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This thread is just a small step in the debate for our 1-A standards. It's a rather specific case, but IMO could hold relevance to which tier being superior to all finite dimensions is, which in turn is relevant for "above all dimensions"-type statements.
The topic is, as the title says, which tier destroying all finite-dimensional spaces at once should get.
Specifically:
Let R be the real number line. That has one coordinate (x).
R^2 shall be the real number plane, with R being embedded in it. (e.g. we identify R as being all points in R^2 that can be written as (x,0) for some x) That space has two coordinates (x,y)
R^3 shall be three-dimensional real number space, with R^2 being embedded in it. (e.g. via R^2 being all points in R^3 that can be written as (x,y,0) for some x and y) That space has three coordinates (x, y, z)
R^4, R^5, R^6... are all defined in the same pattern and embedded in each other. Each space has one coordinate more.
Now, assuming a character destroys the space R^n, for every natural number n simultaneously, which tier would that be?
To be clear, that means R is destroyed simultaneously to R^2 and R^3 and R^4... and all those are parts of each other as specified above.
To decide which tier this equates to I think the only way is to look at which space we wish to assume is destroyed here in total.
I call it the most obvious because it is the straight union of all the above spaces.
With that I mean: Every point that is in this space is also in at least one R^n space. Meanwhile, every point in some R^n space is also in this one.
Specifically, let's say we have the point (3,4,-1) in R^3. That's also embedded in R^∞, where it has the coordinates (3,4,-1,0,0,0,0,...). I.e. if you have a point in some R^n space, you just put infinite 0s at the end and now you have the coordinates of that point in R^∞.
Meanwhile, R^∞ is the space of infinite sequences where all but finitely many coordinates are 0. Meaning there is a last coordinate that is not 0. By deleting all 0s after that coordinate you get the coordinates of a point that is in some R^n.
E.g. (3,2,0,0,0,0,0,0,9,10,3,0,0,0,0,0,....) would be equivalent to the point (3,2,0,0,0,0,0,0,9,10,3) in R^11.
In other words, we can write the points of the R^n space using the coordinates of the space of finite sequences and then the space of finite sequences is the union of all R^n, which sounds pretty much like what we want.
Now, this space has countably infinite dimensions. Meaning that destroying it would be High 1-B.
That sounds pretty good, so what would go against looking at things like this? Well R^∞ is a little weird.
Let's talk about dimensional axes.
The space R is a 1-dimensional space, so its dimensional axis is itself.
The space R^2 is spanned by two axes, as it is two-dimensional. One can be the line described by all points (x,0) (that is what R is when we embed it into R^2) and the other is all points (0,y), which is orthogonal to the first one.
The space R^3 is three-dimensional and hence spanned by 3 axes. The first can be (x,0,0), the second (0,y,0) and the third (0,0,z). Notice how (x,0,0) and (0,y,0) are basically the dimensional axes of R^2 but embedded in R^3 and (x,0,0) in particular is the dimensional axis of R embedded in R^3.
And so on. Each space we go higher we keep the same dimensional axes we have and add one additional one.
Going by this process we would expect that R^∞ has the dimensional axis (x,0,0,0,0....), (0,y,0,0,0...), (0,0,z,0,0....) and so on. And in a sense that is the case. That's because of some mathematical detail where each coordinate has to be able to be written as a finite combination of the coordinates axis, meaning one can only go some distance across a finite selection of axis.
However, if you would like to move some distance along every one axis (like we would imagine it physically possible if you can move in every direction), then that point is not in R^∞ anymore.
That can be seen by the fact that using those axis we could construct points not in the space.
If we go a distance of 1 along the first axis, then a distance of 1 along the second, then 1 along the third and so on for each axis, we get to the point (1,1,1,1,1,1.....). However, that sequence never ends and hence is not part of the space of finite sequences.
So while we could go any distance along any one of these axis in finite dimensions, in infinite dimensions we can't do so. That is in the sense that we can still go an infinite distance along any one of them, but we can't go a little into each direction at once.
This might get even more confusing if one considers distance. Let's say we go a distance of 1 in the direction of the first axis, a distance of 1/2 in the direction of the second one, (1/2)^2 for the third, (1/2)^3 for the fourth, (1/2)^4 for the fifth and so on. Then the total distance walked would be 1 + 1/2 + (1/2)^2 + (1/2)^3 + (1/2)^4 + ... which is a geometric series and hence in total equal to a distance of 2. However, the point (1, 1/2, (1/2)^2, (1/2)^3, ...) is not part of R^∞. However, each step of the way was. That is to say, in a sense, one can walk straight out of the space by just moving a finite distance.
