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Here's a further thought: suppose there's a bunch of stars and associated planets in ∞-space. Then if take a slice of this space in the shape of a Hilbert space, then all the objects within this subspace will be a finite distance from each other, whereas their distances with objects outside of this subspace would be infinite. Which in turn means that the influence of objects outside this subspace on the objects within this space would be at most infinitesimal. IOW, negligible compared to the objects within the subspace.

So we could actually consider objects within this subspace independently of the objects outside. Which reduces the effective space of consideration to a Hilbert space. Since all the stars and planets under consideration are within this subspace, all the matter associated with them would also be in this space, including any inhabitants, objects, buildings, etc.. So we can think of ∞-space as a union of Hilbert spaces that more-or-less do not interact with each other.

The only time we have to consider a space larger than a Hilbert space is when we have objects with infinite outradius, like the ∞-cube. The interesting thing about the ∞-cube is that it has infinite outradius but finite in-radius, so under the influence of self-gravity it ought to collapse into an ∞-sphere of non-finite radius (because volume must be preserved, and the cube has nonzero volume whereas the sphere of finite radius has zero volume). Which means if we restrict ourselves to an ∞-dimensional with finite coordinates, such a sphere would extend outside the space! So it would be an impossible object. Unless we expand our space to one that allows non-finite coordinates.

But if we limit ourselves to a space in which objects exert more than infinitesimal influence on each other, then we're basically limited to a Hilbert space. Which is strangely deficient, geometrically speaking, because objects like the ∞-cross would not have a dual within the space.

So we could actually consider objects within this subspace independently of the objects outside. Which reduces the effective space of consideration to a Hilbert space. Since all the stars and planets under consideration are within this subspace, all the matter associated with them would also be in this space, including any inhabitants, objects, buildings, etc.. So we can think of ∞-space as a union of Hilbert spaces that more-or-less do not interact with each other.

The only time we have to consider a space larger than a Hilbert space is when we have objects with infinite outradius, like the ∞-cube. The interesting thing about the ∞-cube is that it has infinite outradius but finite in-radius, so under the influence of self-gravity it ought to collapse into an ∞-sphere of non-finite radius (because volume must be preserved, and the cube has nonzero volume whereas the sphere of finite radius has zero volume). Which means if we restrict ourselves to an ∞-dimensional with finite coordinates, such a sphere would extend outside the space! So it would be an impossible object. Unless we expand our space to one that allows non-finite coordinates.

But if we limit ourselves to a space in which objects exert more than infinitesimal influence on each other, then we're basically limited to a Hilbert space. Which is strangely deficient, geometrically speaking, because objects like the ∞-cross would not have a dual within the space.

- quickfur
- Pentonian
**Posts:**2988**Joined:**Thu Sep 02, 2004 11:20 pm**Location:**The Great White North

quickfur wrote:take a slice of this space in the shape of a Hilbert space, then all the objects within this subspace will be a finite distance from each other, whereas their distances with objects outside of this subspace would be infinite. Which in turn means that the influence of objects outside this subspace on the objects within this space would be at most infinitesimal. IOW, negligible compared to the objects within the subspace.

Except, there are infinitely many subspaces which intersect yours but have the odd dimension different, and objects within each of which are a finite distance apart. Moreover, since the two n-subspaces now inhabit a finite-dimensional (n+1)-subspace, the distance between any two objects in intersecting subspaces will still be finite.

If you vary that same dimension for each subspace and leave the others the same, you can in the limit construct an infinite-dimensional space in which every object is a finite distance from every other object.

- steelpillow
- Trionian
**Posts:**67**Joined:**Sat Jan 15, 2011 7:06 pm**Location:**England

quickfur wrote:Here's a further thought: suppose there's a bunch of stars and associated planets in ∞-space. Then if take a slice of this space in the shape of a Hilbert space, then all the objects within this subspace will be a finite distance from each other, whereas their distances with objects outside of this subspace would be infinite. Which in turn means that the influence of objects outside this subspace on the objects within this space would be at most infinitesimal. IOW, negligible compared to the objects within the subspace.

I don't think this works. Infinite number of objects each with infinitesimal influence, anything could happen. I think I could use a similar argument to "prove" that the whole inf-dimensional space has a volume of zero.

- PatrickPowers
- Tetronian
**Posts:**481**Joined:**Wed Dec 02, 2015 1:36 am

steelpillow wrote:quickfur wrote:take a slice of this space in the shape of a Hilbert space, then all the objects within this subspace will be a finite distance from each other, whereas their distances with objects outside of this subspace would be infinite. Which in turn means that the influence of objects outside this subspace on the objects within this space would be at most infinitesimal. IOW, negligible compared to the objects within the subspace.

Except, there are infinitely many subspaces which intersect yours but have the odd dimension different, and objects within each of which are a finite distance apart. Moreover, since the two n-subspaces now inhabit a finite-dimensional (n+1)-subspace, the distance between any two objects in intersecting subspaces will still be finite.

If you vary that same dimension for each subspace and leave the others the same, you can in the limit construct an infinite-dimensional space in which every object is a finite distance from every other object.

I think we're talking past each other here. The whole point I was trying to make was to take a subset of R

- quickfur
- Pentonian
**Posts:**2988**Joined:**Thu Sep 02, 2004 11:20 pm**Location:**The Great White North

PatrickPowers wrote:quickfur wrote:Here's a further thought: suppose there's a bunch of stars and associated planets in ∞-space. Then if take a slice of this space in the shape of a Hilbert space, then all the objects within this subspace will be a finite distance from each other, whereas their distances with objects outside of this subspace would be infinite. Which in turn means that the influence of objects outside this subspace on the objects within this space would be at most infinitesimal. IOW, negligible compared to the objects within the subspace.

I don't think this works. Infinite number of objects each with infinitesimal influence, anything could happen. I think I could use a similar argument to "prove" that the whole inf-dimensional space has a volume of zero.

Hmm. It's an interesting conundrum. Does the influence of objects in an infinitude of infinitely-distant Hilbert subspaces of R

- quickfur
- Pentonian
**Posts:**2988**Joined:**Thu Sep 02, 2004 11:20 pm**Location:**The Great White North

quickfur wrote:steelpillow wrote:quickfur wrote:take a slice of this space in the shape of a Hilbert space, then all the objects within this subspace will be a finite distance from each other, whereas their distances with objects outside of this subspace would be infinite. Which in turn means that the influence of objects outside this subspace on the objects within this space would be at most infinitesimal. IOW, negligible compared to the objects within the subspace.

Except, there are infinitely many subspaces which intersect yours but have the odd dimension different, and objects within each of which are a finite distance apart. Moreover, since the two n-subspaces now inhabit a finite-dimensional (n+1)-subspace, the distance between any two objects in intersecting subspaces will still be finite.

If you vary that same dimension for each subspace and leave the others the same, you can in the limit construct an infinite-dimensional space in which every object is a finite distance from every other object.

I think we're talking past each other here. The whole point I was trying to make was to take a subset of R^{∞}that fits within a Hilbert space, i.e., a subset in which every object is a finite distance from every other object. My postulate is that if you take any two disjoint Hilbert (sub)spaces of R^{∞}, the objects in each Hilbert space would be infinitely distant from the objects in the other space, so their influence on each other would be infinitesimal, i.e., negligible.

I just followed up some logical consequences of your postulate. Nothing "talking past" about that.

- steelpillow
- Trionian
**Posts:**67**Joined:**Sat Jan 15, 2011 7:06 pm**Location:**England

Here are some further thoughts about this.

