Living in infinite-dimensional space

Ideas about how a world with more than three spatial dimensions would work - what laws of physics would be needed, how things would be built, how people would do things and so on.

Re: Living in infinite-dimensional space

quickfur wrote:W

Now consider a house in the shape of a dome, i.e., half of an N-sphere. Assuming the diameter is a unit long, it will also take just one unit of time to get from one end to the other. There are no corners, of course, but assuming you start at the point of the N-sphere in the same direction as the corner of the N-cube, and try to walk to the opposite side diagonally. In the case of the dome, it will always only take 1 unit of time, no matter how large N is. In the limiting case where N diverges to infinity, this remains invariant. However, something strange happens: the point at the N-sphere's diagonal, i.e., the intersection of the N-sphere with <1,1,1,1...>, is <1/N, 1/N, 1/N, ...>. The corresponding opposite side is <-1/N, -1/N, -1/N, ..>. In the limiting case, both points converge to <0,0,0,...>, and you get the paradoxical situation where walking from the origin to the origin takes a non-zero amount of time, because the distance between the origin and the origin is 1 rather than 0.

This apparent contradiction goes away if you admit infinitesimals into your number system, in which case the two opposite corners are <ε, ε, ε, ...> and <-ε, -ε, -ε, ...> where ε is an infinitesimal.

Let's say that the unit sphere has radius one. Then that intersection is actually [1/sqrt(N), 1/sqrt(N), 1/sqrt(N)...]. The distance between the positive and negative intersections is always 2 regardless of N, so that's the limit as well.
PatrickPowers
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Re: Living in infinite-dimensional space

PatrickPowers wrote:
quickfur wrote:W

Now consider a house in the shape of a dome, i.e., half of an N-sphere. Assuming the diameter is a unit long, it will also take just one unit of time to get from one end to the other. There are no corners, of course, but assuming you start at the point of the N-sphere in the same direction as the corner of the N-cube, and try to walk to the opposite side diagonally. In the case of the dome, it will always only take 1 unit of time, no matter how large N is. In the limiting case where N diverges to infinity, this remains invariant. However, something strange happens: the point at the N-sphere's diagonal, i.e., the intersection of the N-sphere with <1,1,1,1...>, is <1/N, 1/N, 1/N, ...>. The corresponding opposite side is <-1/N, -1/N, -1/N, ..>. In the limiting case, both points converge to <0,0,0,...>, and you get the paradoxical situation where walking from the origin to the origin takes a non-zero amount of time, because the distance between the origin and the origin is 1 rather than 0.

This apparent contradiction goes away if you admit infinitesimals into your number system, in which case the two opposite corners are <ε, ε, ε, ...> and <-ε, -ε, -ε, ...> where ε is an infinitesimal.

Let's say that the unit sphere has radius one. Then that intersection is actually [1/sqrt(N), 1/sqrt(N), 1/sqrt(N)...]. The distance between the positive and negative intersections is always 2 regardless of N, so that's the limit as well.

Correct, which means that when N=∞, the coordinates of the intersection must be infinitesimal, because otherwise the intersection would be <0,0,0,0...> in both cases and you get the contradiction that the distance between <0,0,0,...> and <0,0,0,...> is non-zero.
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Re: Living in infinite-dimensional space

More thoughts about houses in N-dimensional space, including the limiting case where N→∞:

For simplicity, we assume that houses are in the shape of an N-cube. You can add a sloping roof with various triangular shapes if you want (e.g., a triangular polyprism, or an (N-1)-simplex prism, or a pyramid, whatever, it's not really that important).

Let's say the vertical dimension is the first coordinate. This is somewhat unusual since people usually assign the Y or Z axes to the vertical, but standardizing on the 1st coordinate makes the discussion simpler for general N. Furthermore, let's say the front of the house is the side of the N-cube facing the <0,1,0,0,..> direction, i.e., the +Y direction, the 2nd coordinate. So the back of the house would face the <0,-1,0,0...> direction. Again, this choice is to simplify the discussion for general N.

Now we'd put the front door somewhere in the front of the house, generally in center of this wall. Well, in actual houses we usually offset it for aesthetic purposes, but I think it's safe to say that it's somewhere around the center of the wall -- it'd be strange for the front door of a house to be located right at the corner of the wall (even though this isn't entirely unheard of). But for the purposes of this discussion let's keep it simple and say it's at the center of the front wall. So the door would be installed on the point <0,1,½,½,½,...> (the first coordinate is 0 because that's where the ground is; we assume the door extends upwards from there).

In 3D, this gives us the coordinates <0,1,½>, i.e., standing on the ground, in the front wall of the house, and midway between either side. A pretty standard location for a front door. The distance from the door to either corner of the house that touches the front wall, is ½ units.

In 4D, the front door would be located at <0,1,½,½>, i.e., standing on the ground, in the front wall, which is a cube, right in the center of this cube. Now note that a 4D house has 2 different kinds of corners: a ridge corner where 2 walls meet, and an apical corner where 3 walls meet (and where a vertex of the cube is). The distance from the front door to a ridge corner on the front wall (there are 4 of these in 4D, compared to only 2 in 3D) is also ½. The distance to an apical corner, however, is √2/2, which is slightly larger than ½.

In 5D, the front door would be located at <0,1,½,½,½>, standing on the ground, in the middle of the front wall which is now a tesseract. In 5D, there are 3 kinds of corners: a ridge corner where 2 walls meet, an apical corner where 3 walls meet, and a verticial corner where 4 walls meet. The distance to the ridge corner is again ½; the distance to the apical corner is √2/2, and the distance to the verticial corner is √3/2.