In mathematics, one would say the space is not complete. That is not an issue with the choice of space btw. Every countably dimensional vector space, with a concept of distance and for example a concept of angles, is incomplete. (or, scientifically more precise, every Banach space is not countably dimensional)
So yeah, the space is a little weird and perhaps not what we would a physical space to be like.
Now, this space isn't just the union of all the R^n spaces. We see that by it containing the point (1,1,1,1,1,1,1...) which can not be contained in any R^n as it goes a distance of 1 along each of infinite dimensional axis.
However, in exchange, in this space you can go any distance along any of the dimensional axis that were in R^n at once and not leave the space.
One can also define a metric, i.e. distance, that is complete on it or also some norm. The norm would be weird in that it is discontinuous, i.e. points that are close to each other can have vastly different distances from 0, though.
l^2 would have nicer distance definitions and be complete, but in return would neither be the R^n union nor include all points one can get by going along every one of the dimensional axis.
Furthermore, notice that while (x,0,0,0,0....), (0,y,0,0,0...), (0,0,z,0,0....) and so on can be included in R^N as axis in the space and moving any distance along each of them keeps you inside the space, these axes would not be dimensional axes of R^N. As mentioned earlier, mathematically any point has to be reachable by only going some distance along a finite selection of the dimensional axis of the space. Meaning, R^N needs other ones to reach a point like (1,1,1,1,1...) and, in fact, more of them.
As it so happens, R^N would be more than countably infinite dimensional. I.e. destroying this space would be Low 1-A. One critical level higher than the prior option.
I was split on the issue myself and frequently went back and forth between positions.
I would say my current stance is that we should rank the feat as High 1-B. Why? Well, ultimately the consideration on which space to take is just to decide the tiering. That the sum makes sense as a physical space is not so important. It's more a theoretical construct here, after all, as the spaces mentioned in the feat are just finite-dimensional spaces of any dimension.
To that comes that R^N doesn't have nice definitions of distance and angles and hence is not exactly what we imagine as a proper physical space either.
So overall High 1-B is more appropriate in my opinion.
What do you think, everyone?
The topic is, as the title says, which tier destroying all finite-dimensional spaces at once should get.
Specifically:
Let R be the real number line. That has one coordinate (x).
R^2 shall be the real number plane, with R being embedded in it. (e.g. we identify R as being all points in R^2 that can be written as (x,0) for some x) That space has two coordinates (x,y)
R^3 shall be three-dimensional real number space, with R^2 being embedded in it. (e.g. via R^2 being all points in R^3 that can be written as (x,y,0) for some x and y) That space has three coordinates (x, y, z)
R^4, R^5, R^6... are all defined in the same pattern and embedded in each other. Each space has one coordinate more.
Now, assuming a character destroys the space R^n, for every natural number n simultaneously, which tier would that be?
To be clear, that means R is destroyed simultaneously to R^2 and R^3 and R^4... and all those are parts of each other as specified above.
To decide which tier this equates to I think the only way is to look at which space we wish to assume is destroyed here in total.
The Space of Finite Sequences
The most obvious candidate is the space of all finite sequences... in a sense. (See here for wikipedia's description of it) We will call it R^∞.I call it the most obvious because it is the straight union of all the above spaces.
With that I mean: Every point that is in this space is also in at least one R^n space. Meanwhile, every point in some R^n space is also in this one.
Specifically, let's say we have the point (3,4,-1) in R^3. That's also embedded in R^∞, where it has the coordinates (3,4,-1,0,0,0,0,...). I.e. if you have a point in some R^n space, you just put infinite 0s at the end and now you have the coordinates of that point in R^∞.
Meanwhile, R^∞ is the space of infinite sequences where all but finitely many coordinates are 0. Meaning there is a last coordinate that is not 0. By deleting all 0s after that coordinate you get the coordinates of a point that is in some R^n.
E.g. (3,2,0,0,0,0,0,0,9,10,3,0,0,0,0,0,....) would be equivalent to the point (3,2,0,0,0,0,0,0,9,10,3) in R^11.
In other words, we can write the points of the R^n space using the coordinates of the space of finite sequences and then the space of finite sequences is the union of all R^n, which sounds pretty much like what we want.
Now, this space has countably infinite dimensions. Meaning that destroying it would be High 1-B.
That sounds pretty good, so what would go against looking at things like this? Well R^∞ is a little weird.
