[CP5670]

If I'm understanding you correctly, there is something wrong here.  The continuum C is the set of real numbers, and the hypothesis is that C is À₁.  The question is whether there is some endless set between ¥ and C to which À₁ would correspond.  If there is such a set, then it would be À₁, and C might be À₂.  

There are actually two (related) continuum hypotheses out there; one is simply the contiuum hypothesis, that À₁ = C, and the other one is called the generalized continuum hypothesis, which says that x^À_n = À_n+1 for all finite x. I was referring to the generalized one here, since the other one immediately follows from that as a special case.

If you are asking me whether I think C is À₁, I'd say it makes sense to me running on intuition.  Removing any endless series of numbers from C, by whatever pattern, should produce two sets correspondent to ¥.  How anything else could be true defies my imagination, but I cannot prove it, and neither, it seems, can any better minds.

Interesting; I personally think it actually should not be true. It has been proven that the transcendental numbers are what make up the continuum, since the algebraic numbers make up a À₀ set while the transcendentals are a À₁ set, which means that there are infinitely more transcendentals than algebraics. Unlike algebraics, transcendentals have not really been classified into more subgroups (i.e. integers, rationals, etc.), so it is quite possible that the À₁ cardinality of the continuum comes from several of these subgroups and not just one.

Um, is that a typo?  C^C should give the set of functions on a two-dimensional plain, not just C again.  I presume that that set of functions raised by itself gives the set of three dimensional functions, but I don't know.

ack...I actually meant À₁ = C there. It was 2 in the morning and I had only gotten five hours of sleep the previous night, so I probably made some errors...yeah, uh, that's my excuse!

Well, the big problem in this question is that N and S are not mutually exclusive at all; S is a subset of N.  Thus they are not equinumerable, and never were claimed to be.  S and -S will be (where -S is the set of naturals that are not squares of naturals), but N contains them both.

I also think that S should be only a subset of N, but apparently Cantor (the same guy who proposed the continuum hypothesis) proved that all sets of the same cardinality are equinumerable. This would mean that there are just as many squares, cubes or primes, etc. as natural numbers. This also implies that the set of points on any continuum is equal to that of any other continuum, so something like, say, a line segment of length 1 will have the same number of points as an infinite-dimensional cube with infinite edges. While this could well be true, I personally find the proof of it a bit dodgy, and the implications of it would be quite far-reaching; for example, all of analysis, one of the largest and most developed branches of pure mathematics, would be rendered completely meaningless, since all infinite sums and integrals would be equal.

[saturn114]

if some one  could find that master equation to einsteins unifed core  theory it would explain lots of things including
time travel einsteins unifed theory  is like a manual for the universe
think about it

[CP5670]

if some one  could find that master equation to einsteins unifed core  theory it would explain lots of things including
time travel einsteins unifed theory  is like a manual for the universe
think about it

yeah, that's been one of the major pursuits in physics for the last 40 or so years: to find a unified equation that combines quantum theory and general relativity. (the Dirac equation comes close, but that is only special relativity)

[Kellan]

Meh, they'll probably find at some point that Einstein was wrong and an0n was by far the more correct.

[Sesquipedalian]

Originally posted by Kellan
Meh, they'll probably find at some point that Einstein was wrong and an0n was by far the more correct.

I'm not even going to ask...

[Sesquipedalian]

Originally posted by CP5670
Interesting; I personally think it actually should not be true. It has been proven that the transcendental numbers are what make up the continuum, since the algebraic numbers make up a À₀ set while the transcendentals are a À₁ set, which means that there are infinitely more transcendentals than algebraics. Unlike algebraics, transcendentals have not really been classified into more subgroups (i.e. integers, rationals, etc.), so it is quite possible that the À₁ cardinality of the continuum comes from several of these subgroups and not just one.

But the transcendentals are still a subset of the real numbers, which are what C is supposed to be the totality of.  If it were proven that the transcendentals were of a higher transfinite cardinality than the algebraics, that would also be a proof that C¹À₁, for it would bump C up to at least À₂.