When we go to 6D, we find that the front wall has 4 kinds of corners. The nearest kind, where 2 walls meet, remains at a distance of ½ units from the door. The farthest kind, where 5 walls meet, is now 1 unit away: twice the distance to the ridge corner.

If we now go to 7D, we find that the distance from the front door to the nearest ridge corner is still ½ units away, but the distance to the verticial corner is now √5/2 units away. When we get to 11D, the distance to the verticial corner is 3 times the distance to the ridge corner. In 18D, the distance is 4 times the distance to the ridge corner. In 27D, the distance is 5 times the distance to the ridge corner: you can now walk around the entire house faster than it takes to walk from the front door to one of the house's far corners.

In general, in N dimensions, the front wall has (N-2) kinds of corners; the nearest kind is where 2 walls meet, and is always ½ units from the front door. The farthest kind, where (N-1) walls meet, is √(N-2)/2 units away. As N grows without bound, this distance grows without bound. When N diverges to ∞, we find ourselves in the strange situation where we can walk around the house in only 4 units of time; however, we can never reach the corner of the house from the front door in finite time.

So in a ∞-dimensional city, if you arrive at a house in the direction of its front wall, you will be able to get into the house no problem. But if you arrive at the house from one of its corners, you'll never reach the front door.
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Re: Living in infinite-dimensional space

Furthermore, if you own a ∞-dimensional house, and your house is wide enough to accomodate 2 rooms along each wall (a pretty reasonable size IMO!), then you have more than enough space to run a Hilbert Hotel. (In fact, much more than that. But we'll get to that.) Here's how you do it: suppose again that your house is in the shape of a cube, between <0,0,0,...> and <1,1,1...>. Divide the interior of the house into rooms of dimensions <1,½,½,½,...> (remember, the first coordinate is the vertical; we want at least adequate head room in each room so that our guests don't have to crawl -- that would not be very nice). How many of these rooms can we fit into our house?

A lot. For example, we can fit a room between <0,0,0,...> and <1,½,½,½...>, and another room between <0,½,0,0,0...> and <1,1,½,½,½,...>. And a room between <0,0,½,0,0,...> and <1,½,1,½,½,½,...>. And another room between <0,0,0,½,0,0...> and <1,½,½,1,½,½,...>. For any natural number n, we can fit a room between <0, ... {n zeros}, ½, 0, 0, ....> and <1, ... {(n-1) ½'s}, 1, ½,½,½,...>. So we have at least 1 room per natural number, and so we rent them out and turn our house into a Hilbert Hotel.

But actually, we can run a hotel much bigger than a Hilbert Hotel, all within the same house! To see this, let's return to a finite N-dimensional space and look at our floor plan. The floor is basically an (N-1)-cube with 2N-1 vertices. Since we can fit 2 rooms per wall, that means we can fit 1 room per vertex of the floor plan. (We're simply subdividing our floor in half along each axis.) Now let N diverge to ∞. Since the floor of our ∞-dimensional house is still an ∞-cube, that means we can fit as many rooms in our house as there are vertices in the ∞-cube, which is an uncountable number. So not only we can run a single Hilbert Hotel; we can even run an infinite number of them all within our humble little house!

And in fact, much more than that. A countably infinite number of Hilbert Hotels still would not fill up all our rooms. We can run an uncountable version of the Hilbert Hotel -- let's call it the Cantor Hotel, where we can accomodate an uncountable number of guests!! By comparison, the set of guests in a Hilbert Hotel has measure zero compared to the number of guests in our Cantor Hotel, meaning that our hotel is basically still empty after receiving all the guests from the nearby Hilbert Hotel when they closed down because of a nasty water leak that made all their rooms unoccupiable. We can take in the guests from all the Hilbert Hotels in our town, and we'd still be able to accomodate all of them -- assuming there are a countable number of Hilbert Hotels in town.

Of course, this same argument can be applied to any ∞-dimensional house that's big enough to be subdivided into smaller rooms, so essentially every house in town can run a Hilbert Hotel.

But by the same argument, if our town consists of a map with blocks of N×N×N×... houses, for some small number N>1, that means there are an uncountable number of houses in town, and therefore an uncountable population of townsfolk. So a Hilbert Hotel wouldn't be able to accomodate the entire population of the town, but we don't really expect the number of visitors to match the number of local townsfolk anyway, so we may reasonably assume that we'd only get a countable number of visitors, and they can all stay at the local Hilbert Hotel. But in the event that a disaster destroys the next town over, there'd be an influx of an uncountable number of fleeing refugees. In which case we just have to designate one Safe House -- which has an uncountable number of rooms -- to accomodate them all, without upsetting the rest of the local townsfolk. No problem.
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Re: Living in infinite-dimensional space

Is this why all infinite-dimensional buildings have curved walls - and floors? Any flat bits can have a cube erected on them and take forever to cross!
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Re: Living in infinite-dimensional space

A surface usually has N-1 dimensions. In infinite dimensions there is no fixed N so a surface might seem not possible. Not!

Let's start by constructing an infinite dimensional rectangeloid with finite positive volume. We know that the series 1/n^2 converges. So all we have to do is have the series e^(1/n^2) for the widths of the rectangleoid. The sum of the logarithms of this series converges. So the product of the series also converges to a positive number. To get any positive volume you like use the series e^(x/n^2).

Now that we have this we know the values of all the vertices. Given that, we can determine whether any point in our space is inside, outside, or on the surface of the rectangloid. The surface is well-defined.

