adam ∞ wrote:[...]
If a notion of infinity is used which is an exact, definite, distinct number, one that doesn't change when you rearrange the elements
Suppose I take a point <1,2,3,4,...> from an ∞-dimensional space. I can, ostensibly, construct a lower-dimensional point by dropping one coordinate, obtaining the point <2,3,4,...>. This point ought to belong to (∞-1)-dimensional space, right? Well, not really. Suppose I translate this point by subtracting <1,1,1,1...> from it (<1,1,1,1...> is certainly in ∞-dimensional space, it's one of the vertices of the ∞-dimensional hypercube). What do I get? I get <1,2,3,4...> again
adam ∞ wrote:You are equivocating between different types of infinities.
1+1+1+1...
is not the same as
1+1+1...+1
[...]
Geometry gives us a clear view of the properties of infinity, set theory leads us astray. In set theory, the axiom of infinity states that every finite number has a finite successor, meaning 1 added to any finite number will remain in the closed group of "N", but that there is at least one infinite number that is defined as being "not finite", which is not a useful definition.
The amount of finite numbers in set theory is considered an infinite amount, which is a contradiction.
There is nothing new or different in set theory that happens, to indicate that adding one should eventually reach some special number that is different from the others, and the nature of any kind of transition between the finite and infinite, if there even is one in set theory, is not clearly defined and different sources give contradictory information on the matter.
A limit cardinal is added in axiomatically without evidence, with no explanation of its properties other than it being "not finite". An endless array of higher orders of infinity are represented by symbols, but with no defined properties other than which is larger than another, and an unfounded 2^n operation which cannot be used for much of anything since an exact value of infinity is never defined to begin with.
[...]
Actually, it is exactly the same, you just rearranged the terms a little.
You appear to be confused because of reading unreliable sources on the subject. There is no mathematics that defines an infinite number as being "not finite"
Of course, we will never reach infinity by repeating this process, because such a process will never terminate. Instead, we construct a new set in a somewhat different way: we take all finite numbers as stuff them all into a new set N. That is, we define this set to include every set we can possibly obtain by repeating the above process.
Such a set is guaranteed to contain every finite number there is, and is therefore larger than any finite number (since it contains all of them).
Since this is the smallest set with such a property (if you removed anything from it, there would be at least 1 finite number it does not contain, so we can no longer say this set is larger than that number), we can consider this set to be the first infinite number.
How is that a contradiction? Are you saying that the set of finite numbers is finite?
You cannot add one to some finite number to get an infinite number, no matter how many times you do it. That's obvious, because given any finite number X, adding 1 merely gives you another finite number. To say otherwise is the real contradiction, because if X is finite and X+1 is infinite, then it means that you can reduce infinity to 0 in a finite number of steps (namely, X+1). So it's not infinite after all.
Klitzing wrote:This is why the infinite dimensional Hilbert space once was introduced, consisting only of those vectors with finitely many non-zero components.
--- rk
adam ∞ wrote:Actually, it is exactly the same, you just rearranged the terms a little.
Well, there is a distinction between "1+1+1+1..." and "1+1+1...+1", they mean something different. It does not indicate a rearrangement in any way. It is used as a way to indicate two different types of series, one potentially infinite and one actually infinite.
"1+1+1+1..." = potential infinity
"we will never reach infinity by repeating this process, because such a process will never terminate"
This is the definition of potential infinity. It is synonymous with "such a process will never terminate". That is what potential infinity means. When a distinction between the two is made, the "..." without a 1 at the end is used to make the distinction.
"1+1+1...+1" = actual infinity. Such a process that does terminate. The totality of this process with a last successor. The "..." does not continue, it comes to an end. The previous 1's lead to the last 1 with no discontinuities.
[...]
Have you ever asked a set theorist to define infinity in a way that is not just a negation of being finite? This is what they tell you. This is what the symbols mean. They define "counting" using a bijection function, which is their new definition of "equals" and they say if a set does not equal a finite set, it is an infinite set, and that that is what an infinite set is defined as. As simply being a set that is not finite.
[...]
Why do you presume repeating the process will never reach infinity? Why do you assume the process will never terminate? Geometry informs us otherwise. And you have a major contradiction here: if the process never terminates, meaning it is only potentially infinite, there is no such thing as "all" of them to "stuff" into a new set "N". If the process does not terminate, there is no set of all of them. Logically, it is impossible to have a set that includes "every" set obtained in the process above. There is no such thing as "every" one of something that is not an exact, definite completed amount.
