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I'm currently reading this article and in the proof of Lemma 4.3 the authors write the following (here $R^D$ is just the notation for an equivalence relation over $D$):

"... compactness of $R^D$ easily implies that $D/R^D$ is Hausdorff."

I have tried to prove this with no success at all, but my approach is trying to prove it without the aditional hypothesis given in the Lemma, like this: If $X$ is a compactum (i.e. compact and Hausdorff) and $R$ is an equivalence relation over $X$ which is compact on $X\times X$ then the quotient $X/R$ is Hausdorff. (?)

What I know so far is that each equivalence class must be compact, the quotient map is closed and I've found this result which solves the problem when te quotient map is open. Any help would be aprreciated.

Elma
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  • $R^D$ would indeed be closed in $X \times X$ (because $X \times X$ is Hausdorff and a compact subset is closed in a Hausdorff space) and then the linked question would solve your issues. But I see that that assumes $q$ is open as well, and this is not clear. – Henno Brandsma Jun 24 '21 at 22:18
  • Indeed, if it were the case that $q$ is open, then there is nothing to do but in general we can't say that $q$ will be open. – Elma Jun 25 '21 at 17:14

1 Answers1

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Wow this is indeed a true statement. I was surprised by how this is not mentioned in basic topology textbooks.

Proof: Suppose $[x],[y]$ are distinct equivalence classes. Since the relation is compact, the classes are also compact and since $X$ is $T_4$, $[x]$ and $[y]$ can be separated by two disjoint open sets, say $U$ and $V$, respectively. Now, we already know that the quotient map $q\colon X \to X/R$ is closed and Proposition 2.4.9 of Enegelking's book on general topology states the following:

For an equivalence relation $E$ over $X$ the following statements are equivalent:

1. The quotient map is closed.

2. For every open set $A\subseteq X$ the union of all equivalence classes that are contained in $A$ is open in $X$.

We can take the union of all equivalence classes contained in $U$ and the union of all equivalence classes contained in $V$ to be $R-$saturated disjoint sets that contain $[x]$ and $[y]$ respectively. This gives that $X/R$ is Hausdorff.

Elma
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