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I know that an inaccessible is, and I also know aleph-1 is not weakly inaccessible. What I am confused by is how can a character by strongly inaccessible? Characters are not numbers
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Tiering System Revisions - Part 3
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If we are already back at mathematics I want to point something out.
There are no countably infinite dimensional banach spaces and hence also no Hilbert spaces of that dimension. In particular has the R^N not countably infinite dimensions. So how much sense does it make in the proposed system to even have countably infinite dimensional tier?
That aside I gotta comment on topology as it is something I actually studied.
Math stuff
Real Coordinate Spaces are themselves topological manifolds and are required to be second-countable, hausdorff spaces by default.
I'm curious to see proof of that, since you don't even have to define a topology on a real coordinate space.
Other way around one can also give a real coordinate space a not second-countable topology (E.g. the discrete topology).
If one disregards those (equivalent) conditions entirely, then they cease to be metrizable (meaning they are no longer homeomorphic to a metric space) and to admit a given metric embedded in their structure.
Eh, what? I'm not sure I correctly understand what you want to say.
For a start, there are metrizable topologies that are not second-countable. E.g. the discrete topology on the real numbers is metrizable with the metric belonging to it being the discrete metric.
That aside if a space topological space is metrizable is completely independent of its dimensions. The discrete topology is metrizable on any space, regardless of dimension or cardinality.
In fact, as I already said, there are even real coordinate Hausdorff spaces with arbitrarily high cardinality of dimensions. Obviously these induce metrics, which in turn induce metrizable topologies...
This relates to spaces of dimensionality equal to the real numbers because of the fact that a second-countable space can only have at most continuum cardinality.
Wrong. Counter-example: The trivial Topology is second-countable on every set.
There are no countably infinite dimensional banach spaces and hence also no Hilbert spaces of that dimension. In particular has the R^N not countably infinite dimensions. So how much sense does it make in the proposed system to even have countably infinite dimensional tier?
That aside I gotta comment on topology as it is something I actually studied.
Math stuff
Real Coordinate Spaces are themselves topological manifolds and are required to be second-countable, hausdorff spaces by default.
I'm curious to see proof of that, since you don't even have to define a topology on a real coordinate space.
Other way around one can also give a real coordinate space a not second-countable topology (E.g. the discrete topology).
If one disregards those (equivalent) conditions entirely, then they cease to be metrizable (meaning they are no longer homeomorphic to a metric space) and to admit a given metric embedded in their structure.
Eh, what? I'm not sure I correctly understand what you want to say.
For a start, there are metrizable topologies that are not second-countable. E.g. the discrete topology on the real numbers is metrizable with the metric belonging to it being the discrete metric.
That aside if a space topological space is metrizable is completely independent of its dimensions. The discrete topology is metrizable on any space, regardless of dimension or cardinality.
In fact, as I already said, there are even real coordinate Hausdorff spaces with arbitrarily high cardinality of dimensions. Obviously these induce metrics, which in turn induce metrizable topologies...
This relates to spaces of dimensionality equal to the real numbers because of the fact that a second-countable space can only have at most continuum cardinality.
Wrong. Counter-example: The trivial Topology is second-countable on every set.
Agnaa
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I think I've made my disagreement clear from the start over Ultima's statements of "Dimensions become meaningless past an uncountably infinite number of them", but I didn't have the mathematical background to argue against that point too much.
However, I still think that the system is fine regardless, just that the specific phrase Ultima provided there shouldn't be used.
I think we should start outerversal at R^R^R and equalize "beyond dimensionality" to that tier, because any other starting point we choose has unacceptable outcomes, imo.
If we start outerversal at R^R then characters that simply transcend uncountably infinite dimensions would end up there, there's quite a few characters from verses that would place here. This would also bump some characters who are currently baseline outerversal to far above that, as they were at baseline outerversal for transcending uncountably infinite-D characters.
If we start outerversal at some point far higher than R^R^R, say, a point beyond any extensions of R^R...^R, then we are equalizing many verses up to a ridiculous degree, since we'd still want "Transcends R^omega" characters to be outerversal, equalizing them so far upwards like that is a bit silly to me.
However, I still think that the system is fine regardless, just that the specific phrase Ultima provided there shouldn't be used.
I think we should start outerversal at R^R^R and equalize "beyond dimensionality" to that tier, because any other starting point we choose has unacceptable outcomes, imo.
