8

Suppose $X$ a metric space and $\sim$ an equivalence relation on $X$. Is the space $X/\mathord{\sim}$ metrizable? I think that the answer is no, but I could not arrive at a counterexample.

Taumatawhakatangihangakoauauot
  • 6,613
idm
  • 11,824
  • 3
    Look up the line with two origins, it isn't Hausdorff, but it's the quotient of a metrizable space. – jxnh Aug 21 '14 at 14:39
  • 2
    The cone of $(0,1)$ is not first-countable at the apex, see http://math.stackexchange.com/questions/590205/show-that-the-cone-of-the-open-interval-0-1-can-not-be-embedded-in-any-euclid/594042#594042 – Stefan Hamcke Aug 21 '14 at 14:41
  • In view of the answers which show that the quotient need not be Hausdorff, the more natural question is whether the quotient topology is generated by a pseudo-metric (where $d(x,y)=0$ does not imply $x=y$). This question is answered by the comment of Stefan Hamcke. – Jochen Nov 27 '23 at 16:02

2 Answers2

7

A very simple way in which the quotient can fail to be metrizable is if there are equivalence classes that are not closed. Take your original space to be $\mathbb{R}$ and let $\sim$ have as its equivalence classes $(0,1)$ and all singletons $\{x\}$ with $x\notin(0,1)$. In the quotient topology, you can not separate $(0,1)$ and $1$ by open sets and the quotient space fails to be Hausdorff.

Michael Greinecker
  • 32,841
  • 6
  • 80
  • 137
2

If $X$ is a standard flat torus and the equivalence relation declares two points to be equivalent if they lie on the same line with a (fixed) irrational slope, the quotient is clearly not metrizable.

Mikhail Katz
  • 42,112
  • 3
  • 66
  • 131