How many similar disjoint drawings can be done in $\mathbb{R}^2$? Long ago I encounter the following question:


*

*How many disjoint similar $0$'s can be drawn in $\mathbb{R}^2$?


Answer is simple, it just suffices to draw $0$'s centered at the origin making them with increasing diameter over the real numbers. They all are disjoint and there are not enumerable ones of them.


*

*How many disjoint similar $8$'s can be drawn in $\mathbb{R}^2$?


This case is more interesting. Every disjoint $8$ can be uniquely mapped to a pair of elements in $\mathbb{Q}^2$, where the first one is in the upper half of the $8$ and the second one in the lower half. There are only enumerable of those.
A similar technique could be used to proof that there are enumerable $X$'s or $H$'s. However, I cannot figure out the condition that the letter (or drawing) must have in order for its disjoint similar drawings to be enumerable in $\mathbb{R}^2$.

I have been thinking about this with letters and numbers, but I am sure it can be generalized to more interesting (even discontinuous) figures and more interesting deformations of the figure than just rotating or increasing its size maintaining its proportions.
 A: I suppose that among all path-connected spaces, the letter "Y" is the basic obstacle (and "8", "X", "H" all contain a subspace homeomorphic to "Y"): with each embedding $Y\to\Bbb R^2$, we can associate a finite collection of rationals: If we denote the branch point as $a$, and the three end points as $u,v,w$, then pick a rational $r$ with $r>0$ and $r<\min\{|a-u|,|u-v|,|a-w|\}$.
Then we can pick a rational point $a'\approx a$ such that the $r$-circle around $a'$ intersects all three branches. If we follows the three branches only up to their first intersection with the circle, they partition the interior of the circle into three parts (two branches and an arc each form a Jordan curve). Pick a rational point in each. 
Now one can argue that whenever we have two disjoint Y-embeddings with the same rational data chosen, they lead to a solution of the famous gas-water-electricity problem, which is impossible. Or directly with the Jordan curve theorem: Suppose two Y-shapes lead to the same rational points $a,b,c$ on a rational circle, one having end points $u,v,w$ on the circle with $u$ on arc $bc$, $v$ on arc $ca$, $w$ on arc $ab$. For each pair of branches, by completing them with arcs running slightly outside the given circle, we obtain a simple closed Jordan curve. Any other Y-shape with the same rational points associated then must lie completely inside one of these three closed curves. But this is impossible: If insinde the curve based on the brabches ending in $u$ and $w$, say, the arc $ab$ is completely outside the Jordan curve, but contains one end of the Y-shape.
On the other hand, any path-connected and locally path-connected space not having a "Y-shaped" subspace and that can be embedded at all into $\Bbb R^2$, is an interval or circle (incidentally, also describable as letters: "I" and "O") or at worst a single point (".").
