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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 $T1$? $\endgroup$ Nov 16, 2014 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? $\endgroup$ Nov 16, 2014 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, 2014 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, 2014 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. $\endgroup$ Nov 16, 2014 at 20:10

2 Answers 2

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$\def\RR{\mathbb{R}}$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, 2018 at 23:45
  • $\begingroup$ @DavidESpeyer thanks, I encountered this same problem .. but how does one come up with construction of such an example ? $\endgroup$ Sep 19, 2020 at 9:49
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Theorem. Suppose that $X$ is a compact metrizable space and $R$ is a closed equivalence relation on $X$. Then $X/R$ is also Hausdorff (in fact, even metrizable).

Proof. I will need the following definition:

Definition. The equivalence relation $R$ on a topological space $X$ is usc (upper semicontinuous) if every $R$-equivalence class $[x]\subset X$ is compact and for every $R$-equivalence class $[x]\subset X$ and every (open) neighborhood $U$ of $[x]$ in $X$, there exists a neighborhood $V$ of $[x]$ in $X$ such that whenever an $R$-equivalence class $[y]\subset X$ has nonempty intersection with $V$, then $[y]\subset U$.

Here is how to think of this condition in the case when $(X,d)$ is a compact metric space: First of all, each equivalence class should be closed in $X$. Secondly, for every sequence $[y_n]$ of equivalence classes in $X$, if $d_{min}([x], [y_n])$ converges to zero, then $d_{max}([x],[y_n])$ also converges to zero.

Here $$ d_{min}([x],[y])=\min \{d(x',y'): x'\in [x], y'\in [y]\} $$ and $$ d_{max}([x],[y])=\max \{d(x',y'): x'\in [x], y'\in [y]\} $$
(the latter is also known as the Hausdorff distance between $[x]$ and $[y]$).

The relevant results are Propositions 1 and 2 on page 13 in

Robert J. Daverman, "Decompositions of manifolds." Reprint of the 1986 original. AMS Chelsea Publishing, Providence, RI, 2007.

which state that if $X$ is Hausdorff (resp. metrizable) and $R$ is usc, then $X/R$ is also Hausdorff (resp. metrizable).

[I can add a proof of Proposition 1 since it is an elementary exercise, but Proposition 2 is too long.]

Note that if $X$ is compact and $R$ is closed, then every $R$-equivalence class in $X$ is also closed, hence, compact. It remains to show:

Lemma. If $(X,d)$ is a compact metric space, then every closed equivalence relation $R$ on $X$ is usc.

Proof. Let $[x]$ be an $R$-equivalence class in $X$, and $U\subset X$ its neighborhood. I will consider a sequence $V_n$ of open metric neighborhoods of $[x]$: $$ V_n= \{z\in X: d(z, [x])< 1/n\}. $$ Here $d(z,A)=\inf \{d(z,a): a\in A\}$.

Suppose that for each $n$, there exists an $R$-equivalence class $[y_n]$ such that $y_n\in V_n$ and $[y_n]\cap X\setminus U \ne \emptyset$. Then we find $y'_n\in [y_n]$ such that $y'\in K=X\setminus U$; $K$ is a compact subset. Then, after passing to a subsequence, $(y_n)$ converges to some $y\in [x]$ while $(y'_n)$ converges to some $y'\in K$. Since $K\cap [x]=\emptyset$, it follows that $(y,y')\notin R$. At the same time, $\lim_n (y_n,y'_n)= (y,y')$ and $(y_n, y'_n)\in R$. This contradicts the assumption that $R$ is closed.

Thus, there exists $n$ such that $V=V_n$ has the property that whenever $[y]\cap V\ne \emptyset$, $[y]\subset U$. Hence, $R$ is usc. qed

Combining the lemma with Propositions 1 and 2, we conclude the proof of the theorem. qed

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