Let the ambient ∞-space S be the set of all ∞-dimensional vectors, that is, (countably-long) ordered sequences of real numbers.

The length of a vector v∈S is defined as ∥v∥ := √(v₀² + v₁² + v₂² + ...). If ∥v∥ converges, then we say that it has finite length; otherwise, it has infinite length. Since each term under the √ is positive, the length either converges to some non-negative real number, or diverges to infinity; it cannot be conditionally convergent. So rearrangements of coordinate do not affect its length nor its convergence. I.e., the length, if it's finite, is well-defined and does not change under linear transformations.

Now let b∈S be some arbitrary fixed vector in S with infinite length. Construct the subset T = { v | ∥v - b∥ is finite }. Let's call this a Hilbert subspace of S under b. If b is the zero vector, then T is a Hilbert space as a subspace T₀ of S in which all vectors are of finite length, and the distance between any two vectors also has finite length.

Theorem 1. Let b₁, b₂ ∈S be two infinite-length vectors, and T₁, T₂ their corresponding Hilbert subspaces. If ∥b₁ - b₂∥ is finite, then T₁ = T₂. Proof: let v₁∈S₁ and v₂∈S₂. Then ∥v₁ - b₁∥ is finite, by definition, and so is ∥v₂ - b₂∥. Therefore, ∥v₁ - v₂∥ = ∥(v₁ - b₁) - (v₂ - b₂) + (b₁ - b₂)∥. Since (v₁ - b₁), (v₂ - b₂), and (b₁ - b₂) are finite, this means that ∥v₁ - v₂∥ is finite. By definition of Hilbert subspaces, then, v₂∈S₁ and v₁∈S₂. This holds for any choice of v₁ and v₂, therefore T₁ = T₂. QED.

Theorem 2. Now suppose ∥b₁ - b₂∥ is infinite. Then T₁ and T₂ must be disjoint. Proof: suppose for contradiction that they are not disjoint, that is, there exists some common v such that v∈T₁ and v∈T₂. Then v must be a finite distance from all vectors in T₁, and it must also be a finite distance from all vectors in T₂. But then that would mean that the sum of its distance from any vector in T₁ with its distance from any vector in T₂ is also finite, therefore the vectors is T₁ are a finite distance from the vectors in T₂, so they must be the same space. This contradicts the assumption that ∥b₁ - b₂∥ is infinite. QED.

Lemma 3. Let v be any vector in S. Then there exists some b∈S such that its corresponding Hilbert subspace T_{b} contains v. This is trivial, since we can just take b=v, and it will generate the required containing Hilbert subspace. But the point here is that every vector in S belongs to some Hilbert subspace of S.

Theorem 3. S is a disjoint union of its Hilbert subspaces. Proof: let v₁ and v₂ be vectors in S. It suffices to show that they either belong to the same Hilbert subspace, or belong to two disjoint Hilbert subspaces. By lemma 3, any vector in S has a corresponding Hilbert subspace that contains it. Then the first case follows from theorem 1, and the second case follows from theorem 2. QED.

//

OK, that was the basis of what I was saying. Now let's consider several possibilities for an ∞-dimensional universe whose space is S.

1) Suppose there's only a finite amount of matter and interactions take finite time. Then by theorem 3, this matter must be distributed in at most a finite number of Hilbert subspaces of S. Since these subspaces are disjoint (theorem 2), the distance between matter in one Hilbert subspace from another is infinite, while the matter within the same Hilbert subspace is finite. Therefore, any forces between disjoint subspaces must be zero (or at most infinitesimal). Therefore, the only important interactions between matter happens within the same Hilbert subspace. So we may treat it as multiple separate universes, each corresponding with a Hilbert subspace, which are essentially independent of each other since forces between them are infinitesimal, i.e., negligible, if not outright zero. Therefore, it suffices to reduce the scope of our considerations to a Hilbert subspace of S, not the entire space.

2) Suppose there's a finite amount of matter but interactions may take infinite time. So an infinitesimal interaction between two disjoint subspaces may add up to a non-negligible amount after an infinite amount of time has passed. Nevertheless, interactions within a subspace would require only finite time, so an infinite number of intra-subspace interactions would have taken place by the time a non-zero interaction happens across different subspaces. Therefore again, the most important interactions happen within a Hilbert subspace of S. Inter-subspace interactions are only secondarily important, since they only happen after an infinite amount of time.

3) Support there's an infinite amount of matter. Then it can be distributed either in a finite number of Hilbert subspaces, or an infinite number of Hilbert subspaces. In the first case, since there's an infinite amount of matter within a single subspace, interactions within the subspace are infinitely stronger than any interactions between subspaces. Again, the most important interactions happen within a Hilbert subspace, and any external influences are only secondarily important. In the second case, there could be a finite amount of matter within each subspace, or an infinite amount. If a finite amount, then there's the possibility that matter from other subspaces have a stronger influence on matter within the subspace. In the case of gravity, the overall effect of this is that the matter in that subspace would be scattered, pulled away from each other in the direction of the other subspaces. My guess is that eventually matter would redistribute itself such that the most important interactions happen within a subspace, thus falling in one of the aforementioned cases. In the last case, where you have an infinite amount of matter distributed across an infinite number of subspaces, it could go either way. Either the intra-subspace interaction is strong enough that it dominates external interactions, in which case we again have the most important interactions happening within a Hilbert subspace, or the inter-subspace interactions are stronger, in which case the matter in that subspace would be torn apart and drawn into other subspaces, redistributing itself until it reaches a state where the most important interactions happen within a subspace. This is inevitable, since the distance between matter within a subspace is finite, and the distance between matter in two disjoint subspaces is infinite. So any matter that finds itself within a finite distance to other matter will inevitably interact strongly (and in all likelihood stick to each other), whereas matter that lie in disjoint subspaces will only interact weakly, if at all non-negligibly.

//

The overall picture then emerges, that the most important interactions in an ∞-dimensional space takes place within a Hilbert subspace of S. Interactions outside of a Hilbert subspace are infinitely weaker, and in most cases negligible. (If such is not the case, then the affected matter would simply fly apart, and what's left, if anything at all, will be tightly-interacting, overwhelming any external influences.)

This has profound consequences, because it means that objects like the ∞-cube likely would not arise naturally -- because the matter surrounding any vertex of the ∞-cube are so far apart from the matter surrounding the other vertices that they either don't interact at all, or if they did, would tear the cube apart (e.g., the gravitational influence of matter surrounding the other vertices would be infinite, and would collapse the cube into a ∞-sphere). In contrast, objects like the ∞-cross actually has a chance of being stable. So you have the strange situation where ∞-crosses may exist (in a Hilbert subspace of S) but their duals, being ∞-cubes, cannot!

//

∞-dimensional space just got a lot stranger.

Let the ambient ∞-space S be the set of all ∞-dimensional vectors, that is, (countably-long) ordered sequences of real numbers.

The length of a vector v∈S is defined as ∥v∥ := √(v₀² + v₁² + v₂² + ...). If ∥v∥ converges, then we say that it has finite length; otherwise, it has infinite length. Since each term under the √ is positive, the length either converges to some non-negative real number, or diverges to infinity; it cannot be conditionally convergent. So rearrangements of coordinate do not affect its length nor its convergence. I.e., the length, if it's finite, is well-defined and does not change under linear transformations.

Now let b∈S be some arbitrary fixed vector in S with infinite length. Construct the subset T = { v | ∥v - b∥ is finite }. Let's call this a Hilbert subspace of S under b. If b is the zero vector, then T is a Hilbert space as a subspace T₀ of S in which all vectors are of finite length, and the distance between any two vectors also has finite length.

