Living in infinite-dimensional space

[quickfur]

Starting a new thread here because my comments are starting to veer wayyy off-topic in the other thread.

This thread is dedicated to speculation about an actual infinite-dimensional space, in which there are an infinite number of macroscopic, orthogonal coordinate axes. I realize that such a thing may be controversial, because it implies the acceptance of completed infinites (otherwise the space would be ill-defined -- you can never finish constructing it!), but for the sake of speculation, we'll just take it as a given. What would the consequences of such a space be? What would life be like in such a space (making the HUGE assumption that life can possibly exist in such a space)?

Here are some simple consequences that I've come up with so far:

• The ∞-cube has an uncountable(!) number of vertices. This is a consequence of the fact that there's a bijection between its vertices and the real numbers between 0 and 1, which in standard set theory is an uncountable set.
• A creature that lives in this space either must have an infinite number of sensory organs, or be content to live with the crippling limitation that it can never be fully aware of its surroundings. Why? Because if the creature has eyes similar to ours, i.e., it has directed vision, then it would take an infinite amount of time for it to survey its surroundings -- since it needs to look at least in the direction of each of the coordinate axes, plus or minus a few, but there are an infinite number of axes. So the only way an ∞-dimensional creature can be fully aware of its surroundings at any given moment is that it has an infinite number of eyes (or equivalent sensory organ).
• The ∞-cube has a non-finite outradius. Implying that while it may take only one unit of time to walk from one end of the cube to another, it takes an infinite amount of time to cross from one corner of the cube to the opposite corner.
• The volume of the ∞-sphere is zero, or at most infinitesimal. This means that you can store an infinite number of marbles in your toy box.
• Actually, if your toy box is bigger than your toy, then you can almost certainly store an infinite number of toys in it.
• It takes an eternity to build a house, because you need to stack an infinite number of blocks together -- at least one per axis. In fact, it takes an uncountable amount of time to build such a house. That's infinitely more infinite than infinity.
• However, it only takes a countable amount of time to build a tent -- if the tent is in the shape of an ∞-cross pyramid. Unfortunately, such a tent would have infinitesimal volume, so you wouldn't be able to fit inside it without your limbs sticking out for predatory beasts at night to bite.
• If creatures in ∞-dimensional space lived on a planet (a ∞-sphere), they would be living in total darkness pretty much all of the time, except for maybe one point on the planet that's directly under the sun. This is because with an infinite number of lateral dimensions, any light from the sun would fade to nothing as soon as you leave that single point that receives direct sunlight; so practically 100% of the planet's surface (minus an infinitesimal number) would be in total darkness.
• In fact, light from any light source would almost instantaneously dissipate into nothing as soon as it leaves the source, because there are an infinite number of directions to disperse in, so any finite amount of light would immediately fade to 0 in all except countably few directions. So the only way for the sun to illuminate the planet is if the sun emitted a non-finitely-large amount of light. I.e., the sun is either completely invisible, or must be infinitely bright!

What can you guys come up with about living in an infinite-dimensional space?

[steelpillow]

Such an infinite-dimensional projective space is invoked by certain calculations in the theory of Calabi-Yau spaces, which underlie much string theory.
Projective spaces have no intrinsic infinity, because they just wrap around the farthest edge and come back from the other side. A bit like a spherical surface, but more subtle.

Re the vertices of the ∞-cube, I would draw your attention to the fact that they also have a one-to-one mapping with the natural numbers. This is the definition of a countable infinity.

Consider this list of vertices of a unit n-cube with j-coordinates (0 ≤ j ≤ n) of either 0 or 1 (where * indicates a recurring digit)
00000*
10000*
01000*
11000*
00100*
10100*
01100*
11100*
00010*
...

and so on. This is just a typical logician's (or digital engineer's) construction of a truth table condition by counting up from zero in binary, here extended indefinitely. The ordering in the list is strict, and the position of each binary number generates a one-to-one correspondence with the natural numbers. Indeed, it is no more nor less than a list of the natural numbers, in binary notation.

Any proof that this list is cannot be bijected against itself, and is therefore uncountable, would earn you a Fields medal. But I'm not waiting up.
Personally, I regard it as proof that Wittgenstein was right about Cantor.

[quickfur]

Your list is missing vertices like 101010..., 100100100..., and so on. That is, vertices with an unbounded number of 1's.

