# Prove that the “theta-space” is not homeomorphic to $S^{1}$.

Let $X$ be the "$\theta$-space": \begin{equation*} X = \{ (x,y) \in \mathbb{R}^{2} \colon x^{2} + y^{2} = 1 \} \cup \{ (x,0) \colon -1\leq x \leq 1 \}. \end{equation*} Prove that $X$ is not homeomorphic to $S^{1}$.

My initial thought was to show that $X$ is not compact. I was thinking that the collection of open balls with radius $r>0$, $\{ B(0,r) \}$, is an open cover with no finite subcover. I am not very confident with my level of understanding so can someone tell me if I am on the right track or not.

Unfortunately for that idea, $X$ is compact: it’s clearly bounded, and it’s not hard to check that $X$ is closed in $\Bbb R^2$ as well. Try this idea instead.
• If you remove any two distinct points of $S^1$, how many components does the remaining set have?
• Can you find two points of $X$ that can be removed to leave a different number of components from that?
• Ahh thank you for clarifying what the definition of components meant. I have been reading the definition and trying to put it into play but couldn't figure it out. So if we remove 2 points from $S^{1}$, we have two components, and if we remove the points $(\pm 1, 0)$ from $X$, we have 4 components. Am I on the right track now or am I still way off? – Jack Mar 1 '16 at 5:38
• @Tim: You’re almost right: you miscounted the components of $X\setminus\{\langle 1,0\rangle,\langle-1,0\rangle\}$. Can you fix it? – Brian M. Scott Mar 1 '16 at 5:39
You can also give an alternative proof using $\pi_1$. Since the theta-space is homotopic to the figure eight space i.e $\mathbb{S}^1 \vee \mathbb{S}^1$, it has fundamental group $\mathbb{Z} * \mathbb{Z}$. However, the fundament group of $\mathbb{S}^1$ is $\mathbb{Z}$. Since homeomorphic spaces have isomorphic fundamental groups, your result follows.