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 ?



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