We know all the vertexes of the unit cube so it is also well-defined. The unit sphere, we just calculate the distance from any point to the center, so that's well-defined too. How this is done, not my problem.
Last edited by PatrickPowers on Mon Apr 15, 2024 2:55 pm, edited 2 times in total.
PatrickPowers
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Re: Living in infinite-dimensional space

quickfur wrote:By comparison, the set of guests in a Hilbert Hotel has measure zero compared to the number of guests in our Cantor Hotel, meaning that our hotel is basically still empty after receiving all the guests from the nearby Hilbert Hotel when they closed down because of a nasty water leak that made all their rooms unoccupiable.

I know on good authority that this occurred when infinite dimensional Keith Moon exploded a cherry bomb in his room's toilet.
PatrickPowers
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Re: Living in infinite-dimensional space

PatrickPowers wrote:
quickfur wrote:By comparison, the set of guests in a Hilbert Hotel has measure zero compared to the number of guests in our Cantor Hotel, meaning that our hotel is basically still empty after receiving all the guests from the nearby Hilbert Hotel when they closed down because of a nasty water leak that made all their rooms unoccupiable.

I know on good authority that this occurred when infinite dimensional Keith Moon exploded a cherry bomb in his room's toilet.

Due to the infinite rate of dissipation of an expanding cloud of gas in ∞-dimensional space, an ∞-dimensional bomb requires an infinite amount of explosive power in order to destroy anything outside a radius of 0. Therefore, cherry bombs are inherently safe in ∞-dimensional space.
quickfur
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Re: Living in infinite-dimensional space

quickfur wrote:
PatrickPowers wrote:
quickfur wrote:By comparison, the set of guests in a Hilbert Hotel has measure zero compared to the number of guests in our Cantor Hotel, meaning that our hotel is basically still empty after receiving all the guests from the nearby Hilbert Hotel when they closed down because of a nasty water leak that made all their rooms unoccupiable.

I know on good authority that this occurred when infinite dimensional Keith Moon exploded a cherry bomb in his room's toilet.

Due to the infinite rate of dissipation of an expanding cloud of gas in ∞-dimensional space, an ∞-dimensional bomb requires an infinite amount of explosive power in order to destroy anything outside a radius of 0. Therefore, cherry bombs are inherently safe in ∞-dimensional space.

What makes you think they don't have an infinite amount of explosive power?
PatrickPowers
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Re: Living in infinite-dimensional space

PatrickPowers wrote:
quickfur wrote:[...]
Due to the infinite rate of dissipation of an expanding cloud of gas in ∞-dimensional space, an ∞-dimensional bomb requires an infinite amount of explosive power in order to destroy anything outside a radius of 0. Therefore, cherry bombs are inherently safe in ∞-dimensional space.

What makes you think they don't have an infinite amount of explosive power?

Because I'm assuming they're approximately in the shape of a ∞-sphere of some finite radius, which has zero volume (or at most infinitesimal). Therefore they only contain at most a negligible amount of firepower.
quickfur
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Re: Living in infinite-dimensional space

More strange characteristics of ∞-space:

1) CD diagrams break down. While it is theoretically possible to extend CD diagrams to contain a non-finite number of nodes, it's not clear how to interpret such a diagram, and it runs into the problem that the resulting infinite diagrams cannot distinguish between the ∞-cross and the ∞-simplex. There's also the interesting question of what a mesotruncated ∞-cube might be (there are no obvious answers; there are two distinct limits that one might construct that truncate to each other, but it's not clear if these limits actually exist within ∞-space, and in any case they are unreachable via a countable number of operations starting from either the ∞-cube or the ∞-cross).

2) If we assume that a convex ∞-polytope is the convex hull of a set of coordinates (it's not clear this actually holds, because it's not clear how one would go about defining a convex hull in ∞ dimensions, but let's assume it works), then there are (at least) 3 distinct classes of uniform ∞-polytopes with ∞-cube symmetry:

(a) The "small" uniform polytopes: those whose outradius is finite. This includes the ∞-cross and some of its truncates. These have the property that there exists a ∞-sphere of finite radius that contains all of the vertices of the uniform polytope.

(b) The "large" uniform polytopes: those whose outradius is non-finite, but whose vertex coordinates are bounded. This includes the ∞-cube, some of its truncates, and some of the truncates of the ∞-cross. These have a finite bounding box: you can scale the ∞-cube by a finite factor such that it contains all the vertices of the uniform polytope. However, they do not have an inscribing sphere, because no finite scaling of the ∞-sphere can contain all of their vertices.

(c) the "very large" uniform polytopes: those whose vertex coordinates are unbounded, including some of the truncates of the ∞-cube, for example, the postulated omnitruncated ∞-cube whose vertex coordinates are all permutations of coordinate and changes of sign of <1, 1+√2, 1+2√2, ... 1+n√2, ...>. These also have infinite outradius, but since they have unbounded coordinates, they do not have a finite bounding box (no finite scaling of the ∞-cube can contain all the vertices of the uniform polytope). Obviously, no finite scaling of the ∞-sphere is able to bound them either.

The polytopes in category (a) exist in Hilbert space, and are relatively well-behaved. The polytopes in category (b) are outside Hilbert space, and have some pathological properties such as a non-finite outradius and the consequences of that (such as a house whose front door is unreachably far from its corners).

