheless admits (and can prove the existence of) uncountable sets. (The catch is that, within the model you cannot tell that the countable set that stands in for the real numbers is actually countable; the model will insist that it satisfies the proposition P(x) : x is uncountable. But externally, it has a countable representation -- because almost all of the uncountably many members, being indescribably and unnameable, are thereby indistinguishible, and can thereby be replaced by a countable number of stand-ins that pretend that they represent an uncountable number of set members.) (P.S.: it's a proven theorem that axiomatic set theory admits a countable, albeit non-standard, model.)

What I'm trying to get at -- it's not my intention to turn this into a discussion of model theory -- is that even though an infinite-dimensional cube theoretically has an uncountable number of vertices, we can, for the purposes of actual, concrete calculations, actually get away with a countable model of them. Because any specific vertex V that you may try to describe can only have a finite description (otherwise you couldn't finish describing it!), and so belongs to the countable subset of vertices that admit a finite description. Any vertices that you may obtain from V via whatever geometric operations you please will also be among the members of this countable subset (they are the vertices which can be described as V plus some series of geometric operations, which must also be finitely describable -- e.g., the antipode of V, V rotated by some angle, the product of V with some matrix, any combination of the foregoing, etc.). We can collect all of these describable vertices into a set of describable vertices of the ∞-dimensional cube, and obtain a countable set on which we may perform whatever geometric operations we please, and the result will also be contained in this set. The rest of the ∞-dimensional cube's vertices have no finite description, thus even though theoretically they must exist somewhere out there among the infinitude of vertices, we can never actually name them or describe them. Thus, they are of no use to us as far as geometric speculations are concerned, except for bulk arguments that involve treating some subset of them as a collective whole. In which case we can simply treat them as an opaque, black box unit, which ostensibly contain "uncountably many" vertices (but who's to say the black box isn't empty? we just conveniently labelled it to have whatever bulk properties we need as appropriate to the situation).

So we could actually still work with the ∞-dimensional cube's vertices using only countable models, and draw conclusions by manipulating said countable models. (We just have to take care that when we're working "within the model", we pretend that certain countable model members are labelled "uncountable" and treat them as if they were uncountable. But they are opaque and you could never actually name any specific element inside them.) Thus, whether or not Cantor's diagonalization argument "works" or not, doesn't really matter as far as drawing geometric conclusions are concerned. We can only name countably many explicit vertices anyway, so we only ever need to work with countably many distinct vertices. The rest of the anonymous, amorphous "uncountable" vertices out there that we can never name may as well be collapsed into just a single black box that pretends to be uncountably many, and we wouldn't be any wiser.

(And BTW, the above arguments can be applied to any axiomatic theory that contains uncountable elements. Within the model, certain sets are labelled "uncountable", but externally, the model itself needs only represent a countable subset of them -- those sets that can be named or defined within the theory. This subset is necessarily countable, since a theory that admits descriptions of infinite length is not useful -- you can never complete a proof that involves an element with an infinitely-long description. The uncountably many remaining elements do not admit a finite description and therefore can be collapsed into at most countably many stand-ins (mostly just one stand-in, an empty black box that pretends that it contains uncountably many members, that the model treats specially). The model does not allow you to inspect the contents of the black box, so you can never tell, within the model, that your theory is actually countable from the outside. So in this sense uncountable sets are something of a nebulous entity that, in theory, exist, but for all practical purposes might as well not exist and everything would still be consistent as long as you tweak your model to label certain elements (among the countably many elements) as "uncountable" and treat them accordingly.)

[quickfur]

OK, that came out all abstract and imprecise, I don't know if people are totally confused by what I'm trying to say. :-D

So here's a concrete example: take the real numbers between 0 and 1. We know that, in theory, there are an uncountable number of them. However, what are these uncountably many numbers? We can't name them all -- because they're uncountably many! We can name a few of them -- like 0 and 1 themselves, 1/2, 1/3, 1/4, etc., and we may also name some irrationals between them like 1/√2, 1/pi, 1/e, or e/(2*pi), etc.. Or if you get creative, you may name some Liouville trancendentals that fall between 0 and 1.

But -- and here's the point -- any such number that you can name must necessarily be describably in a finite amount of space, i.e., a finite number of symbols. Most numbers in this uncountable set have no finite description: they cannot be described in a nice, compact way like √2 or √3 or e or pi. You have to describe them as non-terminating decimal digits: 0.583978510238194392.... The problem is with the "..." here. What digits does the "..." actually represent? Well, you can keep going: ...39804112835703291... but it never ends. You can never finish describing this arbitrary real number that you picked, because if you stop somewhere, then it's a mere rational.

