# Proof that Our Planet is 1D

I have took a discrete mathematics course this summer and there we talked about power of groups and functions,and yesterday I though and realize that if we can map all the 3D coordinates with a one on one and surjective function from $(x,y,z)$ to $(a)$, we can prove that our planet is 1D, that's mind blowing. Is that proof is right?

• – Qmechanic Oct 5 '13 at 15:12
• Certainly you can produce such a map, but that doesn't tell you anything about physics only about math. The dimensionality of stuff is set by the distance element $\mathrm{d}s$ which has three (large) spacial components. If the string theory guys are right it may have more than that, but they don't matter on human scales as they are constrained to be small. – dmckee --- ex-moderator kitten Oct 5 '13 at 16:46

I think that you are referring to space-filling curves and how they can map a line segment to more than one dimension. For example the Hilbert space filling curve can be used to map the interval $[0,1]$ to $[0,1]\times[0,1]$.

I am afraid while a continuous bijection is possible one-way, it is not possible to have a homeomorphism between two different Euclidean spaces of different dimensions. A homeomorphism is a mapping that is continuous, bijective and has its inverse continuous. You cannot construct a homeomorphism. Thus the Earth cannot be 1D!

• How can you prove that there is no bijection function to the both sides? Cantor–Bernstein–Schroeder theorem says if you can find a bijection function from (x,y,z) to (a) So you can find a bijection from (a) to (x,y,z). – Gil Oct 5 '13 at 21:15
• @Gil: That's not what he said. There are bijections going both ways, but not homeomorphisms. – jwodder Oct 5 '13 at 22:08
• This relies on the assumption that space is continuous. – Jon Purdy Oct 6 '13 at 17:57
• I did make that assumption implicitly because the OP mentioned a function from $(x,y,z)$, which is a continuous space. – cutculus Oct 7 '13 at 7:54

Varun's answer basically tells it all, but perhaps it's useful to also explain why we want a homeomorphism rather than just a bijection.

The whole idea of modelling "real physical space" by an Euclidean $\mathbb{R}^3$ is so we can make predictions about physical processes, like how an electromagnetic wave spreads. Typically, we use differential equations for that purpose, or path integrals. Those models are local, i.e. no matter how large a process we describe, it can ultimately be split up in fundamental processes taking place in arbitrarily small segments of space. This is with respect to the Euclidean norm: for each point and each "desired precision", there is some open neighbourhood, a tiny vector space on its own, in which we have some simple equations describing the physics.
It is this norm that also induces the topology of $\mathbb{R}^3$ – a topological space is basically just a set together with the notion of which subsets are local neighbourhoods.

If we map the physical space to some other space using a homeomorphism, then the open neighbourhoods are "preserved" (mapped to open neighbourhoods again). Also, we don't get any new neighbourhoods. So a local physical model on the new, homeomorphic space is equivalent to a local physical model on the original space.

Things look very much differently if you map to a space of different dimensionality using something like a Hilbert space filling curve. Those mappings are strongly non-homeomorphic, i.e. open neighbourhoods in $\mathbb{R}^3$ will generally be mapped to a collection of infinitely many disconnected chunks. Any model that was simple on $\mathbb{R}^3$ is translated to a horrible mess on $\mathbb{R}$, and even if you find a usable algebraic description of it then it will depend strongly on which space-filling curve you used exactly. Any such model will be non-local, i.e. look completely different from the ways we're used to describing physics. To do any useful calculations, you'll probably need to go back to $\mathbb{R}^3$ first, use the simple model there, and transform to $\mathbb{R}$ again. This is just pointless.

It is impossible. Suppose the real line, $\mathbb{R}$, is your 1D space and that $\alpha:\mathbb{R}\to \mathrm{Earth}$ is an homeomorphism (physics is more than only topology, but at least the topolology of both sides should match). Then remove a point on every space, say $\mathbb{R}_\times$ and $\mathrm{Earth}_\times$. The restriction, $\alpha_\times$, induces iso between

$$\mathbb Z^2=\pi_0(\mathbb R_\times) \simeq \pi_0(\mathrm{Earth}_\times)=\mathbb Z,$$

which is a contradiction. Hence there is no such a map.

