A possible paradox about topology and the relation between zero and infinity?

Question

We know that the volume of higher dimensional sphere is inversely proportional to the number of dimensions. Hence, as the we increase the number of dimensions keeping the radius fixed, the volume of the sphere tends to zero.

On the other hand, if we increase the number of dimension in case of cubes or polyhedra (to be specific), the volume increases. Hence, the volume of an infinite dimensional hypercube is infinite.

Now, in the context of topology, a sphere is homeomorphic to a cube or parallelepiped, in the case of infinite dimensions... where the volume of the sphere is infinite and the volume of the hypercube is infinite... Does that imply that an object of volume of the magnitude zero is homeomorphic (that it can be continuously transformed) to something which has the volume of the magnitude Infinity?

Perhaps this is not a new problem to the topologists, or maybe it is not a problem at all to them... But it is confusing me.

Answer 1

There are a couple of different issues going on here:

if we increase the number of dimension ... the volume increases

Here's the first problem: the volume of an $n$-dimensional cube whose side lengths are all $1$ is $1$ - for every $n$. These $1$s are not all the same thing, though, because they're measured in different units: e.g. if the side length is $1\mathrm{cm}$, then the volume is $1\mathrm{cm}^3$, or $1 \mathrm{cm}^4$, or $1 \mathrm{cm}^5$, etc. depending on the dimension.

Hence, the volume of an infinite dimensional hypercube

Here's the second problem: you've jumped from a bunch of finite objects to an infinite object. There's no guarantee that you should be able to do this, or even that "volume" makes sense here any more.

Those two aside, here's the real point:

Does that imply that an object of volume of the magnitude Zero is homeomorphic (that it can be continuously transformed ) to something which has the volume of the magnitude Infinity?

Homeomorphism doesn't preserve volume, or area, or length, or anything like that. It preserves notions like connectedness, openness, closed loops, and so on, but it's completely indifferent to most 'geometric' properties.

Answer 2

There are some problems with your exposition:

1) "hence infinite dimensional bla has volume bla": there are two things that are wrong with this; the first is assuming some sort of continuity between the finite dimensional and the infinite dimensional: a priori you cannot just "take the limit" when going from finite dimensional to infinite dimensional. The second (which is related) is that there's no well-defined/established notion of volume in infinite dimensions.

Indeed, there is no analogue of the Lebesgue measure for infinite dimensional Banach spaces, so what would volume even mean in that context ?

2) "Does that imply that an object of volume of the magnitude Zero is homeomorphic (that it can be continuously transformed ) to something which has the volume of the magnitude Infinity? "

As explained in 1), this is not what we get here. But it sems important to me to mention that in general, volume and topology have very little to do with one another.

Volume is related to the notion of measure, whereas homeomorphism is related to the notion of continuity, and shape of a space. The two are somewhat related in that we often try to come up with measures that in a sense respect the topology: we can ask that they be defined on open sets, closed sets (more generally, Borel sets); we can ask that they be finite on compact sets, etc. These are compatibility requirements, that hold in the most common measures (e.g. the Lebesgue measure on $\mathbb{R}^n$).

But the requirement you seem to ask for, or at least have an intuition for, is that the measure should be preserved (perhaps in some wide sense, e.g. allowing for inequalities, that is a controlled difference) by continuous operations. But this is far from true; and you don't even have to consider infinite dimensional spheres to see thatwe can't ask for this.

Indeed any two open intervals are homeomorphic (by a simple affine change when they're finite, using for instance the logarithm or the arctangent when they're infinite, and plenty of other tricks): it wouldn't be sensible to ask that they have the same volume (or in this case, length, which is just $1$-dimensional volume)

So yes, there may very well be a shape with infinite volume (for instance $\mathbb{R}^3$) homeomorphic to a shape with finite volume (for instance $(0,1)^3$); and you don't even need to go to infinite dimensions to see that happening.

This is where geometry and topology break down: in topology you only look at the "overall shape" in a sense, whereas geometry asks for more rigid transformations (for instance, in $\mathbb{R}^n$, a linear transformation $A$ always changes the volume by the same amount : $|\det(A)|$)

Answer 3

There is a lot to unpack here, so let me try to go line by line:

We know that the volume of higher dimensional sphere is inversely proportional to the number of dimensions.

This isn't quite right. First off, I think that you mean the unit ball (not the unit sphere). The sphere is the set $$ S^n(0,r) := \{ x \in \mathbb{R}^n : \|x\| = r \}, $$ whereas the ball is the set $$ B^n(0,r) := \{ x \in \mathbb{R}^n : \|x\| < r \}. $$ Essentially, the ball is solid, while the sphere encloses the solid ball. Typically, when we use the phrase "volume", we are talking about balls, not spheres. I am going to assume that by "sphere," you actually mean the ball.

Note, also, that the volume of the unit ball $B^n(0,1)$ tends to zero as $n$ goes to infinity, but it is not quite correct to say that the volume is inversely proportional to the dimension. This implies that there is some constant $k$ such that $$ \operatorname{vol}(B^n(0,1)) = \frac{k}{n}. $$ No such constant exists. It is actually worse than that.