It's not clear whether the postulated polytopes in category (c) are even polytopes in any meaningful sense, because it's not clear how you'd define their bounding hyperplanes. Their bounding boxes are so large they span the entire ∞-dimensional space(!). Their dual polytopes either do not exist (if we restrict ourselves to ∞-space with finite coordinates), or they must be "extremely large" (the duals have vertices with infinite coordinate values). In the former case, it's impossible to define bounding hyperplanes for the polytope so it can't be a proper polytope in the ordinary sense; in the latter case the bounding hyperplanes lie outside of ∞-space (with finite coordinates), so it's not clear whether this is even well-defined.

This conundrum can only be resolved if the underlying field of coordinate values is extended from the real numbers to something like the hyperreals or the surreal numbers. But if we were to go that route, then that would imply also the existence of infinitesimal polytopes: those who coordinates have infinitesimal, but non-zero, values. Some of them may have non-infinitesimal volume and bounding boxes in ∞-space. And this immediately leads to a lot of very strange, even pathological consequences (and it's not even clear whether the resulting geometry can even be reasonably called a geometry anymore, or whether the whole thing will even be consistent!).
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Re: Living in infinite-dimensional space

Even more weirdness in ∞-space:

In finite-dimensional space, rotations and projections are distinct. In ∞-space, however, this is not necessary true. Consider, for example, a rotation R₁ that rotates the X axis 90° into the Y axis. Let's say we have a point P = <1, 2, 3, ...>. After applying this rotation, we have R₁·P = <-2, 1, 3, 4, ...>. Consider another rotation R₂ that rotates the Y axis 90° into the Z axis. If we apply this to the image of P under R₁, we get: R₂·R₂·P = <-2, -3, 1, 4, 5, ...>. We can invent a series of rotations R_n that rotates the n'th axis 90° into the (n+1)'th axis. If we chain all of these rotations together, we get a transformation that maps <1, 2, 3, ...> to <-2, -3, -4, ..>. What happened to the 1? It's gone! We've effectively dropped it from P. I.e., the limit of an infinite sequence of rotations is equal to a projection. (Don't worry about the minus signs; those can be fixed with a simple reflection.) This is only possible in ∞-dimensional space; in any N-dimensional space of finite N, no matter how large N and no matter how many rotations you chain together, you'll never end up with a projection. In ∞-dimensional space, however, rotations asymptotically approach a projection!

(And BTW, this is one characteristic of ∞-space that you cannot get by studying an N-dimensional space for large N. The limit works only when N=∞; for all finite N there's nothing that would indicate that this is the case.)

Now consider the projection S we've constructed as the limit of ...R₃·R₂·R₁. Applying S twice to P drops 2 coordinates off the front: S·S·P = S²·P= <3, 4, 5, ...>. Applying it n times drops n coordinates off P. Now the million dollar question: what's the limit of Sⁿ·P as n→∞? Note that the coordinates of P are unbounded, and that after n applications of S, the image of P looks like ±<n, n+1, n+2, ...>. So it's not clear what the limiting case would be, or whether there would be a limit at all. But we can get an idea of this by considering the infinite (pseudo)transformation matrix that represents S. It's basically a matrix with -1's down its diagonal, except shifted by one column so that there's a column of 0's on the left. If you compose S with itself, you get an infinite (pseudo)matrix with two columns of 0's on the left, and then the off-diagonal diagonal of 1's. Take that to the limit, and what do you get? The zero matrix. IOW, the result of projecting P along an infinite number of axes collapses it to the origin, i.e., a point in 0D space.

There are problems with this interpretation, though. For one thing, where did those 0 coordinates of P's image come from? Every finite series of applications of S produce a vector with non-zero coordinates, because P does not contain any 0's. Furthermore, each further application of S increases the minimum absolute value of the coordinates of the image of P. By that logic, at the limit we should get a point whose coordinates are infinitely large, rather than 0!

Furthermore, there's the issue of what happens to P just before it got projected to the 0 vector. Intuitively we'd expect that sometime before we dropped all the coordinates of P (via projection), there should be a finite number of non-zero coordinates left, and this number should gradually decrease to 0 at the limit. However, every finite series of compositions of S yields a transformation that maps P to a vector with decidedly non-zero coordinates. The number of non-zero coordinates in P never decreases, in spite of the projection. It's only at the limit that it leaps from an ∞-vector of very large coordinates suddenly into the zero vector. There is no in-between state; something discontinuous happens when we take the limit and it just leaps from a vector of non-zero coordinates to a vector of all zero coordinates.

There's also the issue of a different kind of projection operator: the kind that actually projects a point by setting one of its coordinates to zero. I.e., T₁·P = <0, 2, 3, 4, ...>. We can again, construct a sequence of these projection operators: T₁·T₂·P = <0, 0, 3, 4, ...> and so on. Taking the limit to ∞, we also get the zero matrix that maps P to the zero vector. Except, in this case, there's a gradual reduction of non-zero coordinates in P, until all coordinates become zero. There's still a quantum leap at the end -- any finite iteration of T still yields an image of P with an infinite number of non-zero coordinates, and the limiting transformation gets rid of all of 'em in one fell swoop. But at least there's an increasing prefix of 0's that makes the limit that projects to the zero vector seem more sensible. So T_i is a different kind of projection than S_i, that behaves more like the usual projection operator in finite-dimensional spaces. So we may call T_i the "proper" projection operators, whereas S_i are the "improper" projection operators. Both have the zero matrix at the limit. The "improper" projection operators behave like some weird intermediate class of operators between proper projections and rotation/reflections, that only exist in ∞-space.