I also think that S should be only a subset of N, but apparently Cantor (the same guy who proposed the continuum hypothesis) proved that all sets of the same cardinality are equinumerable. This would mean that there are just as many squares, cubes or primes, etc. as natural numbers.

This is true.  There are as many rationals as irrationals, for another example.  The snag in the problem you proposed before is that we were not comparing two comparable sets: the one was an empty set, while the other was not.  If we were simply comparing squares to naturals, they are of course equinumerable, but we were comparing naturals to squares of naturals that are not themselves naturals, which is comparing an infinite set to one that is obviously not infinite, because it is empty.  Finite (or more generally, non-infinite) sets do not register on the transfinite scale at all, and thus S is an invalid set to use in the equation.

The set of points on any continuum is equal to that of any other continuum, so something like, say, a line segment of length 1 will have the same number of points as an infinite-dimensional cube with infinite edges.

 It is true that there are as many points in an infinitely large cube as in a single line segment, and that makes perfect sense to me.  I don't know if it would have quite the horrible implications that you fear it would, however.  Obviously all operations that involve ¥ end in ¥, and all that involve C end in C, and this has been known for a long time.  The fact that this is true has not impeded the development of analysis.  

Take a square, for example.  Its area is calculated by x*y.  A cube is x*y*z.  The operation involved is multiplication, and as we mentioned previously, the only way to change any transfinite set is to raise it by a transfinite set not less than itself.  So an infinite square is just C*C, which equals C.  An infinite cube is C*C*C, which equals C.  None of these produces any change in the number of points being discussed.  If one one considered an object of infinite size that existed in infinite (C, not merely ¥) dimensions, that would change things I suppose, since it would have C^C many points in it, but so long as we are dealing with a finite number of dimensions, no difference is made.  Thus we have nothing to worry about insofar as the self-destruction of mathematics is concerned, because no matter what, ¥, C, etc. are still ¥, C, etc.

[CP5670]

But the transcendentals are still a subset of the real numbers, which are what C is supposed to be the totality of.  If it were proven that the transcendentals were of a higher transfinite cardinality than the algebraics, that would also be a proof that C¹À₁, for it would bump C up to at least À₂.

Well, that has actually been proven, but the cardinality of the reals would still be À₁ because in this case, À₀+À₁ = À₁.

Finite (or more generally, non-infinite) sets do not register on the transfinite scale at all, and thus S is an invalid set to use in the equation.

Thing is, S is actually infinite, just as N is infinite, but neither one is continuous. Now what this theory says is that they are equinumerable, but that leads to the contradiction I posted before.

Take a square, for example.  Its area is calculated by x*y.  A cube is x*y*z.  The operation involved is multiplication, and as we mentioned previously, the only way to change any transfinite set is to raise it by a transfinite set not less than itself.  So an infinite square is just C*C, which equals C.  An infinite cube is C*C*C, which equals C.  None of these produces any change in the number of points being discussed.  If one one considered an object of infinite size that existed in infinite (C, not merely ¥) dimensions, that would change things I suppose, since it would have C^C many points in it, but so long as we are dealing with a finite number of dimensions, no difference is made.

That sounds fine, but suppose we do this: ¥ - ¥, which is obviously an indeterminate quantity in that form, but we are assuming it is equal to zero if the two terms are truly equal. Basically, the problem here is that ¥ = ¥ is being assumed, and this does not necessarily have to be true; for example, just because the functions 1/x² and e^1/x both have a limit of infinity as x->0 does not mean that they are equal. One thing that might solve this problem is to use a different, "weaker" definition of equality, something like the order O(f(x)) used in analysis.

Although a continously infinite-dimensional object (i.e. existing in 2.31 dimensions, p dimensions, etc.) and having infinite sides, or the C^C you mentioned, would also apparently have an equal number points as any line segment with unit length from what I have read.