The volume of the ball $B^n(0,r)$ is given by $$ \operatorname{vol}(B^n(0,r)) = \operatorname{vol}(B^n(0,1)) r^n = \frac{\pi^{\frac{n}{2}}}{\Gamma(\frac{n}{2} + 1)}r^n, $$ which (clearly?) tends to zero with $n$ if $r \le 1$. It is less clear (perhaps) that this tends to zero if $r > 1$, but $\Gamma(\frac{n}{2}+1)$ is roughly $\lceil\frac{n}{2}\rceil!$ for large $n$, which grows faster than the numerator. In any event, I think that we might want to restrict our attention to the unit ball.

Hence, as the we increase the number of dimension keeping the radius fixed, the volume of the sphere tends to Zero.

This is almost right. If we rewrite this sentence to read "As we increase the dimension, the volume of the unit ball tends to zero," it will be correct.

On the other hand, if we increase the number of dimension in case of Cubes or polyhedra (to be specific), the volume increases. Hence, the volume of an infinite dimensional hypercube is infinite.

The volume of a cube, which I take to be the set $$ C^n(0,r) = [-r,r]^n, $$ is given by $$ \operatorname{vol}(C^n(0,r)) = (2r)^n. $$ This tends to infinity with $n$ if and only if $r > \frac{1}{2}$, and tends to zero if and only if $r < \frac{1}{2}$. This is basically the same behaviour as the ball, with one little caveat: $$ \lim_{n\to\infty} \operatorname{vol}(C^n(0,1/2)) = 1 \ne 0 = \lim_{n\to\infty}\operatorname{vol}(B^n(0,1)). $$ Hence we are, perhaps, most interested in the unit cube, as well.

Now, in the context of topology, a sphere is homeomorphic to a cube or parallelopiped,

This is definitely true in finite dimensional spaces. It is not obvious that it is true in infinite dimensional spaces. Basically, there are many norms that we can define on $\mathbb{R}^n$, two of which are coming into play here. The first, called the Euclidean norm, is given by $$ \|(x_1,\dotsc,x_n)\|_2 := \sqrt{x_1^2 + \dotsb + x_n^2}. $$ The second, called the $\ell^\infty$ or maximum norm is given by $$ \|(x_1,\dotsc,x_n)\|_\infty := \max\{|x_1|,\dotsb,|x_n|\}. $$ Both of these norms induce topologies on $\mathbb{R}^n$, and it turns out that if $n$ is finite then these topologies are equivalent (they have the same open sets). The open balls for the Euclidean norm are the usual open balls, and the open balls for the max-norm are the open cubes, which (after some abstract nonsense) implies that cubes and balls are homeomorphic in finite dimensional spaces.

However, the Euclidean norm and the max-norm do not induce the same topology on $\mathbb{R}^\infty$. Much more delicate arguments need to be made, and it is not at all clear to me that a cube and ball in $\mathbb{R}^\infty$ are homeomorphic (or even what the right topology is).

in the case of infinite dimensions... where the volume of the sphere is infinite and the volume of the hypercube is infinite...

I think that I have already addressed this point.

Does that imply that an object of volume of the magnitude Zero is homeomorphic (that it can be continuously transformed) to something which has the volume of the magnitude Infinity?

Basically, I think that you are asking if an object with infinite volume can be homeomorphic to an object with finite volume, or even zero volume. This is not a surprising result. For example, the unit disk in $\mathbb{R}^2$, i.e. the set $B^2(0,1)$, has "volume" (area, measure, whatever you want to call it) $\pi$. However, the unit disk is homeomorphic to all of $\mathbb{R}^2$, which has infinite volume (area, measure, whatevs).

The issue here is that topologies are very weak structures. The tell us how points relate to each other. Topologies give a very weak notion of "nearness", allow us to define limits, give us a way of talking about connectedness (several ways, in fact), and other such structures. However, there is no intrinsic way to measure volume, and there is no reason to believe that volume should be a topological invariant (i.e. two homeomorphic objects should have the same volume).

If you want to talk about volumes, you need more structure than just a topology. Typically, we talk about volumes in a part of mathematics called measure theory. In this theory, the measure of a set can potentially communicate with the topology (this is, essentially, the idea of a Borel measure on $\mathbb{R}^n$), but the measure gives us more information than the topology alone. Measure spaces have "stronger" structures than topological spaces.

Answer 4

As Billy says, you are not comparing the same things: length, area, volume, hyper-volume, etc and the units would matter.

I see a line magically appear on the ground and I measure it to be $0.5 m$.

It magically extends into a square and I measure it to have an area of $0.25 m^2$.

Next it extends into a cube and I measure it to have a volume of $0.125 m^3$.

I don't notice but my $4d$ alien friend tells me that it has become a hyper-cube and has hyper-volume $0.0625 m^4$.

I think: "that's interesting, as the dimension increases, the volume decreases".

However, at the same time, an American is watching. He measures the line to be $1.64 ft$. He measures the square as having an area of $2.69 ft^2$. The cube with a volume of $4.41 ft^3$. He can't find a $4d$ being who understands the customary system but nonetheless he says: "that's interesting, as the dimension increases, the volume increases".

We are looking at the same objects and yet have reached different conclusions. We could do the same trick with the sphere by choosing our units suitably. Also, if we just talking about the surface of the objects, the same applies.