Let us say for an arbitrary topological space $X$ that it has property $\dagger$ if for any closed equivalence relation $\sim$ on $X$ (closed as a subset of $X^2$), the quotient space $X/{\sim}$ is Hausdorff.

Is there a more "intrinsic" characterisation of property $\dagger$ (in terms of separation axioms of some sort, perhaps)?

I believe compact (possibly non-Hausdorff) spaces have $\dagger$, as in their case for a closed $\sim$, the quotient map is closed, and consequently the product $X^2\to (X/{\sim})^2$ is also closed (and sends $\sim$ to the diagonal).

On the other hand, it is easy to see that if $X$ is Hausdorff and has this property, then it must be normal: otherwise the equivalence relation that identifies all points in two supposedly inseparable closed sets (separately) and leaves all others untouched will be closed will have a non-Hausdorff quotient.

(As a side note, this is true for arbitrary (not neccessarily Hausdorff) topological groups if we restrict $\sim$ to be the relation of lying in the same coset of a subgroup, because the quotient map is open in this case.)

Edit: I've been looking for some references about group actions and stumbled upon a "Lemma" in Duistermat and Kolk's "Lie groups" book which states that $M/{\sim}$ is Hausdorff iff $\sim$ is closed. This is not true (as shown by the above example in a non-$T_4$ space) and I believe I see the mistake made by the authors (they seem to have assumed that the quotient mapping is open), but it has reminded me of this question (about which I'm still curious).

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    $\begingroup$ Closed equivalence? You mean the quotient is $T1$? $\endgroup$ – user2345215 Nov 16 '14 at 16:30
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    $\begingroup$ Is a closed equivalence relation one where $\sim \subset X^2$ is closed? Or one where the quotient map is a closed continuous function? $\endgroup$ – Henno Brandsma Nov 16 '14 at 16:54
  • $\begingroup$ @HennoBrandsma: one where ${\sim}\subseteq X^2$ is closed. The quotient is trivially $T_1$ (actually, it's enough for classes of $\sim$ to be closed for that). It is also trivially Hausdorff if we assume that the quotient map is closed (or open). The point is, if $X/{\sim}$ is Hausdorff, then $\sim$ is obviously closed, but the converse is not always true. $\endgroup$ – tomasz Nov 16 '14 at 20:02
  • $\begingroup$ @user2345215: I'm not sure what you're asking about, but perhaps my previous comment answers your question. $\endgroup$ – tomasz Nov 16 '14 at 20:03
  • $\begingroup$ @tomasz If $X/\sim$ is Hausdorff, then $q^2: X^2 \rightarrow (X/\sim)^2$ is continuous, and $(q^2)^{-1}[\Delta_{X/\sim}] = \sim$, which is indeed closed in that case. $\endgroup$ – Henno Brandsma Nov 16 '14 at 20:10

$\def\RR{\mathbb{R}}$There is no separation condition which will do the job. That's a vague statement, so here is a precise one: There is a subset of $\mathbb{R}^2$ which (equipped with the subspace topology) does not have condition $\dagger$.

Proof: Let $A$ and $B$ be two disjoint dense subsets of $\mathbb{R}$, neither of which contains $0$. (For example, $\mathbb{Q} +\sqrt{2}$ and $\mathbb{Q}+\sqrt{3}$.) Let $$X = (A \times \RR_{\geq 0}) \cup (B \times \RR_{\leq 0}) \cup (\{0\} \times \RR_{\neq 0}) \subset \RR^2.$$ Define $(x_1,y_1)$ and $(x_2, y_2)$ to be equivalent if $x_1=x_2$ and, in the case that $x_1=x_2=0$, that $y_1$ and $y_2$ have the same sign.

Verification that this is a closed equivalence relation: $X^2$ is a metric space, so we can check closure on sequence. Let suppose we have a sequence $(x_n, y_n) \sim (x'_n, y'_n)$ with $\lim_{n \to \infty} x_n=x$, $\lim_{n \to \infty} y_n=y$, $\lim_{n \to \infty} x'_n=x'$ and $\lim_{n \to \infty} y'_n=y'$. We must verify that $(x,y) \sim (x',y')$. First of all, we have $x_n = x'_n$, so $x=x'$ and, if $x=x' \neq 0$, we are done. If $x=x'=0$, we must verify that $y$ and $y'$ have the same sign. But $y_n$ and $y'_n$ weakly have the same sign for all $n$, so they can't approach limits with different signs.

Verification that $X/{\sim}$ is not Hausdorff: We claim that no pairs of open sets in $X/{\sim}$ separates the images of $(0,1)$ and $(0,-1)$. Suppose such open sets exist, and let $U$ and $V$ be their preimages in $X$. Then there is some $\delta$ such that $(A \cap (-\delta, \delta) )\times \RR_{\geq 0} \subset U$ and $(B \cap (-\delta, \delta) )\times \RR_{\geq 0} \subset B$. Then $U \cap \RR \times \{ 0 \}$ is an open set which contains $(-\delta, \delta)$. By the density of $B$, there must be a point of $B \cap (- \delta, \delta)$ in $U \cap \RR$, and then this gives an intersection between $U$ and $V$.

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  • $\begingroup$ That's a nice example, thanks! In hindisght, I should have seen it coming, some sort of compactness/completeness condition is definitely necessary. Otherwise, I wouldn't be surprised if we could prove the following abstraction of your example: if $X$ is any topological space and $E$ is an equivalence relation such that $X/E$ is not Hausdorff, then there is some $X_0\subseteq X$ such that $E$ restricted to $X_0$ is closed and $X_0/E$ is still not Hausdorff. $\endgroup$ – tomasz Mar 13 '18 at 23:45
  • $\begingroup$ @DavidESpeyer thanks, I encountered this same problem .. but how does one come up with construction of such an example ? $\endgroup$ – Noob mathematician Sep 19 at 9:49

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