(And BTW, you can combine the iterated T's with the iterated S's in such a way that the number of 0 coordinates grow at a different rate than the non-zero coordinates are "consumed". AFAIK there is no analogue of such a strange transformation in any finite-dimensional space. Another way to think of "improper projections" is that they behave like "deletion" operators, that delete, or forget, coordinates from an ∞-vector. A kind of weird boundary-case transformation the sits at the limit of infinite sequences of rotations that resembles both rotations and projections but behaves somewhat differently from either, in a way only possible in an ∞-space.)
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Re: Living in infinite-dimensional space

You're right, I don't see how you can do without infinitesimal numbers. I guess ix is the number that when summed together infinite times is equal to x. I imagine there is a better way to do it, but I know zilch about infinitesimals.

As for the infinite Keith Moon cherry bomb, it is in direct contact with water so the infinite explosive dissipation doesn't apply. Can an infinite amount of explosive power can be confined in a finite cherry bomb? No, but the bomb could have infinite volume because any nonzero length of the pipe also has infinite volume. Just make a cylindrical bomb an inch long that fits inside the pipe. Then could water hammer cause countably infinite damage? I guess we require experimental results.
PatrickPowers
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Re: Living in infinite-dimensional space

Here's an interesting question to ponder: does there exist a projection of the ∞-cube whose image is the 2-sphere?

Consider, as motivation, the lower dimensional cases. The 3-cube projects to a hexagon, and the 4-cube projects to a rhombic dodecahedron (among many others, of course -- it depends on the exact projection used). In general, an n-cube projects to some set of zonohedra, from which we can select the one with the maximal area or most evenly-distributed vertices. The question is, does there exist a sequence of zonohedra such that as n→∞ the sequence converges to the 2-sphere?

We know that this is not possible for the ∞-cross, because it only has a countable number of vertices, and therefore cannot completely cover the 2-sphere. The ∞-cube, however, has an uncountable number of vertices, and furthermore these vertices come in antipodal pairs along an uncountable number of diagonals (whereas the ∞-cross's vertices always lie along the coordinate axes, which are only countably many). Thus, in theory it should be possible to map the ∞-cube's vertices onto the 2-sphere. However, it's unclear whether this mapping can be a projection.

Another way to pose the question is, does there exist some point P in ∞-space, such that the viewer, looking at the ∞-cube, would see an object whose 3D shadow is a sphere? Is it possible to compute P (if it exists)?

There are some complications with answering this question. One issue is that the ∞-cube has an infinite outradius, so firstly we need to define what exactly we mean by a projection. It would appear that an orthogonal projection probably would not work, since to maximize coverage of the 2-sphere by the images of the vertices one would likely have to choose a viewpoint from which the projection images are infinitely distant from the origin. So we need some way of "renormalizing" the result to, say, a 2-sphere of unit radius.

//

One can also consider the simpler question of whether the ∞-cube projects to a circle (a 1-sphere). I haven't been able to prove ths one either, but intuition suggests that the answer is in the affirmative, because one can conceive of how one could arrange a projection such that one pair of antipodal points would project to, say, <±1, 0>, and one could arrange for the next pair of antipodal points to project to opposite ends of some line segment which lies at an irrational angle to the previous pair. Each subsequent projection image could be arranged to lie along a line at the same irrational angle to the previous projection image. Now let the number of images diverge to ∞: none of the images will overlap because each is oriented at an irrational division of 360° from the previous, so their angles to the original X axis never coincides. Thus any point on the circle thus traced out could be mapped back to the i'th projection image if it happens to be an integral multiple of said irrational angle, and any other point that isn't an integral multiple of said angle would be, in theory, included at n=∞ (though I don't know how to prove/disprove this).

The question appears much harder to answer in the case of the 2-sphere, however, since there isn't any obvious way to set up the projection such that the images of the vertices would map to the surface of the 2-sphere.
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Re: Living in infinite-dimensional space

quickfur wrote:the ∞-cross ... only has a countable number of vertices ... The ∞-cube, however, has an uncountable number of vertices.

Now there's an interesting claim. The faces of the ∞-cube are dual to the vertices of the ∞-cross, and vice versa. This can be made rigorous via a polar reciprocity about the ∞-sphere, thus invoking the duality theorem of projective geometry. Thus, the claim is that the ∞-cube has countable faces but uncountable vertices - and vice versa for the ∞-cross.
I call the vertices of the ∞-cube to make themselves ac-countable.
steelpillow
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Re: Living in infinite-dimensional space

steelpillow wrote:
quickfur wrote:the ∞-cross ... only has a countable number of vertices ... The ∞-cube, however, has an uncountable number of vertices.

Now there's an interesting claim. The faces of the ∞-cube are dual to the vertices of the ∞-cross, and vice versa. This can be made rigorous via a polar reciprocity about the ∞-sphere, thus invoking the duality theorem of projective geometry. Thus, the claim is that the ∞-cube has countable faces but uncountable vertices - and vice versa for the ∞-cross.
I call the vertices of the ∞-cube to make themselves ac-countable.

Don't forget that each facet of the ∞-cube is, in itself, also an ∞-cube. So each facet has the same cardinality of vertices as the ∞-cube itself, and having a countable number of facets cannot reduce this number. So if the ∞-cube has an uncountable number of vertices to begin with, then each facet must also have an uncountable number of vertices. There is no contradiction here.

