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?
 A: 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.
A: 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.  
A: 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.
