# Subsets of the plane such that there is no embedding between them

Let $X=\mathbb{S}^{1}$ denote the unit circle and let:

$Y=\{(0,y) \in \mathbb{R}^{2}: -1 \leq y \leq 1\} \cup \{(x,0): 0 \leq x \leq 1\}$.

Prove that $X$ cannot be embedded in $Y$ and $Y$ cannot be embedded in $X$.

Well certainly I can see that $X$ and $Y$ are not homeomorphic, remove the origin $(0,0)$ from $Y$ then $Y \setminus \{(0,0)\}$ is not path connected while $X$ minus a point is. However I don't see how to prove $X$ cannot be homeomorphic to any subspace of $Y$ and vicerversa. Any ideas?

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To show that $X$ cannot be embedded in $Y$, perhaps you could use a compactness argument to show that any continuous map $X \to Y$ cannot be injective? The idea is that any such map would achieve a ‘maximum’... – Zhen Lin Aug 7 '11 at 10:24
I would say that every connected subset of $Y$ is contractible while the circle isn't. But this may not be the most elementary way. Conversely, I'd say that every connected subset of $S^1$ is a topological manifold, while $Y$ isn't. Again with the same proviso on elementary. – t.b. Aug 7 '11 at 10:36
@Theo Buehler: could you please ellaborate more on your answer?, I'm not familiar with those concepts but I can study them on my own and try to understand your answer :). – user10 Aug 7 '11 at 10:39
See contractible space and topological manifold on Wikipedia. Note that no neighborhood of $0$ in $Y$ is homeomorphic to an open subset of $\mathbb{R}$ essentially by the argument you give. Every connected subset of $S^1$ is either a point, homeomorphic to an interval in $\mathbb{R}$ or $S^1$ itself. – t.b. Aug 7 '11 at 10:44

I am thinking along these lines: First, show that $X$ is not homeomorphic to any subspace $Z\subseteq Y$ when $Z$ is contained in of any of the three "arms" $$\{(0,y) \in \mathbb{R}^{2}: 0 <y \leq 1\},\qquad\{(0,y) \in \mathbb{R}^{2}: -1 \leq y <0\},\qquad\{(x,0): 0 \leq x \leq 1\}.$$ Also, because $X$ is connected, we must also have that $Z$ is connected. A union of any non-empty subspaces of two or more arms is necessarily disconnected, and so such $Z$'s can be ruled out.

Therefore, if $X$ were homeomorphic to any subspace $Z\subseteq Y$, then $Z$ must contain the origin.

But any $Z$ that is connected and contains the origin is a star domain (this is certainly true, but I don't see a slick way of proving it), and therefore is contractible, while $X$ is not.

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Continuing Zev Chonoles answer, if $X$ were homeomorphic to a subspace $Z\subset Y$ containing the origin then removing the image of $(0, 0)$ from the circle we should obtain a disconnected space, but we don't.

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+1 Aha, a much better finish to my approach! I'm not sure how elementary the non-contractibility of the circle is considered, but the classic remove-a-point-to-disconnect observation is certainly more elementary than that. – Zev Chonoles Aug 7 '11 at 11:05

First let us prove that $Y$ has no subspaces homeomorphic to $\mathbb{S}^1$. In order to reach a contradiction suppose $A\subseteq Y$ is homeomorphic to $\mathbb{S}^1$. Notice that $Y\setminus\{(0,0)\}$ has three connected components, which are $Y_1=\{(0,y):0<y\leq 1\}$, $Y_2=\{(0,y):-1\leq y<0\}$ and $Y_3=\{(x,0):0<x\leq 1\}$. This means that $A\setminus(0,0)$, being connected, is contained in exactly one of the sets $Y_i,i\in\{1,2,3\}$. Hence $A$ is contained in exactly one of the sets $Y_i\cup\{(0,0)\},i\in\{1,2,3\}$. Because all of the $Y_i\cup\{(0,0)\}$ are homeomorphic to a (closed) interval and any connected subset of an interval is an interval, $A$ is homeomorphic to an interval containing more than one point. This is a contradiction, because removing any point from $A$ does not disconnect it, but any interval containing more than one point has a point such that the interval minus this point is disconnected (such a point is called a cut point).

Then let us prove that $\mathbb{S}^1$ has no subspaces homeomorphic to $Y$. Again, to reach a contradiction suppose $B\subseteq\mathbb{S}^1$ is homeomorphic to $Y$. Notice that $B\neq \mathbb{S}^1$, since $\mathbb{S}^1$ does not have cut points but $B$ has. This means that there is a point $p\in\mathbb{S}^1$ such that $B\subseteq\mathbb{S}^1\setminus\{p\}$. But $\mathbb{S}^1\setminus\{p\}$ is homeomorphic to an (open) interval, so we can deduce that $B$ is homeomorphic to an interval as before. This is a contradiction because $B$ has a point $q$ (corresponding to $(0,0)\in Y$) such that $B\setminus\{q\}$ has three components (corresponding to $Y_1$, $Y_2$ and $Y_3$), but an interval which has a point removed has at most two components.

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