TITLE: G/H is Hausdorff implies H is closed (General topology, Volume 1 by N. Bourbaki)
QUESTION [9 upvotes]: 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.

REPLY [5 votes]: $G/H$ 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 $G$.