The answers so far give a good answer to the question, but just adding a general point:

When a system is described axiomatically, one gives a set of possible states, plus a set of valid rules for transforming those states into others. Therefore, having the set of valid states for two systems be equivalent doesn't mean the systems are equivalent, since it doesn't say anything about equivalence of the second parts of the descriptions.

Since your question came up based on discrete mathematics, and "our planet" may well be considered a set of discrete items (such as atoms) ...
... it may be useful to consider a notion of "dimension D" which allows to distinguish discrete spaces of different D values; for instance:

if the set $\mathcal{S}$ of suitable discrete items is a metric space $(\mathcal{S}, d)$, with given distance values ("$d$") between pairs of elements (or at least given distance ratios between any three distinct elements) then either

1. All elements of $\mathcal{S}$ are straight to each other; or at least: the closer any three elements are to each other, the straighter they are to each other (by comparison), too.
(Explicitly, for instance:
for any element $A \in \mathcal{S}$ there are two elements $B, C \in \mathcal{S}$ such that the radius of the circumcircle of triangle $ABC$ is larger than the circumference of triangle $ABC$, $\frac{AB \cdot AC \cdot BC}{\sqrt{2 AB^2 \cdot AC^2 + 2 AB^2 \cdot BC^2 + 2 AC^2 \cdot BC^2 - AB^4 - AC^4 - BC^4}} \gt AB + AC + BC$;
and for any other two elements $X, Y \in \mathcal{S}$ for which the circumference of triangle $AXY$ is less than the circumference of triangle $ABC$ the radius of the circumcircle of triangle $AXY$ is even larger by comparison,
$\frac{AX \cdot AY \cdot XY}{\sqrt{2 AX^2 \cdot AY^2 + 2 AX^2 \cdot XY^2 + 2 AY^2 \cdot XY^2 - AX^4 - AY^4 - XY^4}} / (AX + AY + XY) \gt$
$\frac{AB \cdot AC \cdot BC}{\sqrt{2 AB^2 \cdot AC^2 + 2 AB^2 \cdot BC^2 + 2 AC^2 \cdot BC^2 - AB^4 - AC^4 - BC^4}} / (AB + AC + BC)$
.)
Then the metric space $(\mathcal{S}, d)$ is 1-dimensional. Or else:

2. All elements of set $\mathcal{S}$ are plane to each other; or at least: the closer any four elements are to each other, the planer they are to each other (by comparison), too.
(This can be made explicit by comparisons involving the radii of circumspheres of any four elements of set $\mathcal{S}$.) Then the metric space $(\mathcal{S}, d)$ is 2-dimensional. Or else:

3. All elements of set $\mathcal{S}$ are flat to each other; or at least: the closer any five elements are to each other, the flatter they are to each other (by comparison), too.
(This can be made explicit by comparisons involving the radii of circum-3-spheres of any five elements of set $\mathcal{S}$.) Then the metric space $(\mathcal{S}, d)$ is 3-dimensional. And so on.

By these (or closely related) criteria, the items that make up "our planet" presumably constitute a 3-dimensional set.

[...] coordinates with a one on one and surjective function from $(x,y,z)$ to $(a)$

Note that in the definition of "dimension D of discrete spaces" suggested above there was no mentioning of any coordinates. It doesn't matter whether or which or how many real numbers are assigned to elements of any particular set $\mathcal{S}$; all that's required are the distances (or at least: distance ratios) between elements.

However, once the "dimension D" of a set has been established, then different coordinate assignements may therefore be distinguished.

For instance: if set $\mathcal{S}$ is 1-dimensional (and with two separate "ends") and if real numbers are assigned to the elements of set $\mathcal{S}$ (one distinct real number to each distinct element) then it can be distinguished whether the assignment is monotonous w.r.t. the order of the elements in $\mathcal{S}$, or not.

This way of drawing distinctions can be generalized to sets of larger "dimension D" if real D-tuples (i.e. sets of D real numbers) are assigned to each element.