Let $C$ be the middle-thirds Cantor set. Obviously $C\times [0,1]$ embeds into the plane. But $C\times D$ does not, $D$ being a closed disc in the plane.


Are there any general results which can be applied to sets like this (Cantor set times a plane set) to see if they do, or do not, embed into the plane?


Is there a 1-dimensional continuum $X\subseteq \mathbb R ^2$ such that $C\times X$ does *not* embed into the plane? How about if $\mathbb R ^2\setminus X$ is path-connected?



Trouble with plane embedding
I can answer the last question.


Let $X$ be a "tripod" in $\mathbb{R}^2$. For concreteness, let
$$ X = ([-1,1]\times \{0\}) \cup (\{0\}\times [0,1]).$$


Then $X$ is a $1$-dimensional plane continuum such that $\mathbb{R}^2 \setminus X$ is path-connected.


But $C\times X$ does not embed into the plane: only countably many disjoint topological tripods fit in the plane. (See [How many tacks fit in the plane?](https://mathoverflow.net/questions/27244/how-many-tacks-fit-in-the-plane))