Let's talk about dimensional axes.
The space R is a 1-dimensional space, so its dimensional axis is itself.
The space R^2 is spanned by two axes, as it is two-dimensional. One can be the line described by all points (x,0) (that is what R is when we embed it into R^2) and the other is all points (0,y), which is orthogonal to the first one.
The space R^3 is three-dimensional and hence spanned by 3 axes. The first can be (x,0,0), the second (0,y,0) and the third (0,0,z). Notice how (x,0,0) and (0,y,0) are basically the dimensional axes of R^2 but embedded in R^3 and (x,0,0) in particular is the dimensional axis of R embedded in R^3.
And so on. Each space we go higher we keep the same dimensional axes we have and add one additional one.
Going by this process we would expect that R^∞ has the dimensional axis (x,0,0,0,0....), (0,y,0,0,0...), (0,0,z,0,0....) and so on. And in a sense that is the case. That's because of some mathematical detail where each coordinate has to be able to be written as a finite combination of the coordinates axis, meaning one can only go some distance across a finite selection of axis.
However, if you would like to move some distance along every one axis (like we would imagine it physically possible if you can move in every direction), then that point is not in R^∞ anymore.
That can be seen by the fact that using those axis we could construct points not in the space.
If we go a distance of 1 along the first axis, then a distance of 1 along the second, then 1 along the third and so on for each axis, we get to the point (1,1,1,1,1,1.....). However, that sequence never ends and hence is not part of the space of finite sequences.
So while we could go any distance along any one of these axis in finite dimensions, in infinite dimensions we can't do so. That is in the sense that we can still go an infinite distance along any one of them, but we can't go a little into each direction at once.
This might get even more confusing if one considers distance. Let's say we go a distance of 1 in the direction of the first axis, a distance of 1/2 in the direction of the second one, (1/2)^2 for the third, (1/2)^3 for the fourth, (1/2)^4 for the fifth and so on. Then the total distance walked would be 1 + 1/2 + (1/2)^2 + (1/2)^3 + (1/2)^4 + ... which is a geometric series and hence in total equal to a distance of 2. However, the point (1, 1/2, (1/2)^2, (1/2)^3, ...) is not part of R^∞. However, each step of the way was. That is to say, in a sense, one can walk straight out of the space by just moving a finite distance.
In mathematics, one would say the space is not complete. That is not an issue with the choice of space btw. Every countably dimensional vector space, with a concept of distance and for example a concept of angles, is incomplete. (or, scientifically more precise, every Banach space is not countably dimensional)
So yeah, the space is a little weird and perhaps not what we would a physical space to be like.
The Space of All Sequences
The other space I want to debate in detail is the space of all sequences R^N.Now, this space isn't just the union of all the R^n spaces. We see that by it containing the point (1,1,1,1,1,1,1...) which can not be contained in any R^n as it goes a distance of 1 along each of infinite dimensional axis.
However, in exchange, in this space you can go any distance along any of the dimensional axis that were in R^n at once and not leave the space.
One can also define a metric, i.e. distance, that is complete on it or also some norm. The norm would be weird in that it is discontinuous, i.e. points that are close to each other can have vastly different distances from 0, though.
l^2 would have nicer distance definitions and be complete, but in return would neither be the R^n union nor include all points one can get by going along every one of the dimensional axis.
Furthermore, notice that while (x,0,0,0,0....), (0,y,0,0,0...), (0,0,z,0,0....) and so on can be included in R^N as axis in the space and moving any distance along each of them keeps you inside the space, these axes would not be dimensional axes of R^N. As mentioned earlier, mathematically any point has to be reachable by only going some distance along a finite selection of the dimensional axis of the space. Meaning, R^N needs other ones to reach a point like (1,1,1,1,1...) and, in fact, more of them.
As it so happens, R^N would be more than countably infinite dimensional. I.e. destroying this space would be Low 1-A. One critical level higher than the prior option.
Conclusion
So, with that, I have given some insight into my mathematical considerations on the subject.I was split on the issue myself and frequently went back and forth between positions.
I would say my current stance is that we should rank the feat as High 1-B. Why? Well, ultimately the consideration on which space to take is just to decide the tiering. That the sum makes sense as a physical space is not so important. It's more a theoretical construct here, after all, as the spaces mentioned in the feat are just finite-dimensional spaces of any dimension.
To that comes that R^N doesn't have nice definitions of distance and angles and hence is not exactly what we imagine as a proper physical space either.
So overall High 1-B is more appropriate in my opinion.
What do you think, everyone?