Now you may get smart and say, well I can't name all of the digits in a finite amount of space, but here's a function (or computer algorithm) that will print out an arbitrary number of digits of this number, you can ask it for as many digits as you want. That's a step forward; armed with such functions, we can name a lot more numbers that are impractical to name with decimal digits alone. Included among them, actually, are numbers like e or pi, whose digits have no obvious pattern. But nevertheless, the description of the algorithm must necessarily be finite. An infinitely-long algorithm is useless for naming your number, because you can never finish running it to get any answer! But if the algorithm is of a finite length, that means it consists only of a finite sequence of symbols. Which means that the set of all such algorithms is countable: you just enumerate every possible combination of symbols of increasing length, and collect them all into a set. This set is necessarily countable (we just enumerated its members). And therefore it only contains a countable subset of the uncountably many real numbers between 0 and 1.

That is, the set of numbers corresponding to these algorithms is only a countable subset of the real numbers between 0 and 1, and therefore it has measure zero: it has hardly scratched the surface of most of the numbers that ostensibly exist between 0 and 1!

We can go further. There are numbers that can be defined but for which there are no algorithms that can compute them. Chaitin's constant, for example. Or monsters like the smallest integer not definable in Peano arithmetic (it's an extremely huge number: take its reciprocal to get a real number between 0 and 1). Would such numbers bridge the gap to include the uncountably many numbers that remain between 0 and 1, that we haven't described yet? Nope -- any valid definition of a number in any theory of arithmetic must, again, admit a finite description -- otherwise it's not a valid definition because we can never finish defining the target number! So again, we have a collection of definitions of finite length. And again, if we collect all of these definitions into one big set, we find that, alas, the set is still only countable. It still leaves uncountably many real numbers left that it is unable to describe.

And we can go on and on and invent more and more clever schemes of naming individual numbers, but unless we admit descriptions of infinite length, we can never span all the uncountably many real numbers between 0 and 1!

Now, we can of course be cunning and include them by defining ranges of numbers. Like we just did with the real numbers between 0 and 1: notice how we said "between 0 and 1": we didn't name all of them, and in fact, the above arguments show that we can never name all of them -- we just collected them all by fiat into a set that we call "the real numbers between 0 and 1". But have we really collected uncountably many numbers into our set? We can't name all of them; so can we really say that we've collected all of them? What if we haven't actually collected all of them at all, but only a countable subset of them (those among them that can be defined with a finite number of symbols), plus a special "cheat card" that isn't actually a real number, but a stand-in that claims that it represents uncountably many real numbers, but actually it's empty, because none of these purported uncountably many numbers can actually be named. IOW, our set S of real numbers between 0 and 1 actually looks like this:

S = { n | n can be defined in a finite number of symbols } ∪ { BlackBox }

where BlackBox is a fake number that pretends that it's uncountably many, but when you ask it for an actual number among its contents is unable to provide you with a concrete description -- because its members are, by definition, those numbers that require an infinite amount of symbols to describe. S is obviously countable, even though our model will insist that it's uncountable because it contains the fake number BlackBox, that the theory treats as a marker of an "uncountable" set.

This BlackBox element is a cunning, slippery creature; we may, for example, try to expose its trick by, say, dividing S into two sets: the set T of numbers from 0 to 1/2, and the set U of numbers from 1/2 to 1. But what our model gives us is:

S = T ∪ U
T = { n | n can be defined in a finite number of symbols and n ≤ 1/2 } ∪ { BlackBox }
U = { n | n can be defined in a finite number of symbols and n > 1/2 } ∪ { BlackBox }

Again, T and U are countable. Only, our model, seeing that the trickster BlackBox is among its elements, insists that they are uncountable sets. But again, if we ask the model for an explicit example of a number in T, it will only ever give us a number from the left-hand side of the union, never any element from the right-hand side that it claims contains an uncountable number of elements. (Of course, it never gives us a direct handle on BlackBox; that's its ace up the sleeve that it's not going to reveal to us.) Again, we have a countable model that ostensibly represents an uncountable set, but we can never actually see a concrete instance of a number in the uncountable portion of the set!