Such a set is guaranteed to contain every finite number there is, and is therefore larger than any finite number (since it contains all of them).
No, the definition is illogical and self-contradictory. There is no such thing as "every finite number" if you have defined finite numbers to be a potentially infinite, non terminating process without a last term! I don't mean to sound like a broken record, but I don't think you comprehend what is meant by "potential infinity" and "actual infinity".
[...]
A set of finite numbers cannot be infinite. That is self-contradictory. Any number in the set of finite number must be a finite number, and any highest finite number will be the total amount of finite numbers.
If you claim some infinite number is in the set of finite numbers, it is very clearly a contradiction. How can you say that if you removed the first infinite number from the set, there would be one finite number the set didn't contain?
How is that a contradiction? Are you saying that the set of finite numbers is finite?
Think about this for a minute.
Take any finite number.
The amount of finite numbers up to that point is exactly the same as that number.
The amount of numbers up to some number is the number itself.
The nth finite number is n.
If it is claimed that the nth finite number is infinite, this cannot be true-
it cannot be a finite number because infinite numbers are not finite.
If the amount of numbers up to some number is the number itself,
there cannot be an infinite amount of finite numbers, because infinity cannot equal the number itself if it is not finite.
If you believe there can be infinite numbers, there has to be some kind of transition between the finite and the infinite.
If you don't believe that adding 1 over and over again can ever reach infinity, how can you claim that there are an infinite amount of finite numbers in the "set" "N", relying on the axiom of infinity which uses the successor function of continually adding 1 to finite numbers to produce the set "N"?
You cannot add one to some finite number to get an infinite number, no matter how many times you do it. That's obvious, because given any finite number X, adding 1 merely gives you another finite number. To say otherwise is the real contradiction, because if X is finite and X+1 is infinite, then it means that you can reduce infinity to 0 in a finite number of steps (namely, X+1). So it's not infinite after all.
This is not what I am proposing, but a last finite number that, when you add 1 to it gives the first infinite number is much closer to what I think is actually happening than having no definition of the transition between the finite and the infinite, and claiming that it is logically possible to have an infinite amount of finite numbers.
PatrickPowers wrote:Klitzing wrote:This is why the infinite dimensional Hilbert space once was introduced, consisting only of those vectors with finitely many non-zero components.
--- rk
No, it's vectors with a well-defined magnitude. That is, vectors of finite length.
OK, so please tell me, what's the last term in the set of finite numbers? What's its value?
Klitzing wrote:This is why the infinite dimensional Hilbert space once was introduced, consisting only of those vectors with finitely many non-zero components.
Then no contradictions would occur and vector geometry works out well.
Eg. adding vectors, scalar multiplication, etc. And you don't even need the choice axiom.
Within this space an infinite orthoplex well can be defined, as its vertex coordinates (up to a global scalling) could be taken to be just those which are everywhere zero and at a single position +/-1.
Likewise thie infinite dimensional simplex can be defined, as a simplex always is a facet of the orthoplex, ie. its coordinates could be taken to be everywhere zero and at a single position +1.
But the infinite dimensional hypercube cannot be defined within that Hilbert space! As it would require coordinates which are non-zero throughout.
[...]
Klitzing wrote:PatrickPowers wrote:Klitzing wrote:This is why the infinite dimensional Hilbert space once was introduced, consisting only of those vectors with finitely many non-zero components.
--- rk
No, it's vectors with a well-defined magnitude. That is, vectors of finite length.
That comes out to be the same: a vector can have finitely many non-zero components only if it is required to have a finite length.
--- rk
PatrickPowers wrote:
Consider the infinite series 1, 1/2, 1/4, 1/8.... The sum converges to 2.
Klitzing wrote:This is why the infinite dimensional Hilbert space once was introduced, consisting only of those vectors with finitely many non-zero components.
Then no contradictions would occur and vector geometry works out well.
Eg. adding vectors, scalar multiplication, etc. And you don't even need the choice axiom.
Within this space an infinite orthoplex well can be defined, as its vertex coordinates (up to a global scalling) could be taken to be just those which are everywhere zero and at a single position +/-1.