If we start outerversal at R^R then characters that simply transcend uncountably infinite dimensions would end up there, there's quite a few characters from verses that would place here. This would also bump some characters who are currently baseline outerversal to far above that, as they were at baseline outerversal for transcending uncountably infinite-D characters.
If we start outerversal at some point far higher than R^R^R, say, a point beyond any extensions of R^R...^R, then we are equalizing many verses up to a ridiculous degree, since we'd still want "Transcends R^omega" characters to be outerversal, equalizing them so far upwards like that is a bit silly to me.
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I wonder how powerful Downstreamers will be after the revision. The Manifold is uncountably infinite and contains all mathematically possible stuff.
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Outerversal or not?
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I am not sure. It might take too long/be too inefficient for our purposes.
Anyway, I would appreciate if Ultima would be willing to start (yet) another staff thread that summarises what we currently need to do at this point of the discussion.
Anyway, I would appreciate if Ultima would be willing to start (yet) another staff thread that summarises what we currently need to do at this point of the discussion.
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Is that assuming we are having a tier for countably dimensional characters or not? Asking given the main part of my comment above.Agnaa said:
Agnaa
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I think my reasons for starting outerversal at R^R^R are independent of that question. They'd apply whether we had a tier for countably infinite characters or not.DontTalkDT said:
However, I personally prefer having a tier for countably infinite, and a tier for uncountably infinite, both below outerversal.
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I'm fine with outerversal starting at |R^R^R| many dimensions, I guess.
However, as long as we actually go with a R^x formulation, instead of just saying large manifolds/spaces of x dimensions, I don't see how we could have a countably infinite tier as there just is no R^x with countable dimensions.
However, as long as we actually go with a R^x formulation, instead of just saying large manifolds/spaces of x dimensions, I don't see how we could have a countably infinite tier as there just is no R^x with countable dimensions.
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Please, no more threads.
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Someone can summarise what was have been done here?
Agnaa
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To clarify, it would start at R^R many dimensions.
That does sound like a plausible issue, but I didn't originally come up with the idea, I'll wait for Ultima's thoughts on it.
That does sound like a plausible issue, but I didn't originally come up with the idea, I'll wait for Ultima's thoughts on it.
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@Matthew
This one almost has 500 posts, and is falling into inactivity due to its length. We regrettably need a new, likely final, thread in order to make the final decisions regarding what we should do.
This one almost has 500 posts, and is falling into inactivity due to its length. We regrettably need a new, likely final, thread in order to make the final decisions regarding what we should do.
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>432 posts
Good grief.
Anyways just as a heads up: yeah, I can make a new thread so we can decide upon this more fluidly later on. Although I should preferably address DontTalk's post above before thatprovided I don't fall asleep first that is
Good grief.
Anyways just as a heads up: yeah, I can make a new thread so we can decide upon this more fluidly later on. Although I should preferably address DontTalk's post above before that
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aaand here we go...
Maffs
Although, yeah, you are right in that R ^ N doesn't have a countably infinite number of dimensions. In which case, what do you think of using unbounded n-manifolds / multiplications of R/whatever when dealing with finite-dimensional stuff, while stuff corresponding to infinite-dimensions is just denoted by some arbitrary space with dimension corresponding to a given cardinal number? That'd be more straightforward, I believe.
Yeah, but by doing that, you'd be constructing another Topological Space entirely that doesn't fit the requirements of a Topological Manifold, such as being second-countable and hausdorff (Under the definition used here, that is, I am aware those properties are not universally required), since you'd be taking the power set of its members, and thus end up with a space with greater cardinality than c.
Anyways, my point is that, if a given topological space is not hausdorff nor second-countable, then it is neither a metrizable space (and by extension not metric in the first place) nor a Manifold that is interesting or suitable for the purposes of tiering stuff below 1-A.
I say that because spacetime as modelled in physics needs a metric through which it can be defined, and a space that can admit such a metric in the first place, and the space which I just described is most certainly not able to do such. For example, our spacetime is mathematically modelled as a Riemannian Manifold in physics, which is essentially a Manifold embedded with a riemannian metric in its structure; As seen here, such a manifold must necessarily be second-countable, and it also needs to be a metric space for it to be compatible with... well, a metric, in the first place.
