Can $\mathbb{R}^2$ be covered by disjoint sets homeomorphic to the union of the segments $[(0,0), (0,1)], [(0,0), (1,1)], [(0,0), (1,0)]$? [duplicate]

Question

This question was asked at the french ENS oral examination. I do not really know how to approach it. I think the answers no.

What I've gathered so far :

Lets call $T$ the subset of $\mathbb{R}^2$ in the title (for obvious reasons). If the union exists, by Baire's theorem it must be uncountable. Let $E$ be the set of maps $T \to \mathbb{R}^2$ of which the images disjointly cover $\mathbb{R}^2$.

Since the domain is compact every map is uniformly continuous : by uncountability, for all $\epsilon > 0$ there is an $\eta$ such that an uncountably infinite number of maps $f$ verify the property $|x - y| \leq \eta \implies |f(x) - f(y)| < \epsilon$.

By uncountability, I also think an uncountably infinite number of the maps above should also have their image in a well chosen compact region of the plane, denoted $K$.

Let $E'$ be an uncoutably infinite subset of $E$ where all the maps verify the two properties ("uniform" uniform continuity and image in a given compact set $K$). For any finite subset of $T$, by using successive extractions I should be able to find a sequence of maps $(f_n)$ in $E'$ such that for any point $x$ in this finite subset the images $f_n(x)$ converges to a point $y \in K$. By using uniform uniform continuity I hoped to find that the maps $f_n$ would be arbitrarily close to one another and necessarily cross.

However, the more I think about this approach the less likely I think it is to work.

Would anyone have an idea ?

Answer 1

Such sets are called triods. R. L. Moore (Concerning triods in the plane and the junction points of plane continua, Proceedings of the National Academy of Sciences USA, vol. 14, 1928, pp. 85-88) proved that every set of pairwise disjoint triods in the plane is countable.

https://www.pnas.org/doi/abs/10.1073/pnas.14.1.85

Answer 2

One sketch is as follows: For each Y-set, we can put a small open disk in the vertex, such that the three prongs divide the disk in three sectors. In each sector, we can find a rational coordinate, since the disk is open, so for each Y-figure, we can associate a triple of rational points.

Now, no two Y-figures can produce the same triple of rational points, so the set of Y-figures must produce an injection to sextuples of rational numbers. Hence, we can only have a countable many disjoint Y-figures in the plane.

Answer 3

The question been answered in How many tacks fit in the plane?.

In fact Moore proved a multidimensional version of the theorem: only countably many sets homeomorphic to the $n$-dimensional disc with an orthogonal segment attached in the center can be embedded in $\mathbb{R}^{n+1}$ as pairwise disjoint sets.

While it sounds like an abstract topological problems, it has apllications to Sobolev spaces: https://mathoverflow.net/a/297607/121665.