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> Let $G$ be a graph with disjoint copies of $K\_{1,3}$. Prove that if there are uncountably many copies of $K\_{1,3}$ in $G$, then $G$ is not planar.
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I have a proof of this statement by contradiction i.e. assuming it is planar with uncountably many copies of $K\_{1,3}$ in $G$, but I am not satisfied. I am wondering if anyone has a proof that proves the contrapositive i.e. if $G$ is planar, then there is only countably many copies of $K\_{1,3}$ in $G$.


My idea is to find an embedding of $G$ onto $\mathbb{R}^2$ (or $\mathbb{C}$), and use the fact that $K\_{1,3}$ has $3$ edges, and hence if we find disjoint open sets which contain each $K\_{1,3}$, they will have positive measure.


Thanks a lot!



Planarity of infinite graphs
This is a 1928 theorem of R. L. Moore "[Concerning Triods in the Plane and the Junction Points of Plane Continua](http://www.jstor.org/stable/85527)" PNAS Vol. 14, No. 1 (Jan. 15, 1928), pp. 85-88. Greg Kuperberg gave a [nice proof](https://mathoverflow.net/a/27248/297) of it here on MO. 