This ties into what I said in a previous comment of mine: The reason we choose |R| as a line between High 1-B and 1-A is because, while a point above it doesn't really showcase "tRansCendZ oVer thE cAwnCept oF dImEnShunS", it does allow us to solidly approximate a state similar to what Outerverse level entails, since you won't be able to tangibly represent physical (this being the keyword here) spacetime under a manifold if it is too big and devoid of certain key properties.
Sure, you could have a space with basis equal to, say, beth-omega or whatever, for example, but I don't think you would actually be capable of reproducing physical spacetime under it, at least not as far as I've seen. Although I can see you defining some arbitrary dimension apart from spatio-temporal ones at this scale, which is why, after giving some thought to the matter, I think 1-A is more aptly described as "metadimensional" or something, rather than straight up "beyond-dimensional".
EDIT: Nevermind, after giving some thought to the matter I realized you're indeed correct; I was thinking more about Second-Countability in the case of Fréchet (T1) Spaces, where there is indeed a restriction in terms of cardinality.
Although, that counterpoint is irrelevant anyways, since we are talking about Manifolds, which are never defined under the indiscrete topology and must satisfy the axioms of separation (T0, T1, T2 and yada yada). If you attempted to define one under it, then it'd turn out you would not be able to define any singular point in the space (which by extension would eliminate the notion of a neighbourhood from the book), as the only open sets in it would be the Empty Set and the entire Set itself, and by consequence the separation axioms would all fail unless additional assumptions are made.
This, by extension, would mean that a Trivial Topology is not metrizable either, a indiscrete space defined by it would fail to be metric in the first place, and the whole castle of cards collapses. So... even though I commited that error, the bulk of my argument still holds.
Maffs
As far as I recall, that is only if you adopt the usual notion of a basis (the Hamel Basis, specifically), which indeed doesn't really allow for such a thing as an infinite-dimensional space with countable dimension, but there are other notions which do allow for the construction of such spaces, such as the Schauder Basis, for example.DontTalkDT said:
Although, yeah, you are right in that R ^ N doesn't have a countably infinite number of dimensions. In which case, what do you think of using unbounded n-manifolds / multiplications of R/whatever when dealing with finite-dimensional stuff, while stuff corresponding to infinite-dimensions is just denoted by some arbitrary space with dimension corresponding to a given cardinal number? That'd be more straightforward, I believe.
That is true, but a real coordinate space always has a standard topology defined on its structure anyways, as do most mathematical spaces, hence what makes it a Topological Manifold.DontTalkDT said:
Yeah, but by doing that, you'd be constructing another Topological Space entirely that doesn't fit the requirements of a Topological Manifold, such as being second-countable and hausdorff (Under the definition used here, that is, I am aware those properties are not universally required), since you'd be taking the power set of its members, and thus end up with a space with greater cardinality than c.
Yeah, when I wrote the comment, I was mostly focusing on the part regarding a metrizable space being necessarily Hausdorff, while it satisfying 2nd Countability was more due to that property's natural relation to Topological Manifolds in the first place. Should've given more emphasis to the former in this context, my bad.DontTalkDT said:
Anyways, my point is that, if a given topological space is not hausdorff nor second-countable, then it is neither a metrizable space (and by extension not metric in the first place) nor a Manifold that is interesting or suitable for the purposes of tiering stuff below 1-A.
I say that because spacetime as modelled in physics needs a metric through which it can be defined, and a space that can admit such a metric in the first place, and the space which I just described is most certainly not able to do such. For example, our spacetime is mathematically modelled as a Riemannian Manifold in physics, which is essentially a Manifold embedded with a riemannian metric in its structure; As seen here, such a manifold must necessarily be second-countable, and it also needs to be a metric space for it to be compatible with... well, a metric, in the first place.
This ties into what I said in a previous comment of mine: The reason we choose |R| as a line between High 1-B and 1-A is because, while a point above it doesn't really showcase "tRansCendZ oVer thE cAwnCept oF dImEnShunS", it does allow us to solidly approximate a state similar to what Outerverse level entails, since you won't be able to tangibly represent physical (this being the keyword here) spacetime under a manifold if it is too big and devoid of certain key properties.
Sure, you could have a space with basis equal to, say, beth-omega or whatever, for example, but I don't think you would actually be capable of reproducing physical spacetime under it, at least not as far as I've seen. Although I can see you defining some arbitrary dimension apart from spatio-temporal ones at this scale, which is why, after giving some thought to the matter, I think 1-A is more aptly described as "metadimensional" or something, rather than straight up "beyond-dimensional".
