Show that an attractor of an IFS is not totally connected The question asks to show that  the atttractor, A, of the IFS 
$$F=\bigg\{\mathbb R^2: \bigg(\frac{x}{2},\frac{y}{2}\bigg),\bigg(\frac{x+1}{2},\frac{y}{2}\bigg),\bigg(\frac{x}{4},\frac{y+3}{4}\bigg)\bigg\} $$
is neither totally connected nor totally disconnected. 
I have already shown that it is not totally connected by using the theorem which says each point on a totally disconnected attractor has at exactly one address and finding a point with multiple addresses.
Now I'm stuck on the proof that it is totally disconnected. I realize the attractor looks something like a Sierpinski triangle, where the top triangle is disconnected from the bottom two triangles. By observation, I can see this is clearly not connected since you could draw a small enough open set around the top triangle and another open set around the bottom two triangles which would contain the attractor but whose intersection would be empty. I'm just struggling to figure out how to put this into a rigorous proof. Thanks!
 A: An alternative way to show that the set is not totally disconnected is to focus on the sub-IFS consisting of the first two functions. These are just contractions of ratio $1/2$ about the points $(0,0)$ and $(1,0)$. The invariant set is, therefore the line segment between these points. This line segment forms the "base" of the invariant set.
To show that the invariant set $E$ consists of infinitely many connected components, we could first use the fact that (since the functions in the IFS are all strict contractions with no rotation or reflection), $E$ is contained in the convex hull of the fixed points of the functions. Let's call this convex hull $T$, as it's a triangle. Since the third function in the IFS pulls the base up to the height $3/4$ and the first two pull the top down to height $1/2$, the horizontal line at $y=5/8$ misses
$$\bigcup_{i=1}^3 f_i(T),$$
where the $f_i$s are the functions in the IFS. By iterating the first two functions in the IFS on the point $(x,5/8)$, we see that $E$ misses countably many lines. This illustrated in the figure below:

