Let $D$ be the disk in the complex plane, $D = \{ z : |z|<1\}$. Let $\sim$ be an equivalence relation on $D$ defined by $z_1\sim z_2$ iff $|z_1| = |z_2|$. Is the quotient space $D/{\sim}$ is Hausdorff?

I am looking for two open neighborhoods near $0$ and I think these two intersect. I guess it is not Hausdorff. But I can't prove it mathematically. Hints and ideas are welcomed.

  • 1
    $\begingroup$ Isn't this quotient just homeomorphic to $[0,1) \subset \mathbb{R}$, and subspaces of Hausdorff spaces are Hausdorff? $\endgroup$
    – Josh Chen
    Jun 11, 2015 at 3:08
  • $\begingroup$ @JoshChen Yes, an alternative is to prove the homeomorphism. $\endgroup$
    – Eoin
    Jun 11, 2015 at 3:09

1 Answer 1


The space is Hausdorff.

The equivalence classes of this equivalence relation are the circles $C=\{z : |z|=r\}$ for some $r<1$. So you need to think about whether or not you can separate two distinct circles by open sets. Let $C_i=\{z: |z|=r_i\}$ for $i=1,2$, and let $r_1<r_2$, without loss of generality. Then consider the image under the quotient map of the sets

$U=\{z \in \mathbb{C} : |z|<\dfrac{r_1+r_2}{2}\}$

$V=\{z \in \mathbb{C}: \dfrac{r_1+r_2}{2}<|z|<1\}$.


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