Theorem 1. Let b₁, b₂ ∈S be two infinite-length vectors, and T₁, T₂ their corresponding Hilbert subspaces. If ∥b₁ - b₂∥ is finite, then T₁ = T₂. Proof: let v₁∈S₁ and v₂∈S₂. Then ∥v₁ - b₁∥ is finite, by definition, and so is ∥v₂ - b₂∥. Therefore, ∥v₁ - v₂∥ = ∥(v₁ - b₁) - (v₂ - b₂) + (b₁ - b₂)∥. Since (v₁ - b₁), (v₂ - b₂), and (b₁ - b₂) are finite, this means that ∥v₁ - v₂∥ is finite. By definition of Hilbert subspaces, then, v₂∈S₁ and v₁∈S₂. This holds for any choice of v₁ and v₂, therefore T₁ = T₂. QED.

Theorem 2. Now suppose ∥b₁ - b₂∥ is infinite. Then T₁ and T₂ must be disjoint. Proof: suppose for contradiction that they are not disjoint, that is, there exists some common v such that v∈T₁ and v∈T₂. Then v must be a finite distance from all vectors in T₁, and it must also be a finite distance from all vectors in T₂. But then that would mean that the sum of its distance from any vector in T₁ with its distance from any vector in T₂ is also finite, therefore the vectors is T₁ are a finite distance from the vectors in T₂, so they must be the same space. This contradicts the assumption that ∥b₁ - b₂∥ is infinite. QED.

Lemma 3. Let v be any vector in S. Then there exists some b∈S such that its corresponding Hilbert subspace T

Theorem 3. S is a disjoint union of its Hilbert subspaces. Proof: let v₁ and v₂ be vectors in S. It suffices to show that they either belong to the same Hilbert subspace, or belong to two disjoint Hilbert subspaces. By lemma 3, any vector in S has a corresponding Hilbert subspace that contains it. Then the first case follows from theorem 1, and the second case follows from theorem 2. QED.

//

OK, that was the basis of what I was saying. Now let's consider several possibilities for an ∞-dimensional universe whose space is S.

1) Suppose there's only a finite amount of matter and interactions take finite time. Then by theorem 3, this matter must be distributed in at most a finite number of Hilbert subspaces of S. Since these subspaces are disjoint (theorem 2), the distance between matter in one Hilbert subspace from another is infinite, while the matter within the same Hilbert subspace is finite. Therefore, any forces between disjoint subspaces must be zero (or at most infinitesimal). Therefore, the only important interactions between matter happens within the same Hilbert subspace. So we may treat it as multiple separate universes, each corresponding with a Hilbert subspace, which are essentially independent of each other since forces between them are infinitesimal, i.e., negligible, if not outright zero. Therefore, it suffices to reduce the scope of our considerations to a Hilbert subspace of S, not the entire space.

2) Suppose there's a finite amount of matter but interactions may take infinite time. So an infinitesimal interaction between two disjoint subspaces may add up to a non-negligible amount after an infinite amount of time has passed. Nevertheless, interactions within a subspace would require only finite time, so an infinite number of intra-subspace interactions would have taken place by the time a non-zero interaction happens across different subspaces. Therefore again, the most important interactions happen within a Hilbert subspace of S. Inter-subspace interactions are only secondarily important, since they only happen after an infinite amount of time.

3) Support there's an infinite amount of matter. Then it can be distributed either in a finite number of Hilbert subspaces, or an infinite number of Hilbert subspaces. In the first case, since there's an infinite amount of matter within a single subspace, interactions within the subspace are infinitely stronger than any interactions between subspaces. Again, the most important interactions happen within a Hilbert subspace, and any external influences are only secondarily important. In the second case, there could be a finite amount of matter within each subspace, or an infinite amount. If a finite amount, then there's the possibility that matter from other subspaces have a stronger influence on matter within the subspace. In the case of gravity, the overall effect of this is that the matter in that subspace would be scattered, pulled away from each other in the direction of the other subspaces. My guess is that eventually matter would redistribute itself such that the most important interactions happen within a subspace, thus falling in one of the aforementioned cases. In the last case, where you have an infinite amount of matter distributed across an infinite number of subspaces, it could go either way. Either the intra-subspace interaction is strong enough that it dominates external interactions, in which case we again have the most important interactions happening within a Hilbert subspace, or the inter-subspace interactions are stronger, in which case the matter in that subspace would be torn apart and drawn into other subspaces, redistributing itself until it reaches a state where the most important interactions happen within a subspace. This is inevitable, since the distance between matter within a subspace is finite, and the distance between matter in two disjoint subspaces is infinite. So any matter that finds itself within a finite distance to other matter will inevitably interact strongly (and in all likelihood stick to each other), whereas matter that lie in disjoint subspaces will only interact weakly, if at all non-negligibly.

//

The overall picture then emerges, that the most important interactions in an ∞-dimensional space takes place within a Hilbert subspace of S. Interactions outside of a Hilbert subspace are infinitely weaker, and in most cases negligible. (If such is not the case, then the affected matter would simply fly apart, and what's left, if anything at all, will be tightly-interacting, overwhelming any external influences.)

This has profound consequences, because it means that objects like the ∞-cube likely would not arise naturally -- because the matter surrounding any vertex of the ∞-cube are so far apart from the matter surrounding the other vertices that they either don't interact at all, or if they did, would tear the cube apart (e.g., the gravitational influence of matter surrounding the other vertices would be infinite, and would collapse the cube into a ∞-sphere). In contrast, objects like the ∞-cross actually has a chance of being stable. So you have the strange situation where ∞-crosses may exist (in a Hilbert subspace of S) but their duals, being ∞-cubes, cannot!

//

∞-dimensional space just got a lot stranger.

- quickfur
- Pentonian
**Posts:**2988**Joined:**Thu Sep 02, 2004 11:20 pm**Location:**The Great White North

At least your conclusion "So you have the strange situation where ∞-crosses may exist (in a Hilbert subspace of S) but their duals, being ∞-cubes, cannot!" not at all is surprising.

As both ∞-simplices' and ∞-orthoplexes' vertices have a single 1 and an infinite amount of 0s in its coordinates (or some scaling thereof). Thus those tuples belong to Hilbert spaces. But an ∞-cube will not only have an infinite amount of 1s (or -1s) in its coordinates, rather all its coordinates are such and hence such a vector never belongs to any Hilbert space.

--- rk

As both ∞-simplices' and ∞-orthoplexes' vertices have a single 1 and an infinite amount of 0s in its coordinates (or some scaling thereof). Thus those tuples belong to Hilbert spaces. But an ∞-cube will not only have an infinite amount of 1s (or -1s) in its coordinates, rather all its coordinates are such and hence such a vector never belongs to any Hilbert space.

--- rk

- Klitzing
- Pentonian
**Posts:**1641**Joined:**Sun Aug 19, 2012 11:16 am**Location:**Heidenheim, Germany

Klitzing wrote:At least your conclusion "So you have the strange situation where ∞-crosses may exist (in a Hilbert subspace of S) but their duals, being ∞-cubes, cannot!" not at all is surprising.

As both ∞-simplices' and ∞-orthoplexes' vertices have a single 1 and an infinite amount of 0s in its coordinates (or some scaling thereof). Thus those tuples belong to Hilbert spaces. But an ∞-cube will not only have an infinite amount of 1s (or -1s) in its coordinates, rather all its coordinates are such and hence such a vector never belongs to any Hilbert space.

--- rk

It is more correct to say that vertices of a ∞-cube that are in Hilbert space form a set of measure zero.

