Is the two-dimensional Koch curve space-filling? Say, we'd like to make a Koch curve with self-similarity dimension of two. A Koch curve with the following generator seems to be two-dimensional,

since if we double its size by scaling we'll find four of the original size inside, so
$$\text{self-similarity dimension}=\log_2 4=2$$
Therefore I was expecting the curve to be space-filling. After my initial mistake of failing to make the segments equal, words that end in GRY provided an image of the limit set:

$\ $
Is the curve filling any space? If yes, what space? If no, why not? 
 A: Two observations :
First, your fractal is bounded. If the original segement is $[0;1] \times \{0\}$, the fractal it engenders is bounded in the box $[0,3/2] \times [-1/4 ; 1]$
Second, if you start with a grid of segments (or arrows, to make the orientation more clear), and do one iteration, you obtain another grid four times as dense : 
From this picture, we can deduce that the monoid of the 4 transformatinons is free, and in particular if you iterate any number of times on one arrow, every arrow obtained is distinct, there are no duplicates.
The iterations define a continuous application $\phi_0 : [0;1] \cap \Bbb Z[1/2] \to \Bbb Z[1/2]^2$.
By continuity this extends to a unique continuous function $\phi : [0;1] \to \Bbb R^2$, where the image of $\phi$ is the closure of the image of $\phi_0$ :
given a sequence $(\phi_0(x_n))$ of elements in the image of $\phi_0$, since $[0;1]$ is compact, we can extract a subsequence of $(x_n)$ converging to $x$, and then by continuity, $\phi(x) = \lim \phi(x_n)$.
Now, if you pave the plane with such fractal pieces as indicated by the grid of arrows, you end up filling the plane :
Suppose $P \in \Bbb R^2$. There is a sequence of points with dyadic coordinates converging to $P$. Since every fractal piece is bounded, there is a finite number of fractal pieces involved in that sequence. So we can extract a subsequence that stays in one fractal piece, and since every piece is closed, $P$ is in that fractal piece.
Assuming the pieces are measurable (I think they are since we don't use the axiom of choice), in the grid of arrows, the intersection of any two pieces has zero measure :
Without loss of generality we can suppose that we are looking at two equal-sized pieces generated by arrows that are close enough so that the bounding boxes meet. Then we can further suppose they are two sub-pieces of one original fractal piece, both of the same size
(all the possible ways of having two close arrows happen naturally in the fractal).
Let $A$ be the original fractal piece, and let $(A_k^n)_{0\le k<4^n})$ be the sequence of pieces obtained after $n$ interations. Now suppose we have two intersecting subpieces are $A_i^n$ and $A_j^n$. Then we get $\mu(A) \le \sum \mu(A_k^n) - \mu(A_i^n \cap A_j^n) = \mu(A) - \mu(A_i^n \cap A_j^n)$. Then $\mu(A_i^n \cap A_j^n) \le 0$, so that $\mu(A_i^n \cap A_j^n) = 0$
Since the grid of fractal pieces fill the plane we can deduce : $\mu(A) = 1/4$.
The fractal has non-empty interior if and only if there is an iteration where it has a grid of $16$ arrows as described in the picture above :
If it has nonzero interior, there is an area that will contain all the possible arrows. Conversely, if you draw the bounding boxes of all the other arrows from the plane around those $16$ arrows, they won't reach into a square area at the center of the diagram. This means that in that square we reach all the arrows from the original $16$, so that the whole square is in the image of the fractal.
This happens at the $9$th iteration (it may happen before I'm not sure), so $A$ has non-empty interior.