Is it? As far as I am aware, you can't have a countable base on an uncountable set under a trivial topology, as the only possible base the generated topological space could have would be the very set upon which it is defined, which in this case clearly eliminates the possibility of it being countable.DontTalkDT said:
EDIT: Nevermind, after giving some thought to the matter I realized you're indeed correct; I was thinking more about Second-Countability in the case of Fréchet (T1) Spaces, where there is indeed a restriction in terms of cardinality.
Although, that counterpoint is irrelevant anyways, since we are talking about Manifolds, which are never defined under the indiscrete topology and must satisfy the axioms of separation (T0, T1, T2 and yada yada). If you attempted to define one under it, then it'd turn out you would not be able to define any singular point in the space (which by extension would eliminate the notion of a neighbourhood from the book), as the only open sets in it would be the Empty Set and the entire Set itself, and by consequence the separation axioms would all fail unless additional assumptions are made.
This, by extension, would mean that a Trivial Topology is not metrizable either, a indiscrete space defined by it would fail to be metric in the first place, and the whole castle of cards collapses. So... even though I commited that error, the bulk of my argument still holds.
"Transcending math" is like, reeeeeeally vague and nebulous when it comes to tiering, as nothing can be practically defined as being truly beyond it under a fictional context. At most, you would have to quantify that based on how far mathematics extends in the verse itself.JackJoyce said:
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1-A, as you transcend a platonic concept.JackJoyce said:
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I can, yeah. Just waiting for DontTalk to reply to my comment up there before a new thread can be made.
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Typed the response yesterday and wanted to look at some stuff closer before posting... didn't get to it today. But to not keep stalling, here's the math stuff:
Math Stuff
As far as I recall, that is only if you adopt the usual notion of a basis (the Hamel Basis, specifically), which indeed doesn't really allow for such a thing as an infinite-dimensional space with countable dimension, but there are other notions which do allow for the construction of such spaces, such as the Schauder Basis, for example.
Hamel basis allows for countably infinite dimensional spaces, just not Banach spaces.
Schauder basis is not important, as you don't define dimensions over it, as far as I am aware.
Although, yeah, you are right in that R ^ N doesn't have a countably infinite number of dimensions. In which case, what do you think of using unbounded n-manifolds / multiplications of R/whatever when dealing with finite-dimensional stuff, while stuff corresponding to infinite-dimensions is just denoted by some arbitrary space with dimension corresponding to a given cardinal number? That'd be more straightforward, I believe.
We could just define all as power equal to destroying large arbitrary c-dimensional manifold. Though I'm not familiar with the different kinds of infinite dimensional manifolds, so not sure if it's a good idea.
Should be an arbitary space over R at least, otherwise you end up with some weird space over F2 or something.
I guess we could just say manifolds for finite dimensions and large infinite dimensional structures beyond that.
That is true, but a real coordinate space always has a standard topology defined on its structure anyways, as do most mathematical spaces, hence what makes it a Topological Manifold.
Yeah, but by doing that, you'd be constructing another Topological Space entirely that doesn't fit the requirements of a Topological Manifold, such as being second-countable and hausdorff (Under the definition used here, that is, I am aware those properties are not universally required), since you'd be taking the power set of its members, and thus end up with a space with greater cardinality than c.
The standard topology is only there for finite dimensional real number spaces. It makes sense for them as any norm induces the same topology. The same isn't true for infinite dimensional spaces, therefore you have no standard topology on them.
Is second countable even required for infinite dimensional manifolds? Not an expert on that topic, but by the general definitio I find on the infinite dimensional case it isn't.
I would find it strange if Hilbert space structure is not as good as Manifold structure, as it is generally much more well behaved...
I say that because spacetime as modelled in physics needs a metric through which it can be defined, and a space that can admit such a metric in the first place, and the space which I just described is most certainly not able to do such.
The important part is that it's Hausdorff for that. As said, many not second-countable topologies are metrizable. You can get a metric on any given number of dimensions.
but I don't think you would actually be capable of reproducing physical spacetime under it, at least not as far as I've seen.
What does "reproducing physical spacetime" mean, though? You can have things like distance and angles.
Is it? As far as I am aware, you can't have a countable base on an uncountable set under a trivial topology, as the only possible base the generated topological space could have would be the very set upon which it is defined, which in this case clearly eliminates the possibility of it being countable.