- PatrickPowers
- Tetronian
**Posts:**481**Joined:**Wed Dec 02, 2015 1:36 am

Here's a little puzzle for you geniuses to solve. We know that the origin <0, 0, 0, ....> is enclosed in the ∞-cross (centered on the origin). Question: is it an interior point or a boundary point?

In finite dimensions, a point p ∈ S ⊆ R^{n} is defined to be an interior point of S if there exists an n-ball B of some radius r>0 centered on p such B ⊆ S. We can extend this definition to ∞-dimensional space in the obvious way: given an ∞-dimensional vector p, we define an ∞-ball B of radius r centered on p as the set of all vectors v such that ∥v - p∥ ≤ r.

So the question is, does there exist an ∞-ball of non-zero radius that contains the origin, and is a subset of the ∞-cross?

As far as I can tell, there isn't. Because in n dimensions, the n-ball that contains the origin and inscribes the n-cross has radius r=1/√n. But when n→∞, r→0, which means that the only ∞-ball centered on the origin that's a subset of the ∞-cross has radius zero. Which means that the origin is not an internal point of the ∞-cross, but a boundary point(!!!!!!).

If I didn't make a stupid mistake somewhere, this would imply that the ∞-cross is a "pinched" shape: its center is not an interior point but, ostensibly, lies on its surface!!!

Now, in n dimensions, the set of points in the origin-centered n-cross of outradius 1 can be defined as the set S of points p = <x1, x2, ... xn> such that |x1| + |x2| + ... + |xn| ≤ 1. This, of course, is trivially extended to ∞ dimensions: we just let S be the set of points p = <x1, x2, ...> such that the infinite series |x1| + |x2| + ... converges to some limit L ≤ 1. If L exists, this series is absolutely convergent, so this is well-defined for all points that belong the ∞-cross. Here's the weird thing. In n dimensions (finite n), if the sum |x1| + |x2| + ... |xn| = 1, then it is a boundary point; if the sum < 1, then it's an interior point. Clearly, the sum of the origin's coordinates = 0 in all dimensions, including for infinite n. So does it lie on the surface of the ∞-cross, or is it an interior point??!

In finite dimensions, a point p ∈ S ⊆ R

So the question is, does there exist an ∞-ball of non-zero radius that contains the origin, and is a subset of the ∞-cross?

As far as I can tell, there isn't. Because in n dimensions, the n-ball that contains the origin and inscribes the n-cross has radius r=1/√n. But when n→∞, r→0, which means that the only ∞-ball centered on the origin that's a subset of the ∞-cross has radius zero. Which means that the origin is not an internal point of the ∞-cross, but a boundary point(!!!!!!).

If I didn't make a stupid mistake somewhere, this would imply that the ∞-cross is a "pinched" shape: its center is not an interior point but, ostensibly, lies on its surface!!!

Now, in n dimensions, the set of points in the origin-centered n-cross of outradius 1 can be defined as the set S of points p = <x1, x2, ... xn> such that |x1| + |x2| + ... + |xn| ≤ 1. This, of course, is trivially extended to ∞ dimensions: we just let S be the set of points p = <x1, x2, ...> such that the infinite series |x1| + |x2| + ... converges to some limit L ≤ 1. If L exists, this series is absolutely convergent, so this is well-defined for all points that belong the ∞-cross. Here's the weird thing. In n dimensions (finite n), if the sum |x1| + |x2| + ... |xn| = 1, then it is a boundary point; if the sum < 1, then it's an interior point. Clearly, the sum of the origin's coordinates = 0 in all dimensions, including for infinite n. So does it lie on the surface of the ∞-cross, or is it an interior point??!

- quickfur
- Pentonian
**Posts:**2988**Joined:**Thu Sep 02, 2004 11:20 pm**Location:**The Great White North

quickfur wrote:So the question is, does there exist an ∞-ball of non-zero radius that contains the origin, and is a subset of the ∞-cross?

Given any radius r > 0, find an integer n > 4/r

Therefore, the origin is a boundary point of the orthoplex, as any ball around it contains some points outside of the orthoplex. (And these points have finitely many non-zero coordinates! No need to consider infinitesimal multiples of (1, 1, 1, 1, ...).)

Here's the weird thing. In n dimensions (finite n), if the sum |x1| + |x2| + ... |xn| = 1, then it is a boundary point; if the sum < 1, then it's an interior point. Clearly, the sum of the origin's coordinates = 0 in all dimensions, including for infinite n.

One explanation (or just another weird fact that's related ) is that the sum of absolute values is a continuous function, only in finite dimensions.

In fact the sum of absolute values is precisely the L

ΓΔΘΛΞΠΣΦΨΩ αβγδεζηθϑικλμνξοπρϱσςτυϕφχψωϖ °±∓½⅓⅔¼¾×÷†‡• ⁰¹²³⁴⁵⁶⁷⁸⁹⁺⁻⁼⁽⁾₀₁₂₃₄₅₆₇₈₉₊₋₌₍₎

ℕℤℚℝℂ∂¬∀∃∅∆∇∈∉∋∌∏∑ ∗∘∙√∛∜∝∞∧∨∩∪∫≅≈≟≠≡≤≥⊂⊃⊆⊇ ⊕⊖⊗⊘⊙⌈⌉⌊⌋⌜⌝⌞⌟〈〉⟨⟩

ℕℤℚℝℂ∂¬∀∃∅∆∇∈∉∋∌∏∑ ∗∘∙√∛∜∝∞∧∨∩∪∫≅≈≟≠≡≤≥⊂⊃⊆⊇ ⊕⊖⊗⊘⊙⌈⌉⌊⌋⌜⌝⌞⌟〈〉⟨⟩

- mr_e_man
- Tetronian
**Posts:**524**Joined:**Tue Sep 18, 2018 4:10 am

Wow. So the ∞-cross is a "pinched" shape after all! Its center, the origin, is a boundary point. That's just bizarre.

Of course, in ∞-dimensional space many things are bizarre. Like the ∞-cube having an uncountable number of vertices and infinite outradius, for example. And many other things already mentioned earlier in this thread.

Hmm, come to think of it, if you built a house in the shape of a bisected ∞-cross, i.e., a pyramid with an ∞-cross base, then you could have the front door be in the center of your house! Since the center is a boundary point, you could enter through the door and access the rest of the house from there -- in all cardinal directions! Question: if you remove the origin from the ∞-cross, does it become disconnected? Somehow I suspect the answer is no, though I haven't worked out the proof; but if so, what's the topology of the surrounding connected region? I suspect it might be some kind of pathological high- (or even infinite-)genus manifold, but it's a bit beyond my imagination to visualize. In any case, one could imagine an entire city made of truncated ∞-cross houses. You'd be able to walk from any house to any other house, front door to front door, without ever needing to walk around the house unlike in lower dimensions! How cool is that?

Of course, in ∞-dimensional space many things are bizarre. Like the ∞-cube having an uncountable number of vertices and infinite outradius, for example. And many other things already mentioned earlier in this thread.

Hmm, come to think of it, if you built a house in the shape of a bisected ∞-cross, i.e., a pyramid with an ∞-cross base, then you could have the front door be in the center of your house! Since the center is a boundary point, you could enter through the door and access the rest of the house from there -- in all cardinal directions! Question: if you remove the origin from the ∞-cross, does it become disconnected? Somehow I suspect the answer is no, though I haven't worked out the proof; but if so, what's the topology of the surrounding connected region? I suspect it might be some kind of pathological high- (or even infinite-)genus manifold, but it's a bit beyond my imagination to visualize. In any case, one could imagine an entire city made of truncated ∞-cross houses. You'd be able to walk from any house to any other house, front door to front door, without ever needing to walk around the house unlike in lower dimensions! How cool is that?