[Vector_Graphics]

Your list is missing vertices like 101010..., 100100100..., and so on. That is, vertices with an unbounded number of 1's.

Yeah. It's just Cantor's diagonal theorem from there.

[quickfur]

Cantor or not, the point is that the list is missing vertices that are obviously part of the ∞-cube. Since 0* is part of the cube, its antipode 1* obviously must also be a part of it. However, 1* is not in this list because every item in this list ends with 0*. The construction, as stated, only includes those vertices that eventually end with 0*. Thus, any vertex that doesn't end with 0* obviously isn't in the list. However, it's certainly a part of the ∞-cube. Therefore, the list is incomplete.

[steelpillow]

Your list is missing vertices like 101010..., 100100100..., and so on. That is, vertices with an unbounded number of 1's.

That is not relevant to countability, as I have tried to explain. What matters is setting up a formal 1-to-1 mapping. As Wikipedia says,

In mathematics, a set is countable if either it is finite or it can be made in one to one correspondence with the set of natural numbers. Equivalently, a set is countable if there exists an injective function from it into the natural numbers; this means that each element in the set may be associated to a unique natural number, or that the elements of the set can be counted one at a time, although the counting may never finish due to an infinite number of elements. (My bold)

That is exactly what I began listing for you. Your counting "never finishes due to an infinite number of elements", that does not prevent it from being "associated to a unique natural number." If you really do not get this, I'd suggest you follow up on Wikipedia's reference sources for a while. If you don't believe me, than maybe you'll believe them.

Or write that Fields-winning thesis.

[steelpillow]

Cantor or not, the point is that the list is missing vertices that are obviously part of the ∞-cube. Since 0* is part of the cube, its antipode 1* obviously must also be a part of it. However, 1* is not in this list because every item in this list ends with 0*. The construction, as stated, only includes those vertices that eventually end with 0*. Thus, any vertex that doesn't end with 0* obviously isn't in the list. However, it's certainly a part of the ∞-cube. Therefore, the list is incomplete.

Firstly, none of these numbers ends in a 0, none ends in a 1, they are infinite - they never end!
But never mind, your concern is is trivially fixable:
Construct a parallel list which is the binary complement of the first.
Now interlace the lists to create a single list:
00000*
11111*
10000*
01111*
01000*
10111*
11000*
00111*
00100*
11011*
...

The mapping to the natural numbers is less direct, but you can always count to the nth entry down.

[quickfur]

That does not fix the problem, it only hides it. Your list still does not have vertices of the form 101010... (where 10 repeats indefinitely), 100100100... (where 100 repeats indefinitely), 110110110... (where 110 repeats indefinitely), etc..

Basically, take any binary string of any length > 1, and have it repeat indefinitely. Clearly, being an infinite string of 0's and 1's, it belongs to one of the vertices of the ∞-cube. However, it is not in your list.

[quickfur]

Your list is missing vertices like 101010..., 100100100..., and so on. That is, vertices with an unbounded number of 1's.

That is not relevant to countability, as I have tried to explain. What matters is setting up a formal 1-to-1 mapping.

Yes, that is what matters. Unfortunately, so far you have failed at demonstrating this, because your list is not a 1-to-1 mapping, neither the original one you posted, nor the one in which you added the entries that have trailing 1's. Vertices like 101010..., 110110110110..., etc., are still not in your list, therefore, it is not a 1-to-1 mapping, and is not a valid proof of countability.

[steelpillow]

because your list is not a 1-to-1 mapping.

Of course it's a 1-to-1 mapping! I even set it up as a count of the natural numbers themselves, in order to make that blindingly obvious.
Or, are you under the naive impression that interleaving two such counts somehow breaks the bijection? Maybe you'd better study Wikipedia's article on bijection too.
Sorry, you are continuing to make such basic and gross errors that this discussion is helping nobody. Others may judge for themselves who is making sense here.

[quickfur]

You're missing my point. Your one to one mapping is certainly a 1-to-1 mapping to the natural numbers, but only from a strict subset of the vertices of the infinite dimensional cube. It fails to cover all the vertices of the infinite dimensional cube. Among the missing vertices are the ones I described above. You're assuming that your list eventually covers all vertices; unfortunately, it does not.

All you have managed to prove is that your list is countable. You have not proven that the vertices of the infinite dimensional cube is countable, because your list does not cover all of them.