The trivial topology on a set S is just {Ôêà, S}. You can take {S} as basis of it. |{S}| = 1 and hence the trivial topology is second-countable.
Math Stuff
As far as I recall, that is only if you adopt the usual notion of a basis (the Hamel Basis, specifically), which indeed doesn't really allow for such a thing as an infinite-dimensional space with countable dimension, but there are other notions which do allow for the construction of such spaces, such as the Schauder Basis, for example.
Hamel basis allows for countably infinite dimensional spaces, just not Banach spaces.
Schauder basis is not important, as you don't define dimensions over it, as far as I am aware.
Although, yeah, you are right in that R ^ N doesn't have a countably infinite number of dimensions. In which case, what do you think of using unbounded n-manifolds / multiplications of R/whatever when dealing with finite-dimensional stuff, while stuff corresponding to infinite-dimensions is just denoted by some arbitrary space with dimension corresponding to a given cardinal number? That'd be more straightforward, I believe.
We could just define all as power equal to destroying large arbitrary c-dimensional manifold. Though I'm not familiar with the different kinds of infinite dimensional manifolds, so not sure if it's a good idea.
Should be an arbitary space over R at least, otherwise you end up with some weird space over F2 or something.
I guess we could just say manifolds for finite dimensions and large infinite dimensional structures beyond that.
That is true, but a real coordinate space always has a standard topology defined on its structure anyways, as do most mathematical spaces, hence what makes it a Topological Manifold.
Yeah, but by doing that, you'd be constructing another Topological Space entirely that doesn't fit the requirements of a Topological Manifold, such as being second-countable and hausdorff (Under the definition used here, that is, I am aware those properties are not universally required), since you'd be taking the power set of its members, and thus end up with a space with greater cardinality than c.
The standard topology is only there for finite dimensional real number spaces. It makes sense for them as any norm induces the same topology. The same isn't true for infinite dimensional spaces, therefore you have no standard topology on them.
Is second countable even required for infinite dimensional manifolds? Not an expert on that topic, but by the general definitio I find on the infinite dimensional case it isn't.
I would find it strange if Hilbert space structure is not as good as Manifold structure, as it is generally much more well behaved...
I say that because spacetime as modelled in physics needs a metric through which it can be defined, and a space that can admit such a metric in the first place, and the space which I just described is most certainly not able to do such.
The important part is that it's Hausdorff for that. As said, many not second-countable topologies are metrizable. You can get a metric on any given number of dimensions.
but I don't think you would actually be capable of reproducing physical spacetime under it, at least not as far as I've seen.
What does "reproducing physical spacetime" mean, though? You can have things like distance and angles.
Is it? As far as I am aware, you can't have a countable base on an uncountable set under a trivial topology, as the only possible base the generated topological space could have would be the very set upon which it is defined, which in this case clearly eliminates the possibility of it being countable.
The trivial topology on a set S is just {Ôêà, S}. You can take {S} as basis of it. |{S}| = 1 and hence the trivial topology is second-countable.
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@DontTalk
Uh, you should probably read the edits I made to my post yesterday, mainly regarding the last point I made. That should clear up some confusion I hope :T
Uh, you should probably read the edits I made to my post yesterday, mainly regarding the last point I made. That should clear up some confusion I hope :T
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Because nothing can be meaningfully said to be beyond mathematics, since even when you describe something in any way or define an hierarchy, you are bringing up maths in some way, shape or form. If something were to be truly beyond it, we wouldn't be able to express it, and this is obviously problematic in the indexing context we aim for.
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I'm not a moderator and this comment might be deleted but from the looks of it I don't see any progress being made :/
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We are waiting for DontTalkDT and Ultima to reach some sort of agreement. After which Ultima can start a new staff thread with a summary of our progress so far in the beginning. After which I will highlight it.
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In regards to the edit you made to the last point I can just say that it was only meant as counter example, not as a suggestion for what to use. The point is that second countable topologies exist regardless of cardinality, meaning that we can't exclude there being suitable topologies due to that criteria at least.Ultima Reality said:
If we are at the point of how to define things similar to physical spacetime: Can we even consider any countably dimensional spaces as such? Most strong assumptions about the nature of such a space would give us something like an incomplete space. So, if we have a character destroy an entire infinite dimensional space, is it plausible to assume it is incomplete?
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2nd option. The 3rd option seems to blur the difference between high 1-A and 0 somewhat and I'd still like Boundless for 0. The first option is 1-A overkill imo.
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