Similarly, with the ∞-cross, there are a countable number of vertices, but an uncountable number of facets. This is easy to see, because we know from lower dimensions that each facet has a normal vector N of the form <±1,±1,±1,...> for some fixed combination of signs. For simplicity, omit the coordinate value (they are all 1's) and write only the signs. So N=+++... would define one facet, N=-+++... would define another facet, N=+-+++... yet another, and so on. Therefore, there are as many facets as there are combinations of signs of N. How many is that? Define a mapping from signs to binary digits: + maps to 1 and - maps to 0. Then each normal vector maps to a non-terminating binary string, which can be interpreted as the binary expansion of a real number between [0, 1). (Some digit strings map to the same real number, e.g., 0111... = 100..., but that doesn't change the result, because it merely means that there are at least as many normal vectors as there are real numbers in [0, 1).) Therefore, there are an uncountable number of normal vectors, and therefore an uncountable number of facets.

As far as the countability of the number of vertices is concerned, this isn't a problem either, because there are an uncountable number of facets incident on each vertex. To see this, fix a vertex v whose coordinates are 0 everywhere except 1 at the i'th coordinate. Which facets are incident on this vertex? Precisely those facets for which the normal vector N has a + at the i'th coordinate. For example, if i=0, then the incident facets are +++..., +-+++..., +--+++..., +-+-+++..., etc.. IOW, all combinations of signs whose first sign is +. How many such combinations are there? Define a mapping that maps each sign to a binary digit, except the i'th sign (the i'th sign is always +, so the mapping remains surjective). Again, we get digit strings that map to real numbers in [0, 1), and we know that there are an uncountable number of them. Therefore, there are as many facets incident on v as there are real numbers in [0, 1), which is an uncountable number.

(Note that each facet of the ∞-cross is a ∞-simplex, which has a countable number of vertices. However, the ∞-cross itself contains an uncountable number of ∞-simplices as facets.)
quickfur
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Re: Living in infinite-dimensional space

quickfur wrote:if the ∞-cube has an uncountable number of vertices to begin with, then each facet must also have an uncountable number of vertices.

Excerpt, your conditional "if" is entirely unproven. I say it is false: see below.

Similarly, with the ∞-cross, there are a countable number of vertices, but an uncountable number of facets.

That is absurd. If the vertices are countable, then so are the facets.

I do note that you persist in your incorrect definition of "countable". Just because you cannot count something to the end does not mean that it is uncountable. All you need is an algorithm - any algorithm - that will go on for ever and not miss any on the way. You really need to remember that.
steelpillow
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Re: Living in infinite-dimensional space

I've made a small step towards projecting the ∞-cube to a sphere. More specifically, I managed to find a(n infinite) matrix that maps the vertices of the ∞-cube to a continuous line segment, which is a 0-sphere.

The key insight is the fact that the columns of a projection matrix J is simply the image of the basis vectors of the ambient space. I.e., the i'th column of J is the image of the i'th basis vector under J. So take a ∞-cube with the coordinates <1/2±1/2, 1/2±1/2, ...>, i.e., the ∞-cube between <0,0,0...> and <1,1,1...>. Form an infinite matrix J such that its i'th column is <1/2^i, 0, 0, 0, ...>. Or equivalently, the first row is <1, 1/2, 1/4, 1/8, ...> and all other rows are 0. Now take any vertex v from the ∞-cube, which would be a vector, each coordinate of which is either 0 or 1. The image of v under J would thus be of the form <v0 + v1/2 + v2/4 + v3/8 + ..., 0, 0, 0...>, where v0, v1, v2, ... are the coordinates of v.

Since the vertices of the ∞-cube cover all combinations of 0's and 1's in each coordinate position, the first coordinate of J·v converges to a real number between [0, 1). It spans all real numbers between [0, 1) as v varies over the vertices of the ∞-cube. (All other coordinates of J·v are 0, so we can just discard them and consider only the first coordinate.) IOW, the image of the ∞-cube under J is the continuous(!) line segment [0, 1), in spite of the fact that the vertices of the ∞-cube are discrete! (And BTW, this is another proof that the ∞-cube has an uncountable number of vertices. Or at least as many as there are real numbers between 0 and 1.)

The next step would be to find a matrix that would map the ∞-cube's vertices to a circle, a 1-sphere. Once that's done, all that remains is to extend it to a mapping to a 2-sphere, and we would have proven that the ∞-cube projects to a 2-sphere!

And if this turns out to be possible, my next conjecture would be that the ∞-cube projects to an n-sphere for all finite n. Not 100% sure this would be the case, but if the mapping to a 1-sphere and 2-sphere is possible, there would be a strong case that such a projection applies to all n-spheres too. This would be a property that no n-cube of finite n shares, since in finite dimensions the n-cube can only project to a zonotope at the most, never to a lower-dimensional sphere.
quickfur
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Re: Living in infinite-dimensional space

Then there's sphere packing in an infinite space. Some people actually study such things.

The percentage of space filled with packed spheres would be zero.
PatrickPowers
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Re: Living in infinite-dimensional space

A somewhat tangential but nevertheless interesting next step from mapping the vertices of the ∞-cube to a continuous line segment: change the transformation J such that the first column is of the form <1, 0, 1/2, 0, 1/4, 0, ... 1/(2^(i/2)), 0, ...>, and the second column is of the form <0, 1, 0, 1/2, 0, 1/4, 0, ... 1/(2^((i-2)/2), 0, ...>. The rest of the columns are zero. Then the image of the ∞-cube's vertices under J would be a vector of the form <x, y, 0, 0, 0, ...> where x is a real number determined by the even-indexed coordinates of the vertex, and y is another real number determined by the odd-indexed coordinates of the vertex. Given an ∞-cube between <0,0,0...> and <1,1,1...>, since its vertices cover every combination of 0's and 1's in its coordinates, their images under J would span all vectors in the 2D square [0,1) × [0,1). (We ignore the trailing zeroes in the image vectors.)