[steelpillow]

You're missing my point. Your one to one mapping is certainly a 1-to-1 mapping to the natural numbers, but only from a strict subset of the vertices of the infinite dimensional cube. It fails to cover all the vertices of the infinite dimensional cube. Among the missing vertices are the ones I described above. You're assuming that your list eventually covers all vertices; unfortunately, it does not.

All you have managed to prove is that your list is countable. You have not proven that the vertices of the infinite dimensional cube is countable, because your list does not cover all of them.

What part of "the counting may never finish due to an infinite number of elements" do you not understand?

Offering a rigorous proof in a short post on a forum is not practicable. I am just trying to give you the gist of well-known results.

Let me try one last time. There is no such thing as the end of an infinite series, it is a meaningless concept by definition, and arguments based on it are vapid (Sure, we can define an ordinal which is bigger than any natural, but it is not the end of the series of ordinals, and the maths is different). Such fallacies of infinity are the very games that Cantor and his colleagues were grappling with - not always successfully at the time. There may or may not be a limit value to some function of the elements as their count tends to infinity, but that is all.

[mr_e_man]

Re the vertices of the ∞-cube, I would draw your attention to the fact that they also have a one-to-one mapping with the natural numbers. This is the definition of a countable infinity.

Consider this list of vertices of a unit n-cube with j-coordinates (0 ≤ j ≤ n) of either 0 or 1 (where * indicates a recurring digit)
00000*
10000*
01000*
11000*
00100*
10100*
01100*
11100*
00010*
...

because your list is not a 1-to-1 mapping.

Of course it's a 1-to-1 mapping! I even set it up as a count of the natural numbers themselves, in order to make that blindingly obvious.

According to your bijection, what natural number corresponds to the cube vertex (1, 0, 1, 0, 1, 0, 1, ...), where every even coordinate is 0 and every odd coordinate is 1?

You're missing my point. Your one to one mapping is certainly a 1-to-1 mapping to the natural numbers, but only from a strict subset of the vertices of the infinite dimensional cube. It fails to cover all the vertices of the infinite dimensional cube. Among the missing vertices are the ones I described above. You're assuming that your list eventually covers all vertices; unfortunately, it does not.

What part of "the counting may never finish due to an infinite number of elements" do you not understand?

Perhaps "your list eventually covers all vertices" was badly worded.

The counting up to any particular vertex must eventually finish. Otherwise that vertex is not in the list.

[mr_e_man]

But never mind, your concern is is trivially fixable:
Construct a parallel list which is the binary complement of the first.
Now interlace the lists to create a single list:
00000*
11111*
10000*
01111*
01000*
10111*
11000*
00111*
00100*
11011*
...

The mapping to the natural numbers is less direct, but you can always count to the nth entry down.

Sure, if you miss one vertex, you can always make a new list by inserting it anywhere. And if you miss a countably infinite list of vertices, you can interlace the two lists.
But no matter how you construct your list, there will always be more vertices missing.

[steelpillow]

According to your bijection, what natural number corresponds to the cube vertex (1, 0, 1, 0, 1, 0, 1, ...), where every even coordinate is 0 and every odd coordinate is 1?

Every recurring binary converges to a rational number.

Your 10* converges to 2/3.

(For a bit more on such series, see for example this StackOverflow Q&A )

The rationals are countable. So this one corresponds to the natural number denoting its position in your count of rationals.

Let me repeat that my scheme using the naturals is illustrative and not a rigorous proof. To count recurring binaries it must be extended to the rationals, just as I extended it to count strings ending in 1*. (Note that I needn't have bothered with that extension, since 01* converges to 1, which may be written as 10*, and so all infinite binaries ending in 1* converge to finite binaries which may be found in my original list. On the other hand, this is certainly not true of ∞-cube vertex coordinates. This offers a counter-example to the original assumption that every vertex of an ∞-cube maps to a unique binary string.

Well, either that, or you guys can now begin arguing that the set of ∞-cube vertices is larger than the set of real numbers, because the mapping is many-to-one.....

[mr_e_man]

Yes, I know about geometric series.

But we don't need to bring rational numbers or real numbers into this discussion at all. They are helpful for comparison, but they also complicate things, as you noticed: A real number may have two different sequences of digits representing it. A cube vertex, on the other hand, is nothing more or less than a sequence of '0's and '1's, by definition.


