# Trouble with plane embedding

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?

• Why do you doubt that $C\times S^1$ embeds into the plane? Are you asking about isometric embedding or about homeomorphism to a closed subset of the plane (or some other notion or embedding)? Commented Apr 11, 2018 at 19:12
• What am I missing here? $C\times S^1$ definitely embeds homeomorphically into the plane: just view $C$ as a subset of $[1,2]$ and rotate it round the origin. Commented Apr 11, 2018 at 19:13
• Yes is does, you are both correct... my apologies. I have edited the question now. Commented Apr 11, 2018 at 19:22
• Do you know if $C\times D$ embed into the plane, where $D$ is the closed disk?
– erz
Commented Apr 11, 2018 at 19:25
• @erz I just added that! I think the answer is no because the boundary of $D$ separates the plane. So no other copy of $D$ can limit to the "inside" of $D$. Commented Apr 11, 2018 at 19:27

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?)

• oh, I like this very much. so I guess my next question would be about atriodic continua for $X$. sorry for all the different questions, I am just trying to get a feel for this. Commented Apr 11, 2018 at 19:29
• do you know if $\mathbb Q\times X$ embeds into the plane? Commented Apr 11, 2018 at 19:32
• @David: As old questions are settled and new questions occur to you, please feel free to accept good answers to your old questions and then to open up new questions (burying new questions as comments to answers to old questions is not a good use of the site). Commented Apr 11, 2018 at 20:11
• @David: No, ${\mathbb Q}\times X$ does not embed into the plane. Reason: If $f$ is an embedding of $X$ into the plane, then there is an $\varepsilon >0$ such that if $g$ is another embedding of $X$ that differs from $f$ by less than $\varepsilon$, then $f(X)\cap g(X)\neq\emptyset$. Commented Apr 11, 2018 at 20:36
• @WlodekKuperberg Nice. This shows that in the space of embeddings of $X$ into the plane (with the uniform distance), every set of pairwise disjoint embeddings is discrete. In particular it has to be countable (as any discrete subset of a separable metric space), which is a proof for mathoverflow.net/questions/27244/… (modulo details of your assertion).
– YCor
Commented Apr 11, 2018 at 20:57