We can try various ways of filtering out elements from the left-hand side to try to coax our model to reveal just what it's hiding behind BlackBox, but we can never succeed, because any concrete number that our model gives us must always be finitely-describable. And therefore it always belongs to the countable subset of the supposedly-uncountable set that the model claims to be representing. We can try to exclude the countable left-hand side altogether, e.g., take the set difference of S with { n | n can be defined in a finite number of symbols }. The result R, however, would look like this:

R = { BlackBox }

Supposedly it contains uncountably many numbers, but the model can never name any one of them -- because none of them, by definition, admits a finite description! Thus when we ask our model for an explicit member of R, it just says "such an element exists, but sorry I can't tell you what it is, because it cannot fit in a finite number of symbols". For all we know, there's only one element in R, and we wouldn't be any wiser! The model claims that there are uncountably many elements, but since we cannot obtain a description of any of them, the only thing we know is that R contains such an element. Does it contain more than one? We can't tell -- we can't obtain any explicit description so we can't compare descriptions to discover that our model is lying to us -- there's only one element that's standing in for uncountably many! Because of this, the model only needs a single element to stand in for uncountably many. As long as we stay within the model, we cannot tell the difference.

The same arguments can be applied to the vertices of the ∞-dimensional cube. There's a countable subset that we can explicitly name and define, but there's a nebulous cloud of undefinable vertices (they require an infinitely-long description) that, for all intents and purposes, can be collapsed into a single BlackBox "fake" element that pretends that it's uncountably many, but actually it's only one. And as far as geometry is concerned, we can't tell the difference -- the difference is immaterial! So we could very well do geometry with a countable subset of the ∞-dimensional cube's vertices and the results we obtain would be indistinguishible from if we had used an actual, uncountable model.

[steelpillow]

Now it's my turn to veer off-topic, but there's an interesting borderline approach that perhaps may appeal more to the likes of steelpillow, who IIRC does not buy Cantor's diagonalization argument.

The more I revisit it now, the more I do not buy it even for the reals. It seems that I am in good company - see for example Wikipedia's article on Controversy over Cantor's theory. No wonder we have seen differences of opinion here!

Suffice to say I broadly support what you follow that up with.

[quickfur]

Personally, I don't really care for arguments for or against Cantor's proof. At the end of the day, it's just a matter of building an axiomatic system from which we can derive useful theorems with which we can manipulate our mathematical objects (or models of mathematical objects). You can build an axiomatic system with the axiom of infinity, or a different system without, and it wouldn't really make that much of a difference as far as practical results are concerned. (It's a different story when you're talking about abstract objects like uncountable sets, of course. But those kinds of differences are immaterial when applied to real-world practical problems.)

Now it may make a difference in terms of ease of proof of certain things -- assuming the existence of infinite sets with certain properties allow us to derive certain results in less complicated ways than if we had to restrict ourselves to finite sets or only countable sets, for instance. You'll still get the same results when it comes to practical problems; the only differences exist in the abstract realm where you have weird entities like uncountable sets, fractals, and other such things. Just like how doing non-standard calculus with infinitesimals lets you prove certain things much more easily than using the mind-numbingly tedious standard epsilon-delta proofs -- but you don't actually obtain any results that are materially different.

Basically, the more you assume, the easier proofs become, but as long as your system is consistent, you won't run into contradictions. Not pertinent ones, anyway. You may have an apparent contradiction in terms of the divergent properties of an ∞-dimensional cube vs. a (∞^∞)-dimensional cube (deliberately abusing notation here), but the discrepancies aren't going to be anything you can practically compute. So as far as practical results are concerned, there isn't any material difference. We regularly use calculus for computing real-world physical problems, for example, the equations of which assume that matter is infinitely divisible and appropriately described by the real numbers (or structures involving the real numbers). But the physical world ain't made of infinitely-divisible matter; you get to the atomic level and things become "grainy". Or if you're modern and go subatomic, eventually you hit the Planck length and you can't divide anything anymore. So the assumptions of calculus breaks down. But for the most part, the difference doesn't really matter; a few tens of atoms off in the results does not negate the usefulness of the results you derived using calculus that assumes infinite divisibility. OTOH if you want to be purist and forego the ostensibly illegitimate assumption of infinite divisibility, then your calculations are going to be bogged down with all kinds of irrelevant details involving Planck lengths and atomic radii where they really aren't relevant. You'll still be able to obtain results -- the same results if the calculations on both sides have physical validity -- but with much more effort.

So AFAIC, I couldn't care less whether Cantor was "right" or "wrong" or whatever. If the axiomatic system built from Cantor's assumptions let me derive my results with much more ease and convenience, or "beauty of proof" (whatever that means), then by all means, I'll take it. If it starts getting in the way, e.g. now I have to start accounting for Cantorian "monsters" like Cantor sets and immeasurable point clouds, then I'll throw it out the window. Or introduce a simplifying axiom that lets me sweep the monsters under the rug and ignore their existence (e.g., "if A, B, C are measurable sets, then ..." -- thereby excluding the troublesome immeasurable objects from my having to deal with them), so that I can go ahead and derive the answers I want for my set of objects (which don't include these pathological entities).