Well then, what natural number corresponds to the cube vertex (1, 0, 1, 0, 0, 1, 0, 0, 0, 1, 0, 0, 0, 0, 1, ...), where the nth '1' is followed by n '0's before the next '1'?

[quickfur]

[...]
Let me try one last time. There is no such thing as the end of an infinite series, it is a meaningless concept by definition, and arguments based on it are vapid (Sure, we can define an ordinal which is bigger than any natural, but it is not the end of the series of ordinals, and the maths is different). Such fallacies of infinity are the very games that Cantor and his colleagues were grappling with - not always successfully at the time. There may or may not be a limit value to some function of the elements as their count tends to infinity, but that is all.

Forget about the "end of an infinite series", whatever that means. I don't care what "games" Cantor has been up to, our dispute here is very simple.

(1) My definition of the vertices of the ∞-dimensional cube is the set of all non-terminating binary strings.

(2) My claim is that this set is uncountable. You disagree with this, and say that it's countable. To prove that it's countable, it is sufficient to demonstrate a 1-to-1 mapping of the members of this set, that is, the set of non-terminating binary strings, to the natural numbers.

(3) You claimed that your original list (0*, 10*, 010*, 110*, ...) is one such mapping. However, this is false, because every member in your list has trailing consecutive 0's, but the vertex 01010101... does not have trailing consecutive 0's. Therefore, it cannot be a member of your list, and therefore your list does not demonstrate a 1-to-1 mapping of vertices to the natural numbers.

(4) Later you claimed that your amended list (0*, 1*, 10*, 11*, ...) is a 1-to-1 mapping. However, this is also false, because the vertex 010101... isn't on this list either. All members on this list have either trailing consecutive 1's, or trailing consecutive 0's, but 010101... does not fall in either pattern. Therefore, this amended list does not demonstrate a 1-to-1 mapping to the natural numbers either.

(5) Then later still you claimed that the vertex 010101... "converges to a rational number", and since the rational numbers are countable, the set of vertices must be countable too. However, this mapping also isn't a 1-to-1 mapping. For example, the vertex 101001000100001... where the number of 0's between each occurrence of 1 increases by 1 each time. Since this is a non-terminating binary string, it is by definition one of the vertices of the ∞-cube. However, this string does not correspond with any rational number. Therefore, your claimed mapping of vertices to the rationals is not a 1-to-1 mapping, and therefore you have still not proven the countability of the set of vertices.

I don't care for what Cantor did or did not do; what I ask is very simple. Show me a 1-to-1 mapping from the set of non-terminating binary strings (which is the same as the set of ∞-cube vertices) to the natural numbers, and I will immediately concede that you are right that they are countable.

And just to be clear, a 1-to-1 mapping means a 1-to-1 mapping: every non-terminating binary string I can think of must be provably a part of your mapping. If I am able to come up with a binary string that is provably not in your mapping, then that can only mean one thing: your mapping is not 1-to-1, and therefore does not prove countability. If indeed the set of vertices is countable, then there must exist some 1-to-1 mapping to the natural numbers, by definition. Show me just one such mapping, one mapping is all I ask, and I will immediately agree that the vertices are countable. Please don't waste my time with hand-waving incomplete mappings that obviously do not include binary strings that I can come up with in 5 minutes. Let's settle this question for real. You show me a 1-to-1 mapping of ∞-cube vertices to the natural numbers, and I will immediately agree they are countable. It's just that simple.

[steelpillow]

Well then, what natural number corresponds to the cube vertex (1, 0, 1, 0, 0, 1, 0, 0, 0, 1, 0, 0, 0, 0, 1, ...), where the nth '1' is followed by n '0's before the next '1'?

I do not know if it is possible to say but, as I asked quickfur, what part of "the counting may never finish due to an infinite number of elements" are you having trouble with?

[mr_e_man]

As I said, the counting up to any particular vertex must eventually finish. Otherwise that vertex is not in the list.


Here's a concrete example of counting the rational numbers:

1: 0/1
2: +1/1
3: -1/1
4: +1/2
5: -1/2
6: +2/1
7: -2/1
8: +1/3
9: -1/3
10: +3/1
11: -3/1
12: +2/3
13: -2/3
14: +3/2
15: -3/2
16: +1/4
17: -1/4
18: +4/1
19: -4/1
20: +3/4
21: -3/4
22: +4/3
23: -4/3
24: +1/5
25: -1/5
26: +5/1
27: -5/1
28: +2/5
29: -2/5
30: +5/2
31: -5/2
32: +3/5
33: -3/5
34: +5/3
35: -5/3
36: +4/5
37: -4/5
38: +5/4
39: -5/4
40: +1/6
41: -1/6
42: +6/1
43: -6/1
44: +5/6
45: -5/6
46: +6/5
47: -6/5
48: +1/7
...

