Fractal curves area I was reading about the Sierpinski curve on the wiki page and it says that, considering the sequence of Sierpinski curves $S_n$ such that $\lim\limits_{n\rightarrow \infty}S_n$ completely fills the unit square,  the limit of the area enclosed by those curves is of $\frac{5}{12}$. 
I cannot think how it is possible that the limit of the areas enclosed by curves which are converging to a curve which fills the unit square can be different from $1$
 A: When we say that 
$$\lim_{n\to\infty} S_n = I^2 \text{ (the unit square)},$$
we are referring to a particular notion of distance. In particular, the set $S$ is "close" to the set $T$ if every point of $S$ is close to some point of $T$ and vice-versa. (This is really an intuitive description of the Hausdorff metric.)
The thing is, this notion of distance simply has nothing to do with area. It's quite possible that two sets are close to one another (with respect to this distance), yet have areas that are not close. Once you see a few simple examples, this makes perfect sense. One simple example, is a sequence of finer and finer checkerboard patterns:

Each of these images lies snugly inside the unit square. If we consider the set $S_n$ to be the region shown in black in the $n^{\text{th}}$ checkerboard pattern, then $S_n\to I^2$, yet the area of $S_n$ is 1/2 for all $n$.
For that matter, we could let $S_n$ denote the set of vertices in the $n^{\text{th}}$ checkerboard pattern. Then each $S_n$ is a finite set (so it has area zero), yet $S_n\to I^2$.
A: It shouldn't be surprising that the area inside the curve is less than $1$.  As the curve wiggles around it leaves lots of area outside.  If you look at the images of the construction on the page you link to there is plenty of area outside the curve.  If the square is $[0,1] \times [0,1]$ the segment $(\frac 12, 0)$ to $(\frac 12, \frac 12)$ including the first point but not the second is outside the curve.  There are lots of branches off it, like $(\frac 14,\frac 14)$ to $(\frac 14, \frac 34)$ as well.  I suspect you can play with the construction to make the area inside be anything you like in the range $(0,1)$.  I don't know if you can get the endpoints.  It is important not to confuse the area inside the curve with the area of the curve.  The area inside is a nice finite number at each stage of the construction.  It looks to me that it approaches the limit from below because at each stage we add more area in squares and trapezoids than we take out in half octagons.  The area of the curve is $0$ at every finite stage, as it is a line of finite length, so the limit of the area of the curve is zero.  On the other hand, the area of the limit curve is $1$ as it covers the whole square.  Taking the limit and computing the area do not commute.
