A simple curve of positive area

Question

Let $\gamma : [0,1]\rightarrow\mathbb R^2$ be a continuous curve whose image has positive Lebesgue measure. Must $\gamma$ have self-intersections? Intuitively this seems like it should be true, but I could not find a proof. An equivalent question is whether there is a positive measure subset of $\mathbb R^2$ which is homeomorphic to $[0,1]$.

Note that it is shown here Can a set of non self-intersection points of a space-filling curve contain an arc? that $\gamma$ must in fact have lots of intersections if the image of $\gamma$ is a rectangle. But, here I am not making any topological assumptions on the image of $\gamma$. The answer to the linked question does not work in this generality.

Answer 1

For completeness, I will sketch a construction here, though it is well-known (e.g., Countererexamples in Analysis by Bernard R. Gelbaum and John M. H. Olmsted).

Let $C\subset [0,1]$ be a fat Cantor set, i.e., Cantor-type set of positive measure. Note that $C\times C$ has positive area.

Claim. There exists a simple curve $\gamma:[0,1]\to\mathbb R^2$ whose image contains $C\times C$.

Proof. The map $\gamma$ is defined point-by-point, following the steps of the construction of $C$. First generation is $C_1=[0,1]$. The set $C_1\times C_1$ is a square:

All we do at this step is say that $\gamma(0)$ will be the lower left corner of the square, and $\gamma(1)$ its upper right corner. We haven't decided yet what $\gamma$ should do in between.

Another way to think of this: we have a "black box", in which we label the lower left corner as IN and upper right corner as OUT.

Second generation is $C_2$, union of two intervals. The set $C_2\times C_2$ looks like this:

Now we have four smaller black boxes here (well, they look more like orchid boxes). On each of them, label the lower left corner as IN and upper right corner as OUT. Then connect OUT to IN, in some way:

In terms of parametrization, we divide $[0,1]$ into $7$ intervals (say, of equal length) $I_1,\dots,I_7$, and assign $\gamma$ on them so:

• $I_2$ is mapped onto the segment CE
• $I_4$ is mapped onto GI
• $I_6$ is mapped onto KM

We haven't decided yet what $\gamma$ will do inside of $I_1,I_3,I_5,I_7$: these parts correspond to four boxes in the picture. But we do know that on each of them $\gamma$ will go in at the lower left corner, and come out at the upper right corner. Thus, we can repeat the preceding step for each of these four boxes. And so on...

As a result, $\gamma$ is defined on a dense subset of $[0,1]$. One has to check that it is uniformly continuous (in fact, Hölder continuity holds), and therefore extends to a continuous map of $[0,1]$. It remains to check that $\gamma$ is injective. Let $x,y\in [0,1]$ be distinct points. If one of them is mapped into an interior point of one of connecting arcs, then $\gamma(x)\ne \gamma(y)$ is clear, because $\gamma$ never comes close to such points again. Otherwise, both $x$ and $y$ are mapped into $C\times C$. Since at some step of construction $x$ and $y$ lie in different 7-adic subintervals, we conclude that $\gamma(x)$ and $\gamma(y)$ lie in different components of $C\times C$. $\quad \Box $

Answer 2

Look up Osgood_curve at Wikipedia, https://en.wikipedia.org/wiki/Osgood_curve

In particular I copy the following picture and text from there to make this answer self-contained. (Note that the construction has some similarity with the construction of a "fat Cantor set", https://en.wikipedia.org/wiki/Smith%E2%80%93Volterra%E2%80%93Cantor_set , but now instead of open intervals we are removing open wedges)

Example of an Osgood curve, constructed by recursively removing wedges from triangles. The wedge angles shrink exponentially, as does the fraction of area removed in each level, leaving nonzero area in the final curve.