It's worth iterating that this linear transformation maps the discrete vertices of the ∞-cube into the continuous square [0,1)×[0,1) having non-zero area. IOW, J projects the discrete vertices of the ∞-cube into a continuous area-filling 2D region(!).

We can go even further. By spacing out the non-zero components of J further, say every 3 rows, and having 3 columns each shifted by a row down, we obtain a linear transformation that maps the ∞-cube's vertices into a space-filling 3D region. In general, by spacing out the non-zero components by every k rows and having k columns, each shifted a row down from the previous, we can map the ∞-cube's discrete vertices into a continuous k-dimensional space-filling bulk, for all natural numbers k.

While this isn't quite the projection to the 2-sphere that I was initially looking for, this is astounding!!! This proves that not only the ∞-cube has an uncountable number of vertices, but that they can be mapped into a k-space-filling region with a linear transformation. No suspicious set-theoretic sleight-of-hand here, this is pure and simple a linear transformation. This means that a hypothetical observer in ∞-space just needs to look at the ∞-cube from a particular angle, and its vertices would form a continuous, k-dimensional space-filling image. That's totally wild, man!! (And we aren't even talking about edges, k-faces, facets, of the ∞-cube yet, this is just its vertices -- discrete vertices!!)
quickfur
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Re: Living in infinite-dimensional space

It'll be a while before I absorb this.
PatrickPowers
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Re: Living in infinite-dimensional space

That is remarkable. Let's go even further!

Code: Select all
`  1/2  0  1/4  0   0  1/8  0   0   0  1/16  0    0    0    0   1/32  0 ...   0  1/2  0   0  1/4  0   0   0  1/8  0    0    0    0   1/16  0    0   0   0   0  1/2  0   0   0  1/4  0   0    0    0   1/8   0    0    0   0   0   0   0   0   0  1/2  0   0   0    0   1/4   0    0    0    0   0   0   0   0   0   0   0   0   0   0   1/2   0    0    0    0    0   0   0   0   0   0   0   0   0   0   0    0    0    0    0    0   1/2  ...`

(I'd put a vector on the right side of the matrix to multiply. I guess quickfur would put a vector on the left side. Our matrices are transposed.)

This maps the ∞-cube's vertices onto the entire ∞-cube!
If the input vector has arbitrary coordinates in {0, 1}, then the output vector has arbitrary coordinates in the interval [0, 1].
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ℕℤℚℝℂ∂¬∀∃∅∆∇∈∉∋∌∏∑ ∗∘∙√∛∜∝∞∧∨∩∪∫≅≈≟≠≡≤≥⊂⊃⊆⊇ ⊕⊖⊗⊘⊙⌈⌉⌊⌋⌜⌝⌞⌟〈〉⟨⟩
mr_e_man
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Re: Living in infinite-dimensional space

most of the problems with an ∞D space having to many directions and everything dissipating immediately can be solved with an infinite amount of stuff to fill the infinite dimensional space. like the light problem could be solved by having a light bulb that has an infinite amount of light producing stuff spread out over the ∞D space. Or the house problem could be solved by having an ∞D box and carving out the middle with an ∞D shovel to clear out the ∞D inside and put the extra material in
an ∞D truck to hall away the ∞D material. And the sun could contain infinite plasma spread over all ∞D to light up the ∞D planet with ∞D light
Frisk-256
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Re: Living in infinite-dimensional space

Over and over we have the problem of infinity divided by infinity giving an undefined result. If we can have infinitesimals we may as well go whole hog and have supertesimals, infinities that have a definite size. I'll bet this was already done a hundred years ago.
PatrickPowers
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Re: Living in infinite-dimensional space

mr_e_man wrote:That is remarkable. Let's go even further!

Code: Select all
`  1/2  0  1/4  0   0  1/8  0   0   0  1/16  0    0    0    0   1/32  0 ...   0  1/2  0   0  1/4  0   0   0  1/8  0    0    0    0   1/16  0    0   0   0   0  1/2  0   0   0  1/4  0   0    0    0   1/8   0    0    0   0   0   0   0   0   0  1/2  0   0   0    0   1/4   0    0    0    0   0   0   0   0   0   0   0   0   0   0   1/2   0    0    0    0    0   0   0   0   0   0   0   0   0   0   0    0    0    0    0    0   1/2  ...`

[...]
This maps the ∞-cube's vertices onto the entire ∞-cube!

Awesome!!!

So if you look at the ∞-cube from a particular angle, its vertices would fill its entire bulk.

Is the above matrix invertible? 'cos if it is, you can go in the reverse direction: given some space-filling ∞-dimensional bulk, you can look at it in such a way that it separates into discrete vertices? That's just mind-bending to a point I can hardly imagine it'd be true, but if it was, it would be a prime showcase of why R is so bizarre. Hilbert space couldn't even hold a candle to this, lol.

(I suspect it's not invertible, though. Pretty sure you'd start getting non-finite quantities somewhere in there.)

PatrickPowers wrote:Over and over we have the problem of infinity divided by infinity giving an undefined result. If we can have infinitesimals we may as well go whole hog and have supertesimals, infinities that have a definite size. I'll bet this was already done a hundred years ago.

Ever heard of the surreal numbers? It's exactly the kind of thing you could use for this purpose. Or the hyperreal numbers.