- quickfur
- Pentonian
**Posts:**2988**Joined:**Thu Sep 02, 2004 11:20 pm**Location:**The Great White North

That is pretty cool. (Why doesn't this site have a like button that I could use instead of a post like this?)

- PatrickPowers
- Tetronian
**Posts:**481**Joined:**Wed Dec 02, 2015 1:36 am

Klitzing wrote:At least your conclusion "So you have the strange situation where ∞-crosses may exist (in a Hilbert subspace of S) but their duals, being ∞-cubes, cannot!" not at all is surprising.

As both ∞-simplices' and ∞-orthoplexes' vertices have a single 1 and an infinite amount of 0s in its coordinates (or some scaling thereof). Thus those tuples belong to Hilbert spaces. But an ∞-cube will not only have an infinite amount of 1s (or -1s) in its coordinates, rather all its coordinates are such and hence such a vector never belongs to any Hilbert space.

As I proved in the original post, any vector of infinite norm can be "renormalized" such that it lies inside a Hilbert (sub)space displaced from the origin by a vector of infinite norm. This is essentially what the T = { v | ∥v - b∥ is finite } construction is all about. (I'm extending the definition of a Hilbert space here to allow for displacement from the origin by a fixed vector of non-finite norm.) Since the vertices of the ∞-cube all lie in Hilbert subspaces displaced by non-zero vectors, none of them lie in the same Hilbert space as the vertices of the ∞-cross, obviously.

//

Now here's an interesting question. Take any vertex of the ∞-cube. How many other vertices lie in the same Hilbert subspace as that vertex? Or, to put it another way, collect all of the Hilbert subspaces that contain the ∞-cube's vertices into a set H of (disjoint) Hilbert subspaces. What's the cardinality of each member h of this set? And what's the cardinality of H?

My initial expectation was that card(h)=1, because every vertex of the ∞-cross is infinitely distant from the origin, so each would exist in its own Hilbert subspace. However, actually card(h) > 1, because consider, for example, the vertices <1,1,1,...> and <-1,1,1,...>. Their difference is <2,0,0,...>, which is clearly a vector of finite norm. Therefore, they must lie in the same Hilbert subspace! However, we also have vertices like <-1,-1,-1,...>, which is infinitely distant from <1,1,1,...> (the difference vector is <2,2,2,...>, clearly a vector of non-finite norm). So H must contain more than just a single Hilbert subspace.

Now, card(h) is at least ℵ₀, because for a vector like <1,1,1,...>, we can flip the sign of any finite number of coordinates and obtain another vertex that's a finite distance from it. Specifically, we can flip the sign of each coordinate in turn: <-1,1,1,1,...>, <1,-1,1,1,...>, <1,1,-1,1,...>, ... etc., which gives us at least ℵ₀ vertices that belong to the same Hilbert subspace as <1,1,1,...>. We can also flip more than 1 sign each time, too. However, only a finite number of signs can be flipped (otherwise the difference vector will have a non-finite norm). So card(h) should be the supremum of ℵ₀

Which in turn implies that H must be uncountable, because there are an uncountable number of vertices in total, but each Hilbert subspace contains only a countable number of them.

//

This result is very interesting, because it means that we can partition the vertices of the ∞-cube into disjoint subsets, such that the vertices within each subset are finitely-distant from each other and lie within the same Hilbert subspace. Each subset would contain a countable number of vertices. Now imagine what happens if we were to place this ∞-cube in ∞-dimensional space (i.e., R

Since the vertices within the same Hilbert subspace are finitely-distant from each other, they would interact with each other strongly, whereas vertices in different Hilbert subspaces would only interact weakly (infinitesimally), if at all. However, since there are an uncountable number of different Hilbert subspaces, this weak interaction may add up to a significant amount. So there seem to be two possibilities: if the interaction between Hilbert subspaces is weak, the ∞-cube would be liable to decompose into an uncountable number of disjoint chunks, each of which contains a countable number of its original vertices. OTOH, if the interaction is strong, it may either retain its present form, or collapse under its own weight into some other shape. (I want to say an ∞-sphere, but this doesn't seem to be right; the ∞-cube has finite ∞-volume but the ∞-sphere has zero ∞-volume. Question: what would be this shape? Would it fit inside a Hilbert subspace, or must it necessarily span multiple Hilbert subspaces?)

//

Another interesting result, in terms of geometry, is that it's possible to embed a fragment of the ∞-cube inside a Hilbert space, such that the result will have only a countable number of vertices, and will have surface elements in the form of k-dimensional cubes for all finite k, but will not contain the entire ∞-cube. For an observer that's confined to this Hilbert subspace, this "incomplete ∞-cube" will be indistinguishible from the entire ∞-cube, because the parts of the ∞-cube that lie outside this Hilbert space also lie in directions (vectors) that are not in this Hilbert subspace. So the observer will be unable to tell whether the rest of the ∞-cube is attached to this observable fragment or not.

- quickfur
- Pentonian
**Posts:**2988**Joined:**Thu Sep 02, 2004 11:20 pm**Location:**The Great White North

quickfur wrote:Since the vertices within the same Hilbert subspace are finitely-distant from each other, they would interact with each other strongly, whereas vertices in different Hilbert subspaces would only interact weakly (infinitesimally), if at all. However, since there are an uncountable number of different Hilbert subspaces, this weak interaction may add up to a significant amount. So there seem to be two possibilities: if the interaction between Hilbert subspaces is weak, the ∞-cube would be liable to decompose into an uncountable number of disjoint chunks, each of which contains a countable number of its original vertices. OTOH, if the interaction is strong, it may either retain its present form, or collapse under its own weight into some other shape. (I want to say an ∞-sphere, but this doesn't seem to be right; the ∞-cube has finite ∞-volume but the ∞-sphere has zero ∞-volume. Question: what would be this shape? Would it fit inside a Hilbert subspace, or must it necessarily span multiple Hilbert subspaces?)

If gravity is equally strong from all directions then it sums to zero.

The cube would take an infinite amount of time to collapse under it's own weight, with a sphere the unreachable limit.

- PatrickPowers
- Tetronian
**Posts:**481**Joined:**Wed Dec 02, 2015 1:36 am

I have discovered a very, very, very strange property of the ∞-cube.

Consider any k-dimensional cube. For simplicity, let's say it's origin-centered, with coordinates either 1 or 0. I.e., its vertices are the set of all k-dimensional vectors whose coordinates are either 1 or 0. Where are its edges? Obviously, there is an edge between every pair of vertices that differ only by 1 coordinate. For example, there's an edge between <1,0,1,0,...> and <0,0,1,0,...>, and an edge between <1,0,1,0,...> and <1,1,1,0,...>, etc.. The set of all pairs of vertices that differ only in 1 coordinate is equal to the set of all edges, because there are no edges between vertices that differ by more than 1 coordinate.

Now let k diverge to infinity and consider the ∞-cube. I.e., we're now considering the set of all ∞-dimensional vertices whose coordinates are either 1 or 0. Where are the edges of this ∞-cube? Obviously, there should be an edge between every pair of vertices that differ only by 1 coordinate. These must be the only edges, because there aren't any edges between vertices that differ by more than 1 coordinate (otherwise there'd be diagonal edges, which there aren't in the ∞-cube).