Continue the list with +x/y, -x/y, +y/x, -y/x, where x increases from 1 to y-1 and has no factors in common with y, and y increases by 1.
I give this complete description, so that I cannot change my list, as you have changed yours.

If you give me any rational number, I could continue this list and find that rational number associated with a natural number. The counting up to that number eventually finishes.

[mr_e_man]

Well then, what natural number corresponds to the cube vertex (1, 0, 1, 0, 0, 1, 0, 0, 0, 1, 0, 0, 0, 0, 1, ...), where the nth '1' is followed by n '0's before the next '1'?

I do not know if it is possible to say but, as I asked quickfur, what part of "the counting may never finish due to an infinite number of elements" are you having trouble with?

Really that seems like a non sequitur. I don't understand why you're saying it. (But I've made an attempt, in writing the previous post.)

[steelpillow]

Well then, what natural number corresponds to the cube vertex (1, 0, 1, 0, 0, 1, 0, 0, 0, 1, 0, 0, 0, 0, 1, ...), where the nth '1' is followed by n '0's before the next '1'?

I do not know if it is possible to say but, as I asked quickfur, what part of "the counting may never finish due to an infinite number of elements" are you having trouble with?

Really that seems like a non sequitur. I don't understand why you're saying it. (But I've made an attempt, in writing the previous post.)

It exposes the fallacy of arguing that a number is not in some countable list just because you cannot count to infinity. You can pull Cantor's diagonal manoeuvre with a list of rationals; you can always define one that you have not listed yet. But it does not make the rationals uncountable. For the reals he reached the right conclusion, but (on that occasion) using a flawed argument. But for the rationals and the ∞-cube, it is the wrong answer.

Tip: if you are working with cardinal infinities, do not ascribe properties of the ordinals to them. If you find such properties in there, you have gone astray somewhere. Which thought might affect how we get on in our space: an infinite-cardinality space does not preclude further transfinite ordinal dimensions. How many infiinities of infinite dimensions might we encounter?

[mr_e_man]

I never said anything about counting to infinity.

I never claimed that the counting of all the cube's vertices (or all reals, or all rationals) must eventually finish.

You can pull Cantor's diagonal manoeuvre with a list of rationals; you can always define one that you have not listed yet.

That is not Cantor's diagonal manoeuvre!

It doesn't matter how far you've gone in the infinite list. What matters is that you can go farther (finitely) and eventually find that one rational number.

Cantor's diagonal manoeuvre is defining something that will not be in any finite part of the list, however far you go.


Countability of the rationals is the following statement. (Note that an infinite list is just a mapping (i.e. a function) from the set ℕ of natural numbers.) Pay attention to the order of quantifiers: ∃∀∃.

There exists a function L : ℕ -> ℚ such that, for any rational number x ∈ ℚ, there exists a (finite) natural number n ∈ ℕ, such that L(n) = x.

Here's an equivalent statement:

There exists an infinite list L such that, for any rational number x, there exists a finite sub-list F of L, such that F contains x.

Your version of "Cantor's diagonal manoeuvre" only disproves the following statement, which is not equivalent, as the order of quantifiers is ∃∃∀.

There exists an infinite list L, and there exists a finite sub-list F of L, such that, for any rational number x, F contains x.

The real "Cantor's diagonal manoeuvre" disproves the following statement:

There exists an infinite list L such that, for any real number x, there exists a finite sub-list F of L, such that F contains x.

equivalently proving the inverted statement:

For any infinite list L, there exists a real number x such that, for any finite sub-list F of L, F does not contain x.

Calculus students often have difficulty with the ε,δ definition of continuity of a function. It has three quantifiers. The logic is complex.
It's understandable if someone has difficulty with the definition of countability of a set, as it also has three quantifiers.

