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?
  • $\begingroup$ 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? $\endgroup$ – Jack Mar 1 '16 at 5:38
  • $\begingroup$ @Tim: You’re almost right: you miscounted the components of $X\setminus\{\langle 1,0\rangle,\langle-1,0\rangle\}$. Can you fix it? $\endgroup$ – Brian M. Scott Mar 1 '16 at 5:39
  • $\begingroup$ oh duh, it's three! Thank you! This is the first time I have posted something I was totally confused about and such a simple answer made it so clear. I appreciate the help! $\endgroup$ – Jack Mar 1 '16 at 5:42
  • $\begingroup$ @Tim: Three it is. You’re welcome! $\endgroup$ – Brian M. Scott Mar 1 '16 at 5:43

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.


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.