# A contractible non-planar continuum

Let $Z=\{0\}\cup\{\pm\frac1n\}_{n\in\mathbb N}$ be the sequence that converges to zero from both sides. Consider the contractible continuum $$A=(Z\times[-1,1]\times\{0\})\cup([-1,1]\times\{0\}\times\{0\})\cup(\{0\}\times\{0\}\times[0,1]),$$ which looks like an antenna. It is obvious that $A$ is 1-dimensional and contractible, so does not contain the forbidden Kuratowski graphs $K_{3,3}$ and $K_5$. It is also "obvious" that $A$ does not embed into the plane. Does anybody know a short and correct proof of the latter fact?

Problem. Give a short and correct proof of the non-planarity of the continuum $A$ (maybe with some references).

Thank you.

After thinking some time on this question I found a relatively simple solution based on the well-known fact that all arcs in the plane are ambiently homeomorphic. Using this fact and assuming that an embedding $f:A\to\mathbb R^2$ exists, we can modify $f$ and assume that $f(0,t,0)=(0,t)$ for all $t\in[-1,1]$. By the continuity of $f$, find $\epsilon>0$ such that $f(x,y,z)\in \mathbb R\times(-\frac12,\frac12)$ for any point $(x,y,z)\in A\cap [-\epsilon,\epsilon]^3$. Then use the connectedness argument to prove that the images of the sets $A_-=A\cap([-\epsilon,0)\times[-\epsilon,\epsilon]\times\{0\})$ and $A_+=(A\cap(0,\epsilon]\times[-\epsilon,\epsilon]\times\{0\})$ are containined in distinct connected components of $(\mathbb R\setminus \{0\})\times[-\frac12,\frac12]$. One of this components should contain also the connected set $f(\{(0,0)\}\times(0,\epsilon])$. Finally, applying a known lemma from Dimension Theorem (saying that two arcs linking opposite sides of a square have a common point) we can prove that the arc $f(\{(0,0)\}\times[0,\epsilon])$ has non-empty intersection with an arc $f(\{(\delta,0)\}\times[-1,1])$ for some small non-zero number $\delta$. But this contradicts the injectivity of $f$. This contradition completes the proof of the non-planarity of $A$.

So, sorry for boring you with this (relatively simple) question. (But the solution was not immediately evident).

Arguing by contradiction, assume that there is en embedding $\iota\colon A\to \mathbb{R}^2$.

Given a large integer $n$ consider the closed curve $\gamma_n$ formed by two arcs $$\iota(\{\pm\tfrac1n\}\times[-1,1]\times\{0\})$$ and two line segments $$[\iota(-\tfrac1n,\pm1,0),\iota(\tfrac1n,\pm1,0)].$$

Let $\Omega_n$ be the connected component of $\iota(0,0,0)$ in the complement $\mathbb{R}^2\backslash\gamma_n$.

Note that for some fixed $\varepsilon>0$ and all large integer $n$ we have $$\iota(\{0\}\times\{0\}\times[0,\varepsilon])\subset\Omega_n.$$

Further note that $$\bigcap_{n>N}\Omega_n\subset \iota([-1,1]\times\{0\}\times\{0\}).$$ That is $$\iota(\{0\}\times\{0\}\times[0,\varepsilon])\subset \iota([-1,1]\times\{0\}\times\{0\}),$$ a contradiction.

• Dear Anton thank you for your answer. I also thought this way (at first), but there are some problems in the above proof: first, the set $\gamma_n$ is not a closed curve. This can be fixed choosing an appropriate closed curve inside of $\gamma_n$. Next, the inclusion $\bigcap_{n>N}\Omega_n\subset \iota([-1,1]\times\{0\}\times\{0\})$ is not so clear: we should prove at first that the component $\Omega_n$ is bounded + check many other unpleasant details. So, the proof should be polished at least. (That is why I put this "obvious" question, realizing that it is not so obvious). – Taras Banakh Jun 12 '16 at 13:38
• @TarasBanakh $\gamma_n$ is closed (read carefully). You need to show that for fixed $\varepsilon>0$ the set $\Omega_n$ lies in $\varepsilon$-neighborhood of $\iota(\{0\}\times[-1,1]\times\{0\})$. If not you get a sequence of points $\iota(\pm\tfrac1n, a_n,0)$ which converges to a point outside of $\iota(\{0\}\times[-1,1]\times\{0\})$, a contradiction. – Anton Petrunin Jun 12 '16 at 13:43
• Anton, indeed, $\gamma_n$ is closed curve (but not necessarily a simple closed curve). The contradiction you derive is correct as soon as you prove that the component $\Omega_n$ (containing the point $\iota(0,0,0)$) is bounded. Why is it bounded? More precisely, at which place of the proof is it established? – Taras Banakh Jun 12 '16 at 13:54
• @TarasBanakh you are right --- I overlooked this part. Should not be hard --- let me think. – Anton Petrunin Jun 12 '16 at 18:07