# G/H is Hausdorff implies H is closed (General topology, Volume 1 by N. Bourbaki)

I am reading General topology, Volume 1 By Nicolas Bourbaki. I refer to the proof of Proposition 13. Could someone kindly explain the G/H Hausdorff $\implies$ H closed part of the proof? I understand that $H$ is an equiv class for the relation $x^{-1}y \in H$ bit, but I am failing to see how the Hausdorffness relates to $H$ being closed. I am also trying to understand the converse part of the proof which I think I'd be more successful in doing so if I understand the first part first. I am trying to self-learn topology, and I apologize for the stupidness of my questions on this site. Thanks in advance.

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## 3 Answers

Let's start with the definitions. If $G/H$ is Hausdorff, then given any two distinct points, I can put open balls around them that don't intersect. Let one such point be the orbit of 1, i.e. $1\cdot H=H$, and let $gH$ be any other point. Then, I can put an open ball around $gH$ that doesn't contain $1\cdot H$. Now, you need to use the definition of quotient topology: a ball in $G/H$ is open if its preimage in $G$ is open. So I can put an open ball around $g$ that does not intersect $H$. That is one characterisation of $G\backslash H$ being open.

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Thanks again, Alex! Grateful for the details. – brian Aug 19 '11 at 13:16
Dear Alex: What do you mean by "ball"? – Pierre-Yves Gaillard Aug 19 '11 at 13:22
@brian: If $G/H$ is Hausdorff, the point $eH$ is closed in $G/H$ and $H$ is the pre-image of that point under the canonical projection. – t.b. Aug 19 '11 at 13:22
@Pierre-Yves: touché – t.b. Aug 19 '11 at 13:33
Dear Alex: [A detail: I think you agree that a topological space whose points are closed is not Hausdorff in general.] What you say makes now a lot of sense to me! I think you prove slightly more than Theo, in the sense that you don’t take for granted the fact that “Hausdorff” implies “points closed”. To do a bit of nitpicking: The work you did in $G$ to prove that $G\setminus H$ is open, had you done it in $G/H$, you would have proved explicitly, and with the same amount of labor, that a point of a Hausdorff space is closed (or, if you prefer, that its complement is open). – Pierre-Yves Gaillard Aug 19 '11 at 19:15

If $G/H$ is Hausdorff then every $x \in G\setminus H$ has a neighbourhood disjoint from $H$. This means $G\setminus H$ is an open set, being the union of open sets, which means that $H$ is closed.

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Thank you, lhf. – brian Aug 19 '11 at 13:16

$G/H$ is Hausdorff implies that $G/H$ is $T_1$ implies that the singleton containing the coset $eH$ is closed and by the definition of the quotient topology, this is true if and only if its preimage under the canonical projection is closed, thus if and only if $H$ is closed in $G$.

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Thanks, Mark. Btw what does $T_1$ mean? – brian Aug 19 '11 at 16:18
@brian: It means that finite sets are closed. It's a separation property weaker than Hausdorffness (which is $T_2$). – Mark Aug 19 '11 at 16:22