Likewise thie infinite dimensional simplex can be defined, as a simplex always is a facet of the orthoplex, ie. its coordinates could be taken to be everywhere zero and at a single position +1.
But the infinite dimensional hypercube cannot be defined within that Hilbert space! As it would require coordinates which are non-zero throughout.
--- rk
PatrickPowers wrote:No, it's vectors with a well-defined magnitude. That is, vectors of finite length.
mr_e_man wrote:But what is a ridge, in infinite dimensions? It would have to be an (∞ - 2)-dimensional face.
<±1, 0, 0, 0, ...>
< 0, ±1, 0, 0, ...>
< 0, 0, ±1, 0, ...>
...
< 1, 0, 0, 0, ...>
< 0, 1, 0, 0, ...>
< 0, 0, 1, 0, ...>
...
<-1, 0, 0, 0, ...>
< 0, 1, 0, 0, ...>
< 0, 0, 1, 0, ...>
...
< 0, 1, 0, 0, ...>
< 0, 0, 1, 0, ...>
< 0, 0, 0, 1, ...>
...
< 1, 0, 0, 0, ...>
< 0, -1, 0, 0, ...>
< 0, 0, 1, 0, ...>
...
<-1, 0, 0, 0, ...>
< 0, -1, 0, 0, ...>
< 0, 0, 1, 0, ...>
...
< 0, 0, 1, 0, 0, ...>
< 0, 0, 0, 1, 0, ...>
< 0, 0, 0, 0, 1, ...>
...
< 1, 0, 0, 0, ...>
< 0, 1, 0, 0, ...>
< 0, 0, 1, 0, ...>
...
< 0, 1, 0, 0, ...>
< 0, 0, 1, 0, ...>
< 0, 0, 0, 1, ...>
...
< 0, 0, 1, 0, 0, ...>
< 0, 0, 0, 1, 0, ...>
< 0, 0, 0, 0, 1, ...>
...
< 0, 0, 0, 1, 0, 0, ...>
< 0, 0, 0, 0, 1, 0, ...>
< 0, 0, 0, 0, 0, 1, ...>
...
quickfur wrote:If we restrict coordinates to be non-negative, we obtain one of its facets, ostensibly an infinitely-dimensional simplex:
- Code: Select all
< 1, 0, 0, 0, ...>
< 0, 1, 0, 0, ...>
< 0, 0, 1, 0, ...>
...
mr_e_man wrote:Relatedly, given the cubic mirror vectors as a skewed basis for Hilbert space,
f₁ = (1, 0, 0, 0, 0, 0, ...),
f₂ = (-1/√2, 1/√2, 0, 0, 0, 0, ...),
f₃ = (0, -1/√2, 1/√2, 0, 0, 0, ...),
f₄ = (0, 0, -1/√2, 1/√2, 0, 0, ...),
:
there is no dual basis. That would require a vector g such that g•f₁ = 1 and g•f₂ = g•f₃ = g•f₄ = ... = 0, which could only be g = (1,1,1,1,...) which is not in the Hilbert space.
For the simplicial mirror vectors, there is a dual basis:
g₂ = (-√2, 0, 0, 0, 0, ...),
g₃ = (-√2, -√2, 0, 0, 0, ...),
g₄ = (-√2, -√2, -√2, 0, 0, ...),
g₅ = (-√2, -√2, -√2, -√2, 0, ...),
:
mr_e_man wrote:quickfur wrote:If we restrict coordinates to be non-negative, we obtain one of its facets, ostensibly an infinitely-dimensional simplex:
- Code: Select all
< 1, 0, 0, 0, ...>
< 0, 1, 0, 0, ...>
< 0, 0, 1, 0, ...>
...
That may be a problem. At least, it's counter-intuitive, that the centre of that simplex is 0, as the limit of (1/n, 1/n, ..., 1/n, 0, 0, ...) which has magnitude 1/√n.
Anyway, if the orthoplex is x3o3o3o3..., and the simplex is x3o3o3o3..., then it's obvious that they should be the same.
So there must be some essential difference between the Wythoff construction and your coordinate construction. Perhaps the Wythoff construction is incomplete, giving only vertices (and finite hulls of vertices), not facets. Perhaps the orthoplex could be constructed dually, by applying the group (cubic or simplicial?) to one facet to generate the other facets.
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