Now consider the following subset of the ∞-cube's vertices: the set F of vertices that only have a finite number of 1's. Let v be any vertex in F. Then any edge that contains v can only end in another vertex in F, because the other end of the edge can only be in a vertex that differs from v by 1 coordinate. (Otherwise it'd be a diagonal edge, and we know there aren't any diagonal edges in the ∞-cube.) Since there are only a finite number of 1's in v, any vertex that has an edge to v must also have only a finite number of 1's. That is, given any vertex v in F, it can only have an edge to another vertex in F.

This means that there are no edges between F and the rest of the ∞-cube's vertices!!!!!!!!

Or, put another way, if we start with the vertex <0,0,0,...> and recursively follow all of the edges incident on this vertex, and trace through the edge outline connected to this vertex, we find that the edge outline obtained is incomplete, and disconnected from the rest of the ∞-cube.

In fact, the ∞-cube's edges consists of (uncountably many) disjoint connected edge outlines!!! This is definitely something very, very different from finite-dimensional cubes.

Now, since F contains only vectors with finite norm, its convex hull is an incomplete ∞-cube (it's missing the antipodes of its vertices, and many other vertices whose antipodes are also not in F -- namely, those vertices with an infinite number of 1's and 0's). This incomplete ∞-cube is missing many vertices of the ∞-cube, but contains all of the edges of the ∞-cube reachable from the vertex <0,0,0,...>.

//

This is not all. Let's now consider the square faces of the ∞-cube. Clearly, there's a square face between every set of 4 vertices that differ from each other in exactly 2 coordinates. Now again, consider the set F of vertices with only a finite number of 1's in their coordinates. Obviously, any square face incident on any vertex v in F can only contain vertices that differ from v in 2 coordinates, i.e., they themselves must also be in F. There can't be any square faces with any other vertices, because that would involve diagonal edges or non-planar tetragons, which don't exist in the ∞-cube. That means that there are no square faces between F and the rest of the ∞-cube!

Similarly, consider the k-dimensional cube surtopes of the ∞-cube, for finite k. Each such surtope involves vertices that differ in exactly k coordinates. And again, if such a surtope contains a vertex in F, then the entire surtope must be contained in F; there are no k-surtopes between F and the rest of the ∞-cube. This holds for all finite k.

This implies that the ∞-cube is "finitely-disconnected": starting from any vertex, say <0,0,0,...>, all surtopes reachable from this vertex will only cover a subset of the ∞-cube: there are no finite-dimensional surtopes that join this subset fo the rest of the ∞-cube!! In fact, our "incomplete ∞-cube" -- the convex hull of F, contains precisely this subset of the ∞-cube: all of the vertices and k-surtopes of the ∞-cube that are connected to <0,0,0,...>.

So the question then is, how is this subset connected to the rest of the ∞-cube?!?! Any such connection, if one exists at all, can only involve ∞-dimensional surtopes. But ∞-dimensional surtopes of the ∞-cube are isomorphic to the ∞-cube itself -- which means they themselves are finitely-disconnected. They, like the ∞-cube itself, consists of uncountably many finitely-connected pieces that have no finite-dimensional surtopes bridging them.

All of this seems to be implying that the surface of the ∞-cube is disconnected! The ∞-cube consists of uncountably many finitely-disconnected pieces, each of which BTW fits inside a Hilbert space. And since each such piece contains all connected finite-dimensional surtopes, an observer confined to this Hilbert space will be unable to distinguish between this piece of the ∞-cube and the entire ∞-cube itself. Since there are no finite-dimensional surtopes that connect this piece with the other pieces, such an observer will not be able to tell the difference between this piece, which is in the shape of an incomplete ∞-cube, from the real ∞-cube, because there are no observable missing surtopes from the POV of the observer. (The ∞-dimensional surtopes that ostensibly connect the pieces together, being ∞-cubes themselves, do not fit in any Hilbert space so they are unobservable from within this Hilbert space. Furthermore, these supposed connecting pieces are themselves disconnected ∞-cubes, so are they really connecting pieces, or merely disconnected collections of pieces just like the ∞-cube itself?)

How's that for bizarre?!

Consider any k-dimensional cube. For simplicity, let's say it's origin-centered, with coordinates either 1 or 0. I.e., its vertices are the set of all k-dimensional vectors whose coordinates are either 1 or 0. Where are its edges? Obviously, there is an edge between every pair of vertices that differ only by 1 coordinate. For example, there's an edge between <1,0,1,0,...> and <0,0,1,0,...>, and an edge between <1,0,1,0,...> and <1,1,1,0,...>, etc.. The set of all pairs of vertices that differ only in 1 coordinate is equal to the set of all edges, because there are no edges between vertices that differ by more than 1 coordinate.

Now let k diverge to infinity and consider the ∞-cube. I.e., we're now considering the set of all ∞-dimensional vertices whose coordinates are either 1 or 0. Where are the edges of this ∞-cube? Obviously, there should be an edge between every pair of vertices that differ only by 1 coordinate. These must be the only edges, because there aren't any edges between vertices that differ by more than 1 coordinate (otherwise there'd be diagonal edges, which there aren't in the ∞-cube).

Now consider the following subset of the ∞-cube's vertices: the set F of vertices that only have a finite number of 1's. Let v be any vertex in F. Then any edge that contains v can only end in another vertex in F, because the other end of the edge can only be in a vertex that differs from v by 1 coordinate. (Otherwise it'd be a diagonal edge, and we know there aren't any diagonal edges in the ∞-cube.) Since there are only a finite number of 1's in v, any vertex that has an edge to v must also have only a finite number of 1's. That is, given any vertex v in F, it can only have an edge to another vertex in F.

This means that there are no edges between F and the rest of the ∞-cube's vertices!!!!!!!!

Or, put another way, if we start with the vertex <0,0,0,...> and recursively follow all of the edges incident on this vertex, and trace through the edge outline connected to this vertex, we find that the edge outline obtained is incomplete, and disconnected from the rest of the ∞-cube.

In fact, the ∞-cube's edges consists of (uncountably many) disjoint connected edge outlines!!! This is definitely something very, very different from finite-dimensional cubes.

Now, since F contains only vectors with finite norm, its convex hull is an incomplete ∞-cube (it's missing the antipodes of its vertices, and many other vertices whose antipodes are also not in F -- namely, those vertices with an infinite number of 1's and 0's). This incomplete ∞-cube is missing many vertices of the ∞-cube, but contains all of the edges of the ∞-cube reachable from the vertex <0,0,0,...>.

//

This is not all. Let's now consider the square faces of the ∞-cube. Clearly, there's a square face between every set of 4 vertices that differ from each other in exactly 2 coordinates. Now again, consider the set F of vertices with only a finite number of 1's in their coordinates. Obviously, any square face incident on any vertex v in F can only contain vertices that differ from v in 2 coordinates, i.e., they themselves must also be in F. There can't be any square faces with any other vertices, because that would involve diagonal edges or non-planar tetragons, which don't exist in the ∞-cube. That means that there are no square faces between F and the rest of the ∞-cube!

Similarly, consider the k-dimensional cube surtopes of the ∞-cube, for finite k. Each such surtope involves vertices that differ in exactly k coordinates. And again, if such a surtope contains a vertex in F, then the entire surtope must be contained in F; there are no k-surtopes between F and the rest of the ∞-cube. This holds for all finite k.

This implies that the ∞-cube is "finitely-disconnected": starting from any vertex, say <0,0,0,...>, all surtopes reachable from this vertex will only cover a subset of the ∞-cube: there are no finite-dimensional surtopes that join this subset fo the rest of the ∞-cube!! In fact, our "incomplete ∞-cube" -- the convex hull of F, contains precisely this subset of the ∞-cube: all of the vertices and k-surtopes of the ∞-cube that are connected to <0,0,0,...>.

