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The cardinality of R[tex]^{1}[/tex] is equal to the cardinality of R[tex]^{n}[/tex]. This means we can make a one-to-one mapping between them. In that case, any physical theory using vector fields could just as well be described by scalar fields. For example, the electric and magnetic fields could be described by scalar fields, and the divergence and curl would become functions of a single variable. For that matter, instead of describing my position in the universe by a vector, I could simply describe it by a humble scalar. To make things even more preposterous, I could map R[tex]^{3}[/tex] onto R[tex]^{15}[/tex] and describe physics in fifteen spatial dimensions. If this is true, then in what sense do we have a right to say the universe is three-dimensional? It appears as if they're telling me that any finite number of dimensions are mathematically equivalent. If so, the universe isn't three-dimensional, it's just n-dimensional, where n is any damn natural number you please.

Here are some further thoughts:

I read that the cardinalities are the same in a little sidebar type thing in a book. It wasn't the main point of the book and the book didn't go into detail describing it. So I don't know exactly how you would map one dimension onto two dimensions, for example. After some worried thought, I came up with two ideas about how this might work: one geometric and one algebraic. Please give me some feedback on whether there make any sense.

Geometric:

I will try to map [0,11] onto a finite two-dimensional surface. The cardinality of the finite regions are the same as of infinite regions of the same dimension, so if this argument works it shows the cardinality of R[tex]^{1}[/tex] = cardinality R[tex]^{2}[/tex].

Here is how I want to map a given point in the 2d surface onto a scalar. Take the finite region and cut it into ten equal sized blocks. Label the blocks 0,1,2,...9. Next, find the block that contains the target point (say block 3), and cut it into ten equal-sized sub-blocks, labeled 3.0, 3.1, 3.2, ... 3.9. Next, find the sub-block that contains the point (say 3.6) and chop it into ten equal-sized sub-sub-blocks label 3.60, 3.61, ... 3.69, etc. Just keep going indefinitely and you can zero in on the point you're after.

Now, I'm not truly convinced by my own geometric argument for two reasons. First, what if at some stage the point ends up lying on the line separating two regions? My answer is that you take all the separation lines and map them onto the leftover [10,11], which I haven't used yet. But I have enough ignorance of and respect for mathematics that I get extremely nervous about infinities, and here I'm trying to map infinitely many division lines onto a segment. I'm not sure if that's fair. The second reason the geometric argument bothers me is also with the infinities. I'm dividing my sub-blocks down infinitely many times, and claiming that this allows me to zero in on a point. But for any given number of sub-divisions, there is still an entire neighborhood of points that have the same decimal expansion under my scheme. This is a bit bothersome to me.

Algebraic argument:

I propose to map the entire plane R[tex]^{2}[/tex] onto the real number line. My procedure is:

1) write out the decimal expansions of the ordered pair. I.E. the point (12.4, pi) becomes (12.4, 3.141592...)

2) if you have a terminating decimal expansion, add infinitely many zeros on the end. Also add infinitely many zeroes at the front of the number. So my point now becomes

(00000....0012.40000000..., 0000...0003.141592...)

3) now I need to turn this ordered pair into a scalar. i'll do it by essentially putting the first coordinate into the scalar's "even" digits (the tens place, thousands place, hundred thousands place, and tenths place, hundredths place, etc.) and the second coordinate into the scalar's "odd" digits (ones place, hundreds place, hundredths place, etc.) The rest of this paragraph is probably horribly confusing. Just skip to the example, I guess. Construct the corresponding point in R[tex]^{1}[/tex] by starting with the decimal, then working step by step to the left and right. The first digit after the decimal of the target number is the first digit after the decimal of the first coordinate you're mapping from. The second digit after the decimal of the target number is the first digit after the decimal of the second coordinate you're mapping from. Keep alternating digits, and do the same thing to the left of the decimal, with the ones digit being the ones digit of the second coordinate, the tens digit of the target being the ones digit of the first coordinate, and so on. So, for the example point, I map it onto (0000...001023.4104010509020...)

Now I have a one-to-one mapping. Any real number maps onto the plane by picking all the "even" digits to be the first coordinate, and all the "odd" digits to be the second coordinate. Similarly, any point on the plane can be mapped onto the real number line.

My own algebraic argument also does not have me 100% convinced that this is mathematically a good way to do the mapping, because again I am a bit nervous about infinities, and saying that I am going to use the decimal expansion of transcendental numbers, for example. One thing that really bugs me is that I know for sure that the cardinality of R[tex]^{1}[/tex] is higher than countably infinite, but just by using a decimal expansion aren't I mapping R[tex]^{1}[/tex] onto countably infinite decimal places? I guess that's not the same, because I'm not mapping it onto countably infinite elements, but it doesn't seem far off.

Regardless of whether my schemes are valid (the book I was reading told me that Cantor took years to discover a mapping that did what I'm trying to do, so I'm guessing the full story is a bit more complicated), I know for sure that such a mapping exists because it was in a book written by a smart person.

So, as I asked before, if one dimension and three dimensions have the same cardinality (would it be appropriate to say they are "isomorphic"?), how can we get off claiming the universe of classical mechanics has three spatial dimensions? Aren't we really just saying, "the universe is most conveniently described, at least from the point of view of us glorified ape-men, as three dimensional", when in fact we could get away with just one dimension if we felt like it.

Now, the dimensionality of the Hilbert space in quantum mechanics is infinite, and I have one more question about this. Suppose we're talking about a one-dimensional harmonic oscillator. Then in the energy basis I can describe the wave function as a superposition of the energy eigenstates. These are n=1, n=2, n=3, ... There are a countably-infinite number of energy eigenstates. But in the position basis, I need a continuous wavefunction, and the basis vectors are delta functions at each x-value. The x-values are a continuum, so there are uncountably many position basis vectors. So is the dimension of this Hilbert space uncountably infinite or countably infinite?

Well, I know that was a rather long post. My hope is that I understood the math not necessarily well enough to ask well-posed questions, but at least well enough that some kind soul can get an inkling of what it is that's confusing me. If you can grant a little insight, or point me in the direction of a good resource for learning more, you'll have my thanks.

Mark