[PatrickPowers]

A creature that lives in this space either must have an infinite number of sensory organs, or be content to live with the crippling limitation that it can never be fully aware of its surroundings. Why? Because if the creature has eyes similar to ours, i.e., it has directed vision, then it would take an infinite amount of time for it to survey its surroundings -- since it needs to look at least in the direction of each of the coordinate axes, plus or minus a few, but there are an infinite number of axes. So the only way an ∞-dimensional creature can be fully aware of its surroundings at any given moment is that it has an infinite number of eyes (or equivalent sensory organ).

The unit sphere is a subset of what's called a Hilbert space, a space of vectors with infinite dimension but finite length.

If the creature has one eye that can see a hemisphere and the other for the other hemisphere then it can see it's whole space. Would an uncountably infinite brain be enough to register all that data?

If gravity has a 1/N^inf law, then there is no gravity except when things collide.

[quickfur]

[...]
The unit sphere is a subset of what's called a Hilbert space, a space of vectors with infinite dimension but finite length.

The unit ∞-sphere is a curious object. In finite dimensions, the intersection of the n-sphere with the line v = t*<1,1,1,1...> is a point whose distance from the origin is equal to the radius of the sphere. However, the ∞-sphere either does not intersect with v = t*<1,1,1,1...> at all, or the intersection is a point with infinitesimal coordinates(!). Hence, you may say that the ∞-sphere is "pinched"; in spite of being convex, it behaves as though it were a non-convex, star-like object with concave "dimples" in certain directions.

If the creature has one eye that can see a hemisphere and the other for the other hemisphere then it can see it's whole space. Would an uncountably infinite brain be enough to register all that data?

It's rare for a creature's vision to span 180° of the field of view in the direction it's looking at. But not impossible. So let's say it does. But even then, this doesn't quite give you the panoramic view of your surroundings as it might first appear. The reason is that there are an infinite number of perpendicular directions to the direction the creature is looking at. Meaning that almost all of the surroundings would only fall on its peripheral vision; what it sees directly ahead of itself is only an insignificant subset of its entire surroundings.

It can try to turn its head from that position to try to catch more of the scene that falls on its peripheral vision, but it can only do this for a small number of axes. Almost all of the rest of the axes continue to fall only on its peripheral vision, so unless it has exceptionally high-resolution peripheral vision, this wouldn't help increase its field of view very much at all!

So it seems that we can't get away from needing infinite time to look around at everything, or having an infinite number of eyes to observe all directions at once (or at least a significant subset of the infinite number of axes).

If gravity has a 1/N^inf law, then there is no gravity except when things collide.

That's an interesting observation. So that means things will either be completely oblivious to each other, or they collide and then stick together forever or bounce off and immediately forget the other exists at all.

Hmm, does that mean that matter in the ∞-dimensional universe will essentially behave like an ideal gas with brownian motion? There'd be no non-trivial gravitational wells to speak of, and objects would basically just fly inertially except when they collide.

Although a 1/N^∞ gravity law may still produce interesting effects if there's an infinite number of objects nearby, which, given how many directions there are, wouldn't be surprising. I mean, the ∞-cube has an uncountable number of vertices, so there has to be at least that many particles that constitute the cube. I.e., an uncountable number of sub-blocks. If you cut the ∞-cube into its constituent blocks, you'd get an uncountable number of objects, the total gravity of which would involve an infinite sum of 1/N^∞ contributions. But how to analyze this situation mathematically without immediately running into intractible problems is beyond my ability to figure out.

(Incidentally, this means that objects of non-zero volume like the ∞-cube must have infinitely strong gravitational force, since it has at least an uncountable number of constituent parts, each of which exerts a bit of gravitational influence. This force may be enough to overcome the vanishing value of a 1/N^∞ gravity law...? This is really going out on a limb here though, because the maths break down with infinite quantities everywhere and no consistent way to deal with them.)

[PatrickPowers]

[...]
The unit sphere is a subset of what's called a Hilbert space, a space of vectors with infinite dimension but finite length.

The unit ∞-sphere is a curious object. In finite dimensions, the intersection of the n-sphere with the line v = t*<1,1,1,1...> is a point whose distance from the origin is equal to the radius of the sphere. However, the ∞-sphere either does not intersect with v = t*<1,1,1,1...> at all, or the intersection is a point with infinitesimal coordinates(!). Hence, you may say that the ∞-sphere is "pinched"; in spite of being convex, it behaves as though it were a non-convex, star-like object with concave "dimples" in certain directions.

