Intuitively, why aren't all points in a circle covered in an Apollonian gasket? An Apollonian gasket is a figured formed by taking three mutually tangent circles, forming their Soddy circles, then recursively adding more circles by grabbing three circles and adding their Soddy circles. Here's an example:

I've read online that not all points in the enclosing outer circle are covered by the circles that make up the Apollonian gasket, though the set of uncovered points has measure zero. While I trust that this is indeed the case, I'm having trouble articulating a coherent intuitive argument that explains

*

*why not all points are covered in the Apollonian gasket, but

*why all points would be covered in other limiting self-similar tilings.

For example, consider the following way of tiling a square: subdivide the square into four congruent, smaller squares. Merge three of those squares into an L-shaped figure. Then recursively subdivide the remaining smaller square in the same way. That's shown here:

This process will eventually cover every point in the original square.
Is there an intuitive explanation for why Apollonian gaskets fail to cover every point in the outer circle while other sorts of tilings do eventually fill all of space?
 A: Why aren't all points are covered in the Apollonian gasket?
The question asks for intuition—my goal here is not to make things overly rigorous.  Let's look at how the image in the original might have been put together.  First, we might start by removing three closed disks from the original disk:[1]

Note that when these closed disks are removed, what remains of the original circle is divided into four non-empty open sets.  In the next step, remove a closed disk from each of these open sets.

Each of the three original open sets is now subdivided into three smaller non-empty open sets.  Indeed, at the $n$-th step of this process, the part of the original disk which has not been removed will consists of some number of non-empty open sets.  To get to the $(n+1)$-st step, we remove a closed disk from each one of those open sets, which leaves $3$ new non-empty open sets.
Roughly speaking, then, every time a disk is removed, three "new" regions are created, each of which must eventually be removed if this scheme is going to eventually remove the entire circle.  From an intuitive point of view, this can never happen—gaps are being created faster than they can be covered.
I will note that this isn't an entirely rigorous intuition, but the general idea can be made rigorous in this situation.
Why are all points covered in other limiting self-similar tilings?
They aren't.
In general, if you are trying to cover a set with other sets in a way that leaves "gaps" at every step, you will never completely cover the set.  There will always be gaps.  For example, in the covering of the square by L-shaped regions, the point in the upper-left corner is never covered.

[1] I created the images here by modifying the image from the original question.
A: One nice property of the Apollonian packing is that it is closed. Thus, if we take a non-repeating sequence of circles with each one tangent to the previous, then they will converge to some point in the packing. That gives us a way to find points in the limit set.
Now, in order for the resulting point to lie on the boundary of one of the circles in the packing, we simply need to arrange to continually choose circles tangent to the desired circle. Here's an illustration of the beginnings of such a sequence constructed to converge to a point on the largest upper left circle from the initial set of circles:

To find a point in the limit set that is not on the boundary of one of the initial circles, we simply need to construct our sequence in a different manner. One possibility is to choose, at each step, a circle that is tangent to the most recently chosen circles in our sequence. Since the sequence of radii of our sequence of circles converges to zero, we are then guaranteed that we won't converge to some single circle.
Here's an illustration of the beginnings of a sequence constructed in this second fashion:

In a lot of ways, this reminds me of finding points in the Cantor set that are not endpoints of the removed intervals. One way to solve both problems is to effectively address the points in the limit set so that you produce points in the set as a limit. In both cases, you actually have more flexibility when choosing addresses that avoid the boundaries of removed parts, indicating that the boundary is actually the smaller part of the limit set.
