# Space filling paths

So I noticed that it was possible to assign a real number to every lattice point in $R^2$ while looking at a wall somewhere. This got me thinking about a path that could "go through" every lattice point, and then making this path arbitrarily dense so that an arbitrary point in $R^2$ could be represented by a limit for the arbitrarily dense function, call it $T(x)$. This would allow us to represent an arbitrary vector $(a,b)$ as $\lim_{n\to\infty}[T(f(a,b,n))]$ where $f$ is an expression that will be determined by the transformation $T(x)$. This shouldn't be possible though because a 1-dimensional structure (here $T(x)$) can never completely "fill" $R^2$ or (any solid region in $R^2$ for that matter). I realize that this probably makes more sense to me than you so I came up with an example of what I mean. The transformation from $R$ to $R^2$ given by $$T(x)=(x-\lfloor x\rfloor, 2^{-\lfloor log_2(x+1)\rfloor}\lfloor x+2-2^{\lfloor log_2(x+1)\rfloor}\rfloor)$$ maps the real line for $x\in[0,\infty)$ onto the open unit square on the Cartesian Plane (vertices $(0,0)$, $(0,1)$, $(1,0)$ and $(1,1)$).

For $x\in[0,1)$ the output is the line y=1, for $x\in[1,2)$ the line $y=1/2$, for $x\in[2,3)$ the line $y=2/2=1$, for $x\in[3,4)$ the line $y=1/4$, for $x\in[4,5)$ the line $y=2/4=1/2$, and so on. If that doesn't make it clear a quick plot on a calculator/computer should clear things up.

Then the "solution" is $$lim_{n\to\infty}[T(a+\lfloor 2^{n}(b+1)\rfloor]=(a,b)$$ as long as the point $(a,b)$ is inside the unit square. That's what I consider paradoxical: that the left hand side has only one vector component while the right has two that are arbitrary (not really arbitrary since they need to be inside the unit square but it is still a solid region). What's more is that paths like this one can be used to represent an arbitrary vector of arbitrary dimension but the only dimension I can really "wrap my head around" is $R^2$, which is why my example is in $R^2$. So what's going on? Is it because the expression inside $T(x)$ doesn't converge? Is that why? I apologize if maybe this question isn't very well conveyed.

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Have you seen the Wikipedia article on space-filling curves? It seems from the body of your question that you haven't, but if so, you made quite a prescient choice of title. – Rahul Nov 12 '12 at 6:21

Notice that the path you describe is not continuous. It is not customary to call such things paths so let's call them shtap instead. Since you don't care about continuity for your shtap you just care about the underlying sets and since the paradoxical situation is a result of some geometric intuition you can't really ignore the topology.

In more detail, there are many many onto (and even bijective) functions from $\mathbb {R}$ to $\mathbb{R}^2$ (or $\mathbb{R}^n$ for that matter) simply because these two sets have the same cardinality. It is evident that there is an injection $f:\mathbb {R}\to \mathbb {R}^2$. To construct an injection in the other direction consider interleaving the two decimal expansions of a pair $(x,y)$ or real numbers to form the decimal expansion of a single real number. The existence of the two injections then implies, by using the Cantor-Schroeder-Bernstein Theorem, that there exists a bijection between $\mathbb R$ and $\mathbb {R}^2$. This result only seems paradoxical if you think of these two sets as spaces, one of a higher dimension than the other. Set-theoretically, there is no paradox.

A well-known situation that is somewhat more paradoxical is the existence of what is known as a Peano space-flling curve. There exists a continuous surjective path $\gamma:[0,1]\to [0,1]^2$, that is a continuous function from the one-dimensional space $[0,1]$ onto the two-dimensional square $[0,1]^2$. This situation is indeed somewhat counter intuitive.

The intuition calms down by the following theorem: There is no homeomorphism (i.e., a continuous bijective with continuous inverse) function $[0,1]\to [0,1]^2$. A topological proof is obtained by consider such a hypothetical homeomorphism and then removing an arbitrary $x\in (0,1)$, and its image in the square. The restriction of the homeomorphism must then be a homeomorphism between the resulting spaces, but $[0,1]-{x}$ is disconnected while the square minus a point is still connected.

More generally, it can be shown that (with the appropriate definition of dimension) two spaces of different dimensionality can't be homeomorphic. Surprisingly, the proof is not trivial nor easy.

I fail to understand what kind of paradox you try to derive about vectors.

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Yes that was my motivation: Cantor's diagonal argument and its implication that there are as many points in a line as in a plane or any other subset of $R^n$. I did not know that onto transformations from $R$ to $R^2$ where possible; I was under the impression that it was impossible, which is why I considered my transformation paradoxical, but if what you claim is true then there really is no paradox. I find it even more surprising that bijective mappings from $R$ to $R^2$ exist. – Hautdesert Nov 12 '12 at 20:04