# Intersection of neighborhoods of 0. Subgroup?

Repeating for my exam in commutative algebra.

Let G be a topological abelian group, i.e. such that the mappings $+:G\times G \to G$ and $-:G\to G$ are continuous. Then we have the following Lemma:

Let H be the intersection of all neighborhoods of $0$ in $G$. Then $H$ is a subgroup.

The proof in the books is the following one-liner: "follows from continuity of the group operations". (this is from "Introduction to Commutative Algebra" by Atiyah-MacDonald)

I must admit that I don't really see how that "follows". If there is an easy explanation aimed at someone who has not encountered topological groups in any extent, I'd be happy to read it.

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If $U$ is a neighbourhood of $0$ then so is $-U=\{-x:x\in U\}$. This shows that if $x\in H$ then $-x\in H$.
To show that $H$ is closed under addition, use the fact that if $U$ is a neighbourhood of $0$ then there is another neighbourhood $V$ of $0$ with $V+V\subseteq U$. The existence of $V$ follows from the continuity of addition at $(0,0)$.
Thanks. So since $+$ is continous, there exists a neighborhood $W \subset G\times G$ such that $+(W)\subseteq U$. Projecting $W$ down to $G$, we get our $V$. Is this correct? – Fredrik Meyer Dec 7 '10 at 11:53
Roughly speaking yes, you need that $W$ contains a set $V\times V$ where $V$ is a neighbourhood of $0$ in $G$. – Robin Chapman Dec 7 '10 at 14:14