What is the topological dimension of the Peano curve?

Question

The Hausdorff dimension of the Peano curve is know to be two. And I assume it to be a fractal since it's on the List of fractals by Hausdorff dimension. Moreover:

According to Falconer, one of the essential features of a fractal is that its Hausdorff dimension strictly exceeds its topological dimension.

So one can conclude that the topological dimension of the Peano curve must be one. But how could it be? I thought that in this context when we say curve, we mean the image of the curve. But the image of the Peano curve is $[0,1]^2$ and the topological dimension of $[0,1]^2$ is two.

But if the topological dimension of the Peano curve is two, why do we consider it to be a fractal?

Added:

And if the topological dimension of the Peano curve is one, how do we prove it? (For example, the topological dimension of the Koch curve is one because it's homeomorph to $[0,1]$ but it won't work with the Peano curve since $[0,1]$ and $[0,1]^2$ aren't homeomorph.)

Answer 1

A "curve" is a function. The "Peano curve" is a function whose domain is $[0,1]$ and whose range is $[0,1]^2$. But "topological dimension" and "Hausdorff dimension" apply only to metric spaces. So when you say "the topological or Hausdorff dimension of $X$", then $X$ must be a metric space.

So the question arises: what is meant by "the topological or Hausdorff dimension of the Peano curve"?

• Do you mean "the topological or Hausdorff dimension of the domain of the Peano curve"? Then the answer in both cases is $1$. Which is boring but sensible.
• Do you mean "the topological or Hausdorff dimension of the range of the Peano curve"? Then the answer in both cases is $2$. Again boringly sensible.
• But if you really, really, really mean exactly what is written, "the topological dimension of the Peano curve", well, the Peano curve is a function $$p=(p_1,p_2) :[0,1] \to [0,1]^2 $$ and a function from $[0,1]$ to $[0,1]^2$ is formally a certain subset of $[0,1] \times [0,1]^2 = [0,1]^3$, in this case $$\{(x,p_1(x),p_2(x)) \,\, | \,\, x \in [0,1]\} $$ Which is sensible and not at all boring. The topological dimension is $1$ because it is homeomorphic to $[0,1]$ (any graph of any continuous function is homeomorphic to its domain). But, the Hausdorff dimension is really interesting, I think. I'm not at all sure what it is equal to. I would need a specific formula or description to investigate this.

Answer 2

Just because it is a space-filling curve, this curve need not have topological dimension 2. In fact, it has topological dimension 1, so it is a fractal under this definition.

There is more to read on the matter here.

Answer 3

This is an addendum to Lee Mosher's answer, taking off where he stopped.

First of all, there is no such this as "the" Peano curve, as there are many different constructions of surjective continuous maps $g:I=[0,1]\to I^2=[0,1]^2$. Let $G$ denote the graph of such $g$. It is immediate that $G$ is homeomorphic to $[0,1]$ and, hence, has topological dimension $1$. As usual, we equip $G$ with the metric restricted from the one on $R^3$. Let $Hd$ denote the Hausdorff dimension.

Lemma. $Hd(G) \ge 2$.

Proof. $G$ is a subset of $I^3=I\times I^2$. Let $\pi$ denote the projection of $I^3$ to the second factor $I^2$. Then $\pi$ is Lipschitz (with Lipschitz constant $1$); the same, of course, applies to the restriction $\pi|G$. The image of the latter map is $I^2$, since $g$ is surjective. Let $H_2$ denote the 2-dimensional Hausdorff measure. It then follows from the definition of $H_2$ (and the 1-Lipschitz property of $\pi$) that $$ 1=H_2(\pi(G))\le H_2(G). $$ Therefore, $G$ has Hausdorff dimension $\ge 2$. QED

In particular, every such $G$ is a fractal where we define a fractal to be a metric space whose Hausdorff dimension is strictly larger than its topological dimension.

Computing the exact value of Hausdorff dimensions for graphs of particular surjective continuous functions $g: I\to I^2$ is an interesting but nontrivial task. I suspect that somebody already done this. If $\pi_1: I^2\to I$ is the coordinate projection to the $x$-axis, then computation of Hd of the graphs of $\pi_1\circ g$ for various "Peano curves" $g: I\to I^2$ are done here. I suppose that following these methods, one can compute precise values of $Hd(G)$ for graphs of various Peano curves $g: I\to I^2$ as well, but I am too lazy (and busy) for such computations.

Conjecture. For every $\alpha\in [2,3]$ there exists a "Peano curve" $g: I\to I^2$ with the graph $G$, such that $Hd(G)=\alpha$.