I see this frequently but think it is correct to see the sphere as smooth and the unit cube as the spiky star-like object.

If the creature has one eye that can see a hemisphere and the other for the other hemisphere then it can see it's whole space. Would an uncountably infinite brain be enough to register all that data?

It can try to turn its head from that position to try to catch more of the scene that falls on its peripheral vision, but it can only do this for a small number of axes.

If it has infinitely many eye muscles then that eye can rotate in infinitely many planes at once. If the rotational speed is the same for each plane then each point in the eye is moving infinitely fast. In for a penny, in for a pound.

If gravity has a 1/N^inf law, then there is no gravity except when things collide.

That's an interesting observation. So that means things will either be completely oblivious to each other, or they collide and then stick together forever or bounce off and immediately forget the other exists at all.

Hmm, does that mean that matter in the ∞-dimensional universe will essentially behave like an ideal gas with brownian motion? There'd be no non-trivial gravitational wells to speak of, and objects would basically just fly inertially except when they collide.

Although a 1/N^∞ gravity law may still produce interesting effects if there's an infinite number of objects nearby, which, given how many directions there are, wouldn't be surprising. I mean, the ∞-cube has an uncountable number of vertices, so there has to be at least that many particles that constitute the cube. I.e., an uncountable number of sub-blocks. If you cut the ∞-cube into its constituent blocks, you'd get an uncountable number of objects, the total gravity of which would involve an infinite sum of 1/N^∞ contributions. But how to analyze this situation mathematically without immediately running into intractible problems is beyond my ability to figure out.

(Incidentally, this means that objects of non-zero volume like the ∞-cube must have infinitely strong gravitational force, since it has at least an uncountable number of constituent parts, each of which exerts a bit of gravitational influence. This force may be enough to overcome the vanishing value of a 1/N^∞ gravity law...? This is really going out on a limb here though, because the maths break down with infinite quantities everywhere and no consistent way to deal with them.)

There's a similar situation with our ostensibly infinite Universe. There's an infinite amount of matter and an infinite amount of space so a density cannot be calculated, only measured.

Quantum particles couldn't have a symmetrical wave function because the probabilities would sum to infinity instead of one.

[steelpillow]

Cantor's diagonal manoeuvre is defining something that will not be in any finite part of the list, however far you go..

Cantor defines a list of card ∞ sets and then posits a card (∞+1) set beyond the list, i.e. an ordinal. His logic is, to say the least, historically controversial.

Attempting to prove either case here is fruitless. Like I say, I am not being rigorous but illustrative. Let us agree to differ.

[quickfur]

The unit ∞-sphere is a curious object. In finite dimensions, the intersection of the n-sphere with the line v = t*<1,1,1,1...> is a point whose distance from the origin is equal to the radius of the sphere. However, the ∞-sphere either does not intersect with v = t*<1,1,1,1...> at all, or the intersection is a point with infinitesimal coordinates(!). Hence, you may say that the ∞-sphere is "pinched"; in spite of being convex, it behaves as though it were a non-convex, star-like object with concave "dimples" in certain directions.

I see this frequently but think it is correct to see the sphere as smooth and the unit cube as the spiky star-like object.

The ∞-sphere is more "pinched" than spiky, though. But it's still weird in the sense that there are lines through the origin that it does not intersect with. (Or intersects with only in a point with infinitesimal coordinates.)

[...]
It can try to turn its head from that position to try to catch more of the scene that falls on its peripheral vision, but it can only do this for a small number of axes.

If it has infinitely many eye muscles then that eye can rotate in infinitely many planes at once. If the rotational speed is the same for each plane then each point in the eye is moving infinitely fast. In for a penny, in for a pound.

Well yes, this is why I eventually came to the conclusion that in order for a "full-bodied" infinite-dimensional space to make sense (full-bodied meaning having objects of non-zero volume -- objects in Hilbert space are all zero-volumed), the underlying number system must include infinitesimals and non-finite numbers. Once you admit infinitesimals and infinities into your number system, weirdness such as the out-radius of the ∞-cube being an infinite quantity ceases to become a problem. The intersection of the ∞-sphere with the line p = t*<1,1,1,...> is simply a point with infinitesimal coordinates; the ∞-sphere ceases being weird or having "holes". A creature having infinitely many eye muscles becomes just a part of the fact of life rather than a strange phenomenon, similarly with the creature turning its head infinitely fast.