So the question then is, how is this subset connected to the rest of the ∞-cube?!?! Any such connection, if one exists at all, can only involve ∞-dimensional surtopes. But ∞-dimensional surtopes of the ∞-cube are isomorphic to the ∞-cube itself -- which means they themselves are finitely-disconnected. They, like the ∞-cube itself, consists of uncountably many finitely-connected pieces that have no finite-dimensional surtopes bridging them.

All of this seems to be implying that the surface of the ∞-cube is disconnected! The ∞-cube consists of uncountably many finitely-disconnected pieces, each of which BTW fits inside a Hilbert space. And since each such piece contains all connected finite-dimensional surtopes, an observer confined to this Hilbert space will be unable to distinguish between this piece of the ∞-cube and the entire ∞-cube itself. Since there are no finite-dimensional surtopes that connect this piece with the other pieces, such an observer will not be able to tell the difference between this piece, which is in the shape of an incomplete ∞-cube, from the real ∞-cube, because there are no observable missing surtopes from the POV of the observer. (The ∞-dimensional surtopes that ostensibly connect the pieces together, being ∞-cubes themselves, do not fit in any Hilbert space so they are unobservable from within this Hilbert space. Furthermore, these supposed connecting pieces are themselves disconnected ∞-cubes, so are they really connecting pieces, or merely disconnected collections of pieces just like the ∞-cube itself?)

How's that for bizarre?!

Last edited by quickfur on Thu Jul 11, 2024 8:05 pm, edited 1 time in total.

- quickfur
- Pentonian
**Posts:**2988**Joined:**Thu Sep 02, 2004 11:20 pm**Location:**The Great White North

quickfur wrote:[...]

Now, card(h) is at least ℵ₀, because for a vector like <1,1,1,...>, we can flip the sign of any finite number of coordinates and obtain another vertex that's a finite distance from it. Specifically, we can flip the sign of each coordinate in turn: <-1,1,1,1,...>, <1,-1,1,1,...>, <1,1,-1,1,...>, ... etc., which gives us at least ℵ₀ vertices that belong to the same Hilbert subspace as <1,1,1,...>. We can also flip more than 1 sign each time, too. However, only a finite number of signs can be flipped (otherwise the difference vector will have a non-finite norm). So card(h) should be the supremum of ℵ₀^{k}for all finite k. Therefore card(h) = ℵ₀^{ω}= ℵ₀. I.e., each Hilbert subspace contains a countable number of vertices.

[...]

Hmm, looks like I made a mistake here. The supremum of ℵ₀

Anybody want to try a hand at proving this?

- quickfur
- Pentonian
**Posts:**2988**Joined:**Thu Sep 02, 2004 11:20 pm**Location:**The Great White North

I don't know much about infinite ordinals, or whether "supremum" is the right notion here. But note that the infinite cardinal ℵ₀^{k} = ℵ₀ (since k is finite), and the supremum of ℵ₀ is of course ℵ₀.

Anyway, we can see directly that a Hilbert subspace has countably many vertices. As you said, only a finite number of signs can be flipped. (If k signs are flipped then the difference vector has norm 2√k.) For each n, you can make a finite list of all vertices which differ from (1,1,1,...) only in their first n coordinates. (To ensure that the lists are disjoint, the n'th coordinate must be different, and the previous coordinates may be different.) Then you can put these lists in order, varying n, to make a countably infinite list.

Anyway, we can see directly that a Hilbert subspace has countably many vertices. As you said, only a finite number of signs can be flipped. (If k signs are flipped then the difference vector has norm 2√k.) For each n, you can make a finite list of all vertices which differ from (1,1,1,...) only in their first n coordinates. (To ensure that the lists are disjoint, the n'th coordinate must be different, and the previous coordinates may be different.) Then you can put these lists in order, varying n, to make a countably infinite list.

Last edited by mr_e_man on Thu Jul 11, 2024 8:19 pm, edited 1 time in total.

ΓΔΘΛΞΠΣΦΨΩ αβγδεζηθϑικλμνξοπρϱσςτυϕφχψωϖ °±∓½⅓⅔¼¾×÷†‡• ⁰¹²³⁴⁵⁶⁷⁸⁹⁺⁻⁼⁽⁾₀₁₂₃₄₅₆₇₈₉₊₋₌₍₎

ℕℤℚℝℂ∂¬∀∃∅∆∇∈∉∋∌∏∑ ∗∘∙√∛∜∝∞∧∨∩∪∫≅≈≟≠≡≤≥⊂⊃⊆⊇ ⊕⊖⊗⊘⊙⌈⌉⌊⌋⌜⌝⌞⌟〈〉⟨⟩

ℕℤℚℝℂ∂¬∀∃∅∆∇∈∉∋∌∏∑ ∗∘∙√∛∜∝∞∧∨∩∪∫≅≈≟≠≡≤≥⊂⊃⊆⊇ ⊕⊖⊗⊘⊙⌈⌉⌊⌋⌜⌝⌞⌟〈〉⟨⟩

- mr_e_man
- Tetronian
**Posts:**524**Joined:**Tue Sep 18, 2018 4:10 am

mr_e_man wrote:[...]

However, we can see directly that a Hilbert subspace has countably many vertices. As you said, only a finite number of signs can be flipped. (If k signs are flipped then the difference vector has norm 2√k.) For each n, you can make a finite list of all vertices which differ from (1,1,1,...) only in their first n coordinates. (To ensure that the lists are disjoint, the n'th coordinate must be different, and the previous coordinates may be different.) Then you can put these lists in order, varying n, to make a countably infinite list.

Ahhh, nice! That makes sense.

So this means that the (surface of the) ∞-cube must have an uncountable number of finitely-disconnected pieces, each of which contains a countable number of vertices connected to each other via k-surtopes for all finite k. The facets of the ∞-cube themselves, of course, span multiple such pieces, but they are also inherently finitely-disconnected in themselves, because they're isomorphic to the entire ∞-cube itself (via a weird projection operation that maps a surtope to the entire polytope, that I've mentioned before).

This is just astounding... the facets of the ∞-cube have multiple, disjoint edge nets, face nets, ... k-surtope nets, and the entire ∞-cube is also like this. This has all sorts of weird implications, like the fact that if you start from a vertex and recursively walk the edges, you will never reach all the edges. Even if you iterate your walk transfinitely, because there are literally no edges between the pieces! (Or is there some kind of sleight-of-hand at limit ordinals where you "leap" from one set of edges to another? Either way, something very weird is going on.)

- quickfur
- Pentonian
**Posts:**2988**Joined:**Thu Sep 02, 2004 11:20 pm**Location:**The Great White North

Been thinking more about the disconnected edge outline of the ∞-cube. Basically, the ∞-cube's surface can be split into "pieces"; each piece consists of the union of all finite Cartesian products of the line segment. Each piece contains a countable number of vertices, of which we can choose one vertex v as the "representative" vertex of that piece, so we can talk about the piece P(v) corresponding to the vertex v. If two vertices v1 and v2 are a finite distance apart, then P(v1) = P(v2). The ∞-cube consists of an uncountable number of pieces, i.e., there exists a set V of vertices such that any two vertices are an infinite distance apart; then each v in V corresponds with a piece of the ∞-cube. Each such v corresponds with a corner of the ∞-cube that's infinitely far away from other pieces. The (finite part of the) face lattice of the ∞-cube is connected within a piece, but disconnected between pieces: you can never reach another piece via a finite walk from a vertex in a piece. There are literally no edges that traverse the pieces.