//

A lot of the trouble involving uncountable sets come from the assumption of completed infinity (a "general" real number involves an infinitely-long digit string in its complete realization, as something distinct from its finite initial segments which approximate it), a concept some struggle with. It really depends on how you view it. If you're approaching it from the physical, materialistic POV, then yeah, it doesn't make any sense because nothing in the material world is truly infinite. Everything is finite and limited, and consequently everything that we observe in the material world can be adequately described by a finitistic (or at most, countable) model. The world of completed infinities is something that's fundamentally different from the material world, a fascinating world to explore in a sense (what some would call "Cantor's paradise"), but if one is going to set foot in that realm of the abstract, then one has to be prepared to face things that fly in the face of what we encounter and know from the physical world. And one has to be prepared to face the fact that our finite minds can never fully grasp what purportedly exist in the world of completed infinities -- and that's why we'll never be able to name an uncountable number of objects. And will never be able to tell the difference between a model that has a fake stand-in that pretends it contains an uncountable number of members, vs. an actual uncountable model that contain an actual, uncountable number of things.

Objects like ∞-dimensional cubes are an example of this: the very existence of such a thing implies the accepting of the axiom of infinity, because otherwise, the ∞-dimensional cube is ill-defined; you can never complete its construction! So right from the get go we're already making the leap out of the finite into the infinite; we're making the huge assumption that we have actually completed the construction of a cube with an infinite number of dimensions, and all the consequences that entails. Among the consequences of which are the existence of unnameable vertices -- because it takes an infinitely long description to fully define such a vertex. Its coordinates consist of a string of 1's and -1's that follows no pattern of finite description, and therefore forever eludes our finite grasp. The only way you can manipulate such things is by fiat: you assume that P is such a vertex, in the abstract (because you can't be specific here), then use its generic properties (its coordinates consist of 1's and -1's, etc.) to draw conclusions. But actually accessing specific coordinates is not possible in general, because by definition they cannot be finitely described. And ultimately, from our finite POV, we will never be able to tell whether we're actually manipulating some platonic ∞-dimensional vertex that exists somewhere out there, or we're just deceiving ourselves and merely manipulating labels on empty black boxes that claim to be infinite. Does the ∞-dimensional cube actually exist somewhere out there in the universe of ideas?

Or is it just a clever ruse, a sleight of hand, a sock puppet that promulgates statements of delusional grandeur but is ultimately empty inside? Just like the rest of the paradise of completed infinities? Of which we can only ever obtain finite glimpses, never to behold in their full glory, if such even exists to begin with. The kicker is, we can't tell the difference. The difference is unknowable by definition.

[steelpillow]

To be honest, I think that set theory is a bad way to understand infinite-dimensional geometry. As you say, it is axiomatic and there is no provable reason to choose any one set of axioms over any other. (I also think that converging series, whether a Taylor series or a binary string or whatever, are a bad way to understand the behaviour of real numbers, but enough already! (For more, see my refutation of the diagonal proof in the thread on Living in infinite-dimensional space)
The old "in the limit as n → ∞" is a much better way. No pesky axioms or set-theoretic constructs to deal with.

So we just develop a formula expressing the ratio we are interested in, from one level to the next, along the lines of f(n+1) = g.f(n), hope that it leads to a converging series, and take the limit as n goes to infinity (whatever that is).

But first things first, back to the original problem. No solution, but a couple of thoughts:

Not sure if it has been stated as such, but the unit n-cube and the n-sphere of unit volume have equal volume.
The volume of the cube outside the sphere therefore equals the volume of the sphere outside the cube.

as the number of dimensions increase, the distance of a unit n-cube's vertices from its center increases without bound. Eventually this shifts the majority of its bulk away from its center and towards its vertices.

Is this not also true of the n-sphere?
I would suggest that as n → ∞ we find that:
* both volumes approach the same distribution, i.e. wholly at the hypersurface.
* both external volumes approach the same limit.

Because each hypersurface is in part inside the other one, the external part is some fraction of the total hypersurface. Since the n-volume remains 1 by definition, the external volume must be some fraction of 1. Definitely not infinite!

Question: so if the volume all lies against the surface and the external volumes are equal, is it valid to deduce that the external surfaces of the ∞-cube and ∞-sphere are equal too, or is infinity tripping us up again? Probably.

There are 2n hyperfaces, with total hyper-area 2n. The volume close to each hyperface is equal to 1/2n. In the limit the surface area goes to infinity and the associated volume for each hyperface becomes infinitesimal.
Something similar must happen with the sphere. My analytical skills begin to fail me here - my calculus especially is long gone - but I would hope that developing some f(n+1) = g.f(n) from all this might offer a productive way forward.