Let $H$ be the intersection of all neighborhoods of $0$ in a topological abelian group.
On page 102 of the book introduction to commutative algebra by Atiyah and Macdonald, the fourth line of the proof of Lemma 10.1, it said that ii) implies that the cosets of $H$ are all closed. From ii) we know that $H$ is closed since $H=\overline{\{0\}}$. But if $y \in G\backslash H$, how can we show that $yH$ is closed?
On page 102 of the book introduction to commutative algebra, the 5-th line of the proof of Lemma 10.1, it is said that the converse is trivial. If $H=0$, then $G/H=G$. Since it is shown that $G/H$ is Hausdorff in iii), we know that $G$ is Hausdorff. But how to show that if $G$ is Hausdorff, then $H$ (the intersection of all neighborhoods of $0$) is $0$?
One equivalent definition of a Hausdorff space is that: Any singleton set ${x} ⊂ X$ is equal to the intersection of all closed neighbourhoods of x. (A closed neighbourhood of x is a closed set that contains an open set containing x.) But here $H$ is the intersection of all neighborhoods of $0$ not the the intersection of all closed neighborhoods of $0$.
Thank you very much.
(1) Multiplication on the left is a homeomorphism (why?), so $\,H\,$ is closed $\;\implies yH\;$ is closed for all $\,y\in G\;$
(2) Suppose $\;G\;$ is Hausdorff yet there exists $\,0\neq y\in H\;$ . But then for any open $\,U\,$ s.t. $\,y\in U\,$ we have that $\,U\cap W\neq\emptyset\;\;\forall\;$ the open neighborhoods $\,W\,$ of zero, and since any open $\,V\,$ s.t. $\;0\in V\;$ is a neighborhood of zero we get $\;G\;$ is not Hausdorff