Let ~ be the equivalence relation on $[0, 1]$ $×$ {$0, 1$} defined by $(x,0)$ ~ $(x,1)$ if $x > 0$ and $x$ ~ $x$ for all $x$. Let $X^*$ be the corresponding quotient space with the quotient topology. Prove $X^*$ is not Hausdorff.

I have found myself hung up on the equivalence relation. If the quotient map $p$:$X$ → $X^*$ by mapping $x$ to its equivalence class under ~, what is $p^{-1}(x^*)$? Is it the two element set {$(x,0), (x,1)$}? I'm having troubles seeing what the open sets of $X^*$ actually are.

Any help is much appreciated.

  • $\begingroup$ Hint: Quotioning is like pasting, so one expects things like $A=\{\{(0,1)\}\}\cup\{\{(x,0),(x,1)\}:x\in ]0,1] \}\subset X^{\star}$ would be open. In fact, $p^{-1}(A)=\{[0,1]\times\{1\},\{0\}\times]0,1]\}$, which is open in $[0,1]\times\{0,1\}$ $\endgroup$ Mar 7, 2017 at 0:49

1 Answer 1


The elements of $X^*$ are the equivalence classes of $\sim$, so an element $x^* \in X^*$ is either of the form $x^* = \{(x,0),(x,1)\}$ for $x > 0$, or $x^* = \{(x,0)\}$ or $x^* = \{(x,1)\}$ for $x = 0$. So these are the three possibilities for the form of $p^{-1}(x^*)$.

For your question about open sets of $X^*$, recall the universal property of the quotient map: $U \subseteq X^*$ is open iff $p^{-1}(U) \subseteq X$ is open.

As an aside, have you tried visualizing $X^*$? You should imagine something like this

$\hspace{4.5cm}$ enter image description here

or maybe the following is a more accurate picture.

$\hspace{4.5cm}$ enter image description here

Do you see how something funny is happening near $x = 0$? Do you see why you can't separate the points $\{(0,0)\}$ and $\{(0,1)\}$?

In response to your question: denote the equivalence class of $(x,y)$ by $[x,y]$. Note that $p^{-1}(\{[0,0]\}) = \{(0,0)\}$. Since this set is not open in $X = [0,1] \times \{0,1\}$, then the singleton $\{[0,0]\}$ is not open in $X^*$ by the universal property I mentioned above.

Suppose $U_0$ and $U_1$ are open subsets of $X^*$ with $[0,0] \in U_0$ and $[0,1] \in U_1$. We aim to show that $U_0$ and $U_1$ are not disjoint. Observe that $p^{-1}(U_0)$ and $p^{-1}(U_1)$ are open subsets of $X$ with $(0,0) \in p^{-1}(U_0)$ and $(0,1) \in p^{-1}(U_1)$.

By definition of the subspace topology, there exist open balls $B_{0}$ and $B_{1}$ centered at $(0,0)$, $(0,1)$, resp., of radii $\epsilon_0$ and $\epsilon_1$, resp., such that \begin{align*} (0,0) \in B_{\epsilon_0} \cap \{y = 0\} &\subseteq p^{-1}(U_0)\\ (0,1) \in B_{\epsilon_1} \cap \{y = 1\} &\subseteq p^{-1}(U_1) \end{align*}

Choose an $x$ with $0 < x < \min\{\epsilon_0, \epsilon_1\}$. Can you show that $[x,0] = [x,1] \in U_0 \cap U_1$?

  • 1
    $\begingroup$ I have tried drawing a picture, which helped me see the form of $x^*$, it unfortunately made it harder for me to see why $X^*$ is not Hausdorff. I'm curious to see your picture to make sure I didn't do something stupid. Also, the case for $x < 0$ should not occur, as $X$ is only on $[0,1]$, no? Is the case at $x=0$ the key to the problem? $\endgroup$ Mar 7, 2017 at 0:19
  • $\begingroup$ Oh, whoops, you're very right. But $x =0$ does still occur, and that's the important point. I'd better change my picture! $\endgroup$ Mar 7, 2017 at 0:22
  • $\begingroup$ I feel like the solution is staring me in the face at this point... I'm confused what happens when we take $p^{-1}(0,0)$ and likewise for $(0,1)$. How do we know these two points are not open sets? The point $x = 0$ is not defined under the equivalence relation. $\endgroup$ Mar 7, 2017 at 0:36
  • $\begingroup$ Ok, so since $[0,0]$ and $[0,1]$ are not open sets, what open set do they belong to? Do they only belong to $X^*$? Thank you very much for all your help by the way! $\endgroup$ Mar 7, 2017 at 0:50

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.