I have always been fascinated by self-similarity, particularly in fractals. I was always wanted to find a simple definition of a self-similar fractal. Of course, saying "is self-similar, and is a fractal (has noninteger fractal dimension)" is a valid definition, but not the one I was looking for.
So here I go. A shape $S$ in euclidean space is a self-similar fractal if:
- It is self similar (this definition seems as suitable as any, but feel free to use another if you prefer).
- It does not tessellate the plane (if it does, it is trivially self-similar).
For example, the square is self-similar (if you remove the right and bottom edge), but it also tessellates the plane. A line segment too is self similar, and, using an uncountable number of lines it can tessellate the plane. An octagon can not tessellate the plane, but it is not self-similar. So these things are not fractals.
Sierpinski's triangle is self-similar but can not tessellate the plane. The Koch Curve is also self-similar and can not tessellate the plane (if you try the same trick with the line, it will necessarily intersect itself).
My question is, is the above definition equivalent to being a fractal and being strictly self-similar in the plane? Does it work in other euclidean dimensions? How about other unbounded metric spaces?
Your definition of self-similarity is solid - essentially, a set $E$ is self-similar if it is the invariant set of an IFS $\{f_i\}_{i=1}^m$: $$E = \bigcup_{i=1}^m f_i(E).$$ For example, the unit square is self-similar since it is the invariant set of the list of four similarities $f_i(x) = x/2 + s_i$, where each $s_i$ is a shift vector: $$ \begin{align} s_0 &= \langle 0,0 \rangle \\ s_1 &= \langle 1/2,0 \rangle \\ s_2 &= \langle 0,1/2 \rangle \\ s_3 &= \langle 1/2,1/2 \rangle. \end{align} $$ Your insistence that the set not tile the plane is more difficult, however. I'd say that the square fits well into the realm of fractal geometry, as it's self-similar. Now, suppose we change the vector $s_3$ above to $$s_3 = \langle -1/2,-1/2 \rangle.$$ Then, we get the following set:
Is this a fractal set? As it's a subset of a plane that contains a solid disk, its dimension is 2. Its boundary has dimension $\log(3)/\log(2)$, however. And, believe it or not, it does tile the plane. If we shift it two units to the left or to the right or up or down or $2\sqrt{2}$ units diagonally Northeast or Southwest, the filled parts match perfectly with the holes:
How about the twindragon?
Solid proofs that these sets tile the plane is tricky. They both fall under a general scheme elaborated by Christoff Bandt in the Proceedings of the AMS in 1991. I published my own small contribution in The Mathematica Journal a few years ago. My paper was based on Bandt's work as well as work by Bob Strichartz and Yang Wang.
That second paper describes how I generated the images here. Also, the boundary of the twindragon is discussed in this answer on MathOverflow.