# Question about Cardinality - Is the Universe One-Dimensional?

1. May 29, 2008

### meichenl

Something has been bothering me. I can't sleep, and neither can the little people inside my head. I'm not a mathematician, and I don't understand what I'm dealing with all that well, but I'm trying to figure out if the following makes sense:

The cardinality of R$$^{1}$$ is equal to the cardinality of R$$^{n}$$. 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$$^{3}$$ onto R$$^{15}$$ 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$$^{1}$$ = cardinality R$$^{2}$$.
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$$^{2}$$ 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$$^{1}$$ 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$$^{1}$$ is higher than countably infinite, but just by using a decimal expansion aren't I mapping R$$^{1}$$ 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

2. May 29, 2008

### Diffy

I am not sure that this will work,

Cosider (1, 1) (-1, 1), (1, -1) and (-1, -1)

in R you only have 1.1 and -1.1, you need your function to be a bijection.

Last edited: May 29, 2008
3. May 29, 2008

### CRGreathouse

Interleaving digits is a perfectly good way to do this: combine

3.141592653589793238462643383 and
2.718281828459045235360287471 into
23.17411852982168523854859970943522338543662062473437813...

Space-filling curves (see for example http://www.math.ohio-state.edu/~fiedorow/math655/Peano.html ) gives a different method to see that the cardinalities are equal.

If you had, say, three decimal places, you could describe 10^3 = 1000 numbers. A real number has a countable number of decimal places, one for each integer.* Thus it can describe $$10^{\aleph_0}$$ numbers, where $$\aleph_0$$ is the 'number' of integers. While $$\aleph_0$$ is countable, $$10^{\aleph_0}$$ is not.

* Ignore, for the time being, the part to the left of the decimal point; there are lots of ways to deal with it, and in the end they don't matter. Just think of the "fractional part" to the right.

"Most conveniently described" may be a good description. But I'm not very knowledgeable in physics, and here you leave the realm of mathematics.

Last edited: May 29, 2008
4. May 29, 2008

### meichenl

CRGreathouse,

Thanks for the replies! Good information. Those plane-filling curves are pretty wild. I had been wondering if there was some sort of fractal-type thing that could do that. Nice.

Diffy,
I see your point, since I can't have "negative decimals". So the mapping described might only work on quadrant I. However, quadrant I has the same cardinality as the plane, so I could then make a new mapping from quadrant I to the rest of the plane.

5. May 29, 2008

### morphism

I was replying to this thread when the power went out and I lost everything I'd written. Here's a condensed repost:

First of all, I think your main issue stems from a misunderstanding of the notion of 'equivalence'. When two sets X and Y are put into a one-one correspondence, we get an equivalence between their cardinalities -- that's all. This equivalence does not take into account any of the additional structure that may be present. Case in point: While R and R^3 are equivalent ("isomorphic") 'as sets' (i.e. they have the same cardinality), they are not equivalent ("isomorphic") as vector spaces. This is why you have to specify the sense in which your objects are "isomorphic".

It's not too difficult to prove that if a Hilbert space is infinite-dimensional, then its dimension is necessarily uncountably infinite. In fact this same statement is true for Banach spaces in general. There are a few proofs of this fact, and my favorite one is a Baire category theorem argument: Suppose X is an infinite -dimensional Banach space with a countable basis given by {e_1, e_2, ...}. Put F_n = span{e_1, e_2, ..., e_n}. Then X=$\cup_n$F_n, while each F_n is nowhere dense (this requires proof), contradicting the Baire category theorem. [A proof that doesn't use the BCT can be found in Morrison, Functional Analysis, Wiley-Interscience, 2001, if memory serves me.]

Last edited: May 29, 2008
6. May 29, 2008

### Hurkyl

Staff Emeritus
It's Hamel dimension (i.e. dimension as a vector space) is uncountably infinite, but its Schauder dimension (i.e. dimension as a topological vector space) may be countably infinite.

7. May 31, 2008

### maze

When you map R3 into R1, you are cutting R3 up and rearranging the pieces mercilessly. The "3D-ness" of R3 is lost in this process. Points "nearby" to each other in R3 get moved "far away" in R.

8. Jun 2, 2008

### chem_master

Dear meichenl,

I can't sleep as well! The same thing has been bothering me.
I've read your post; you noticed that formulating physics in n dimensions gives the same result because space has the same cardinality - the cardinality of a continuum - aleph-1.

I'd like to go a bit further. Since a free patricle in motion requires only one dimension (assuming it gives no central force or any force field), the cardinality of the one-dimensional space should have aleph-1. Any other relations with a particle, including a central force field, uncertainity and particle interactions require a higher cardinality aleph-2 (like Hilbert space, space of wavefunctions etc.).

In my opinion there is something strange going on in our perception of a dimension. We live in 3D, so why the third dimension isn't "blocked" in a cardinality.

Einstein for me was a genius, he discovered a beautiful relation between space and time but in his eqauations just "sticked" those two extra dimensions of space just because they exist and not because they result from his theory.

Any ideas, suggestions or thoughts are very welcome.

9. Jun 2, 2008

### CRGreathouse

The cardinality of the continuum is $$\aleph_1$$ under the Continuum Hypothesis and Axiom of Choice. Without the Continuum Hypothesis, it's possible that $$\aleph_1<\mathfrak{c}.$$

10. Jun 2, 2008

### matt grime

Please can people stop thinking that the fact there is a set theoretic bijection from R to R^n has any bearing on their structure as vector spaces? You're talking about things in completely different categories - SET and MANIFOLD (or VECT), whatever. There are many bijections between many things, that doesn't mean that there is any other kind of relationship at all: there is a bijection between the number of zeros on the line 1/2 + it of the Riemann zeta function and isomorphism classes of finite groups. So what....

11. Jun 3, 2008

### CRGreathouse

But who's to say the OP cares about the universe as a vector space not a set?

12. Jun 3, 2008

### matt grime

He states that he wants to consider R^n with its structure as a differential manifold.... so he needs maps that preserve the structure of being a differential manifold.

13. Jun 3, 2008

### CRGreathouse

The original post uses "cardinality" several times and "differential manifold" (even "manifold") never. It does mention vector fields, but I took that to mean "Why n-dimensional vector fields instead of a scalar?".

14. Jun 3, 2008

### matt grime

Just because the OP doesn't use the terms properly doesn't mean that that isn't what is necessary. If you wish to talk about vector fields, then that is choosing a section of the tangent bundle on R^n.

15. Jun 3, 2008

### morphism

It's clear that the OP doesn't care about the universe only as a set:

16. Jun 4, 2008

### Son Goku

The simple resolution of this is that the position basis kets are not elements of the Hilbert space. Quantum Mechanical wavefunctions are always (Lesbegue) square integrable functions. A position basis ket $$|a\rangle$$ is a dirac delta distribution $$\delta(x-a)$$, which is not square integrable function and hence not an element of the Hilbert space. If you look at a basis that is actually contstructed from square integrable functions, then you'll see the dimensionality is countable. Basically you are using, as basis elements, things which aren't members of the space.