SImilarly, problems with arithmetic involving infinite quantities, at least in theory, become solvable, though I've yet to work out how exactly this all works. Surreal arithmetic comes to mind but there are some fundamental problems, such as the surreals inherently having gaps (and their completion without gaps are no longer a field, which ruins the possibility of using them for general arithmetic in any sane way).

[...]
(Incidentally, this means that objects of non-zero volume like the ∞-cube must have infinitely strong gravitational force, since it has at least an uncountable number of constituent parts, each of which exerts a bit of gravitational influence. This force may be enough to overcome the vanishing value of a 1/N^∞ gravity law...? This is really going out on a limb here though, because the maths break down with infinite quantities everywhere and no consistent way to deal with them.)

There's a similar situation with our ostensibly infinite Universe. There's an infinite amount of matter and an infinite amount of space so a density cannot be calculated, only measured.

What proof do we have that our universe is actually infinite? As opposed to merely unbounded.

[PatrickPowers]

What proof do we have that our universe is actually infinite? As opposed to merely unbounded.

Measurements of the curvature of our Universe are consistent both with an infinite Universe and with a very large but finite Universe. If the curvature is zero to make an infinite "flat" Universe then measurements will get closer and closer to that but it seems it would be impossible to prove via uncertain measurements that it is actually zero. If our Universe is spherical/finite or hyperbolic/infinite then there might some day be a firm conclusion.

[steelpillow]

What proof do we have that our universe is actually infinite? As opposed to merely unbounded.

None. It is a popular assertion by physicists who believe their own simplifying assumptions.

Since the moment of the Big Bang, the universe has expanded at a finite rate for a finite period of time. It can only have reached infinite size if it was infinite in size to start with.

Some theorists suggest that it indeed was, perhaps the latest in an eternal (temporally infinite) sequence of bang-bust cycles (such as the conformal cyclic cosmology of Roger Penrose).
Other theorists retreat hotly (and I am always surprised at their fervour) into defensive wriggles such as "infinite - indefinitely large - what's the difference, it doesn't matter!" or "The multiverse is, even if our little bubble universe isn't - of course that's what is meant. What does it matter!"
People commonly also get confused between infinite space and infinite spacetime. If either space or time is infinite, then spacetime must be too. But it need not be infinite in every dimension. This is where "the geometry of spacetime" is an unsolved problem. General Relativity tells us how to describe it, but cannot tell us what shape to describe. For example if space expands forever and time does not end, then the geometry of spacetime is hyperbolic - and that is so whether space is finite or infinite. But maybe time will end one day, whether in a big crunch or a big rip.
Then again, according to typical string theories, spacetime has ten dimensions, of which six are "rolled up or compactified" smaller than the smallest fundamental particle. It seems a touch grandiose to suggest that the other four are open-ended. Also, an infinite universe is not consistent with Hawking's "imaginary time" model of a finite, closed spacetime manifold such that "the boundary condition of the Universe is that it has no boundary".
I have another gripe: if space was infinite in the first place, then what can it possibly mean to say that space is expanding? Infinity cannot expand. You can expand a metric within an infinite continuum (as Penrose does), but that is not at all the same thing.

So to me, the whole idea of an infinite universe is based on an illogical fantasy. Indefinitely large, i.e. larger than anything we can study scientifically, is quite possible; it is certainly appears that way for us today.

[quickfur]

We're veering far off topic here. To return to the original topic of speculating on life in an infinite-dimensional universe:

The ∞-cube has the strange property that to walk from one end of it to another, i.e., alongside an edge, it takes only a unit of time. However, to walk from one corner to the opposite corner (two antipodal vertices, or if we presuppose gravity along one dimension, two antipodes of a single facet) takes an infinite amount of time.

Well, actually, we see this phenomenon appear already in finite dimensions: suppose in N dimensional space (for some large but finite N) there's a house in the shape of an N-cube. WLOG assume that the front door lies in the -X direction and the back door in the +X direction, and that the cube is 1 unit long in each dimension. Then you can walk from the front door to the back door in a single unit of time. However, say you start at <0,0,0,...>, which is a corner of the house, and want to walk to the the opposite corner diagonally, i.e., to <0,1,1,1,...>. Then it would take N-1 units of time. The larger N is, the bigger this number becomes. In the limiting case where N diverges to infinity, you will never get to the other side in finite time!

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.