Here we see the very subtle but important distinction between union and limit: the limit of the Cartesian product of n copies of the line segment as n→∞ is the entire ∞-cube; however, the union of its finite initial segments is only one piece of the ∞-cube (in the above sense), not the entire ∞-cube. It's missing uncountably many other pieces.

What I haven't figured out yet, is how to characterise the set V. We know that the piece P(<0,0,0...>) contains all vertices with a finite number of 1's in their coordinates. Its antipodal piece, P(<1,1,1,...>), contains all vertices with a finite number of 0's in their coordinates. Besides these two, there are many others, like P(<1,0,1,0,1,0,...>), P(<1,1,0,1,1,0,1,1,0,...>), and so on. But note that P(<1,0,1,0,1,0,...>) = P(<1,1,1,0,1,0,1,0,...>). Any finite modification of v in a piece P(v) will also be contained in P(v). My question is, if we pick one representative vertex for each piece in the ∞-cube, is there an elegant description of these vertices? Obviously, the set of such representative vertices exists, by the axiom of chioce. But that doesn't give us a very useful description. Is there some kind of pattern to generate these representative points, or some kind of algorithm? I doubt the latter is possible, since you could have the vertex v* whose i'th coordinate is 0 if the corresponding Turing machine halts and 1 otherwise; then there can be no algorithm that would enumerate the coordinates of v*.

Originally I thought I could map the representative points to, say, the irrational numbers, but the set of irrationals also includes irrationals that differ from each other in a finite number of (binary) digits, so that doesn't work. Also, some rational numbers have repeating non-zero patterns, which should be included in the set. But not all rationals qualify, obviously, since there are rationals with only a finite number of either 1's or 0's, which all belong to P(<0,0,0...>) or P(<1,1,1...>). So it must be some other kind of set that includes rationals with non-terminating binary digit patterns, and irrationals that differ from each other in infinitely many positions (but excluding irrationals that differ only in finitely many places). Is there any simple description for such a set??

Here we see the very subtle but important distinction between union and limit: the limit of the Cartesian product of n copies of the line segment as n→∞ is the entire ∞-cube; however, the union of its finite initial segments is only one piece of the ∞-cube (in the above sense), not the entire ∞-cube. It's missing uncountably many other pieces.

What I haven't figured out yet, is how to characterise the set V. We know that the piece P(<0,0,0...>) contains all vertices with a finite number of 1's in their coordinates. Its antipodal piece, P(<1,1,1,...>), contains all vertices with a finite number of 0's in their coordinates. Besides these two, there are many others, like P(<1,0,1,0,1,0,...>), P(<1,1,0,1,1,0,1,1,0,...>), and so on. But note that P(<1,0,1,0,1,0,...>) = P(<1,1,1,0,1,0,1,0,...>). Any finite modification of v in a piece P(v) will also be contained in P(v). My question is, if we pick one representative vertex for each piece in the ∞-cube, is there an elegant description of these vertices? Obviously, the set of such representative vertices exists, by the axiom of chioce. But that doesn't give us a very useful description. Is there some kind of pattern to generate these representative points, or some kind of algorithm? I doubt the latter is possible, since you could have the vertex v* whose i'th coordinate is 0 if the corresponding Turing machine halts and 1 otherwise; then there can be no algorithm that would enumerate the coordinates of v*.

Originally I thought I could map the representative points to, say, the irrational numbers, but the set of irrationals also includes irrationals that differ from each other in a finite number of (binary) digits, so that doesn't work. Also, some rational numbers have repeating non-zero patterns, which should be included in the set. But not all rationals qualify, obviously, since there are rationals with only a finite number of either 1's or 0's, which all belong to P(<0,0,0...>) or P(<1,1,1...>). So it must be some other kind of set that includes rationals with non-terminating binary digit patterns, and irrationals that differ from each other in infinitely many positions (but excluding irrationals that differ only in finitely many places). Is there any simple description for such a set??

- quickfur
- Pentonian
**Posts:**2988**Joined:**Thu Sep 02, 2004 11:20 pm**Location:**The Great White North

Start with the binary expansion of pi. Get the first pair of digits. If it is 00 replace it with 01. Get the second pair, do it again, and so forth. With probability one this number will have infinitely many digits different from pi. Call it pi prime.

Do it again replacing all 00s with 10. With probability one this number will have infinitely many digits different both from pi and from pi prime. Replace all pairs of 00 with 11. Now we have four numbers all infinitely distant from one another with probability one.

Start over with pi except sequential triplets and replace 000 with 001. And so forth for all seven possibilities.

Start over with pi and replace 0000 with 0001. And so forth for all fifteen possibilities.

And so forth and so on ad infinitum. Do all this infinitely quickly. This might generate your set. Maybe there could be some duplicate numbers but I would expect that this set is of measure zero.

Do it again replacing all 00s with 10. With probability one this number will have infinitely many digits different both from pi and from pi prime. Replace all pairs of 00 with 11. Now we have four numbers all infinitely distant from one another with probability one.

Start over with pi except sequential triplets and replace 000 with 001. And so forth for all seven possibilities.

Start over with pi and replace 0000 with 0001. And so forth for all fifteen possibilities.

And so forth and so on ad infinitum. Do all this infinitely quickly. This might generate your set. Maybe there could be some duplicate numbers but I would expect that this set is of measure zero.

- PatrickPowers
- Tetronian
**Posts:**481**Joined:**Wed Dec 02, 2015 1:36 am

I guess the real question is, would that algorithm generate the complete set? I dunno and I'm not going to look into it.

- PatrickPowers
- Tetronian
**Posts:**481**Joined:**Wed Dec 02, 2015 1:36 am

PatrickPowers wrote:I guess the real question is, would that algorithm generate the complete set? I dunno and I'm not going to look into it.

Now that I think about it, there can't be any algorithm that generates all pieces, because the number of pieces is uncountable. If there were an algorithm that generates the entire list, that would imply the pieces can be put in bijection with the natural numbers, which means they are countable, contradicting their uncountability.

Any algorithm can at the most generate only an initial segment of the whole set of pieces. And correspondingly, there must be an uncountable number of limit points in the transfinitely-long list of pieces, which means that even if you enumerate the pieces in chunks of countably-many blocks at a time, the list of blocks would still be uncountably long and impossible to enumerate by any algorithm.

- quickfur
- Pentonian
**Posts:**2988**Joined:**Thu Sep 02, 2004 11:20 pm**Location:**The Great White North

quickfur wrote:PatrickPowers wrote:I guess the real question is, would that algorithm generate the complete set? I dunno and I'm not going to look into it.

Now that I think about it, there can't be any algorithm that generates all pieces, because the number of pieces is uncountable. If there were an algorithm that generates the entire list, that would imply the pieces can be put in bijection with the natural numbers, which means they are countable, contradicting their uncountability.

Any algorithm can at the most generate only an initial segment of the whole set of pieces. And correspondingly, there must be an uncountable number of limit points in the transfinitely-long list of pieces, which means that even if you enumerate the pieces in chunks of countably-many blocks at a time, the list of blocks would still be uncountably long and impossible to enumerate by any algorithm.

If you think of an algorithm as a little machine that runs in real time, then one can imagine infinitely many of them running at the same time generating a set. One can specify any aleph you like of machines running that algorithm. Either that or have one machine go infinitely fast in which case it can do just about anything.

We aren't concerned about computability (finite number of machines in real time) in a case like this.

- PatrickPowers
- Tetronian
**Posts:**481**Joined:**Wed Dec 02, 2015 1:36 am

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