How many vertices are in the Koch Snowflake?

Question

EDIT: The question was put on hold because I didn't specify what I meant by vertex. In a comment below by Mark McClure, by "vertex" I mean one of the vertices of the standard, polygonal approximations to the Koch curve.

I have been trying to create a bijection from the integers to the vertices in the Koch Snowflake to show that there are countably infinite vertices. I bijected the vertices of the 0th iteration triangle onto {0,1,2}, and in each stage of iteration biject the newly created vertices into the digits {0,1,2,3}. Then, I concatenate the new digit to the left to encode the location into the integer. This way, we can use the integer to "locate" the theoretical vertex at infinity, and use the location of the vertex to create the integer.

However, I'm worried that the bijection falls apart if the integer is finite, since there will be an infinite amount of 0's on the left. Is my argument still okay?

Thank you in advance.

Answer 1

Let's focus first on the Koch curve. A standard construction for the curve starts with a single line segment, breaks it into thirds, and replaces the middle third with the other two sides of an equilateral triangle.

Now, by "vertex of the Koch curve", I suppose you mean one of the vertices in one of these polygonal approximations. The easiest way to prove that there are countably many of these is to simply point out that there are finitely many at each step and that the total collection of vertices is the countable union of these finite sets. Nonetheless, it is an interesting problem to enumerate them explicitly and potentially useful as there are certainly self-similar sets where these points play a distinguished role in the limit set.

The construction of the Koch curve can be described using an iterated function system $\{T_0,T_1,T_2,T_3\}$, where \begin{align} T_0(x) &= x/3 \\ T_1(x) &= R(\pi/3)\,x/3 + \langle 1/3,0 \rangle \\ T_2(x) &= R(-\pi/3)\,x/3 + \langle 1/2, \sqrt{3}/6 \rangle \\ T_3(x) &= x/3 + \langle 2/3, 0 \rangle. \\ \end{align} In this notation, $x\in\mathbb R^2$ and $$ R(\theta) = \left( \begin{array}{cc} \cos(\theta) & -\sin(\theta) \\ \sin(\theta) & \cos(\theta) \end{array}\right) $$ is the rotation matrix through the angle $\theta$ about the origin.

Now, the segments in the first step of the approximation are exactly the images of the unit interval $((x,0):x\in [0,1])$ under the functions in the IFS. We might represent this like so:

The side labeled $\{i\}$ is the image of the unit interval under the function $T_i$. The process extends naturally to the next level.

The segment labeled $\{i_2,i_1\}$ is the image of the unit interval under the function $T_{i_1}\circ T_{i_2}$. These segments come in a natural order to form a continuous path moving from $(0,0)$ to $(1,0)$. As such, each has an initial endpoint and a terminal endpoint. The labels on the edges can be used to enumerate the vertices. Specifically, each label is the base four expansion of an integer; we'll map that integer to the initial endpoint of the corresponding segment. At level three, this produces an enumeration that looks something like so:

To map onto the vertex of the snowflake itself, you can simply map the multiples of three to the vertices above, the numbers of the form $3n+1$ to one of the other sides and the numbers of the form $3n+2$ to the remaining side. Note that the right most point above is not taken. That's fine, as that can be the initial point of the next side.