The problem with surreal numbers is that there too many of them. So many that it's impossible to fit them into a set; they form a class-sized field, which is the largest ordered field that can possibly exist; every other ordered field is a sub-field of the surreals. Why is this a problem? Because you cannot guarantee that if you take a bunch of surreals, you'll be able to fit them into a set. This gives them some bizarre properties that cause calculus with surreals to break, or at least behave in very odd ways. Also, the surreals are not closed under limits; there exist many subsets of surreals that do not have an upper/lower bound. So doing integration on surreal functions, for example, often either cannot be done, or you have to use a different definition of integration that exhibits weird properties. Furthermore, the surreals are inherently "incomplete": there exist an infinitude of "gaps" in the surreal line, and unlike the case of the rationals, there is no completion of the surreals by filling in these gaps; you can't "plug" them with new elements without breaking the field properties of the surreals. One of these gaps, unfortunately, sit exactly between the finite surreals and ω (the "simplest" infinite surreal). This gap is, for various reasons, called ∞ (not to be confused with ∞ in the sense of ∞-dimensional space that we're using here).

The problem with the gap ∞ is that, being a surreal gap, you cannot add it to the field without breaking field properties. So you cannot do much meaningful manipulations of it, and it does not satisfy field axioms, making calculations difficult and messy. It's tempting to use ω as the limit of 1+1+1+... instead; however, the surreal ω has properties that break some fundamental expectations when we want to work with ∞-dimensional space. The most obvious is that there are an uncountable number of surreal integers between the finite surreals and ω -- this means that we cannot meaningfully use ω to talk about indexing an ∞-dimensional vector. A vector of length ω is very different from a vector in R, because the former has an uncountable number of coordinates (in fact, it has so many coordinates you cannot fit their indices into a set; there are a class-sized number of coordinates!). This is very inconvenient, because firstly, it obviously behaves very different from vectors in ∞-space that we're talking about here. (Surreals of the form n·ω1/u where n is a natural number and u is a non-zero positive surreal are surreal integers. Since u is arbitrary, there are a class-sized number of distinct surreal integers between ω and the finite surreals.) Secondly, because there are a class-sized number of coordinates, it's impossible to define their sum in any meaningful way, so it's impossible, for example, to assign a length to such a vector.

We could bypass the class-size issue with the surreals by using the hyperreals instead. But again, we face a problem: there is no "smallest" infinite hyperreal; there are an uncountable number of infinite hyperreals between the finite hyperreals and any arbitrary positive infinite hyperreal. So again, we have the problem of how to assign a value to 1+1+1+.... There are an infinitude of possibilities to choose from, and it's not clear why any of them would be better than any other. And again, there are an uncountable number of infinite analogues of integers between any finite hyperreal and any infinite hyperreal. So it's hard to justify any particular choice of hyperreal H := 1+1+1+.... If we remove a term from the sum, does that make the sum now (H-1) instead? Unfortunately, from the POV of the series itself, it's still exactly the same series. So the sum should still be H. But then you just broke the transfer principle of the hyperreals, so you have a contradiction.

The underlying problem here is that we're trying to apply field properties (and other properties) to an infinite number of terms (coordinates), but an actual infinite number of coordinates just do not work like that. There's an inherently different behaviour once you're dealing with infinite quantities. You can have infinite quantities like surreals or hyperreals that obey field axioms (and so are nice to manipulate), but you won't be able to map them to a countable number of coordinates in any meaningful way. Or you can deal with the number of coordinates as a countable ordinal / cardinal, but then it will not obey field axioms, and calculations will be messy.

//

This brings me back to my geometry of arbitrarily large dimensions: intuitively speaking, our mental model of an infinite-dimensional space isn't actual infinite-dimensional space. Rather, it's a fixed, finite N-dimensional space for some very large N. (Large enough that from our POV that its magnitude is intractible, and for all normal intents and purposes it behaves "as if" it were infinite.) Geometry in such an N-space would be "nicely behaved" in the way we expect; i.e., it would behave like a finite-dimensional space with all the niceties that come with that. Well, actually it is a finite-dimensional space. Only N is so large that we perceive it as "essentially the same as ∞" for all intents and purposes.

An actual infinite-dimensional space, however, is a whole 'nother beast, as this thread is starting to show. In addition to the bizarre properties we've been talking about, ∞-space also behaves in a fundamentally different way from any finite dimensionality. Certain things we come to take for granted do not hold for the infinite.
quickfur
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Re: Living in infinite-dimensional space

quickfur wrote:So if you look at the ∞-cube from a particular angle, its vertices would fill its entire bulk.

That description is rather loose. I wouldn't call this linear transformation a projection.

Is the above matrix invertible? 'cos if it is, you can go in the reverse direction: given some space-filling ∞-dimensional bulk, you can look at it in such a way that it separates into discrete vertices? That's just mind-bending to a point I can hardly imagine it'd be true, but if it was, it would be a prime showcase of why R is so bizarre. Hilbert space couldn't even hold a candle to this, lol.

(I suspect it's not invertible, though. Pretty sure you'd start getting non-finite quantities somewhere in there.)

It's simpler than that. The output vector (1/2, 0, 0, 0, ...) is produced by two different input vectors, since 1/2 + 0/4 + 0/8 + 0/16 + ... = 0/2 + 1/4 + 1/8 + 1/16 + ... .
So the mapping given above is not invertible.

For the more general question of whether there exists any linear mapping F of the ∞-cube onto its vertices:
Consider vectors x and y in the cube such that F(x) and F(y) are two distinct vertices of the cube. By linearity, (x+y)/2, which is in the cube, gets mapped to (F(x)+F(y))/2, which is not a vertex, being the midpoint of two vertices.
A linear transformation of a convex set is still convex. (Convexity is defined simply as containment of the line segment connecting any two of its points.) The cube is convex, but the set of its vertices is not convex.
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mr_e_man
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