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.
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.
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.
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.
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.
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
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?
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.
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.
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?)
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 ℵ₀k for all finite k. Therefore card(h) = ℵ₀ω = ℵ₀. I.e., each Hilbert subspace contains a countable number of vertices.
[...]
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.
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.
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.
function f(v):
- For each modication m of v:
- If m is canonical, return m.
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