Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

Let $G$ be a topological abelian group. Let $H$ be the intersection of all neighborhoods of zero.

How is $H = \mathrm{cl}(\{0\})$? Isn't the closure of a set $A$ the smallest closed set containing $A$ which is the same as the intersection of all closed sets containing $A$? But neighborhoods in $G$ are not necessarily closed. Thanks.


To see that $H$ is a subgroup:

First note that by construction $H$ contains $0$.

Furthermore, $f: x \mapsto -x$ is continuous and its own inverse so that $f$ is also open. Hence $U$ is a neighborhood of $0$ if and only if $-U$ is. Now let $x \in H$. Then $x$ is in every neighborhood $U$ of $0$. Hence $x$ is in every neighborhood $-U$ of $0$. Hence $-x$ is in $U$ and hence in $H$.

Alternatively one can verify it as follows: $$x \in H \iff x \in \bigcap_{U \text{ nbhd of } 0} U \iff x \in \bigcap_{U \text{ nbhd of } 0} -U \iff -x \in \bigcap_{U \text{ nbhd of } 0} U \iff -x \in H$$

To see that $x+y$ is in $H$ if $x,y \in H$, note that $g: (x,y) \mapsto x+y$ is continuous. Now let $V$ be an arbitrary neighborhood of $0$. Then since $g$ is continuous there exists a neighborhood $N \times M$ of $(0,0)$ such that $g(N \times M) \subset V$. Since $G \times G$ has the product topology, $N \times M$ is a neighborhood of $0$ if and only if $N$ and $M$ are neighborhoods of $0$. Hence $x,y \in N$ and $x,y \in M$ and hence $g((x,y)) = x + y \in V$ since $g(N \times M) \subset V$.

share|improve this question
I'm sorry: when I was looking for a duplicate, the question about the closure of $\{0\}$ wasn't there (and it isn't answered in the thread I linked to which only addresses the first part of the question), so I retract my vote for closure. –  t.b. Jul 25 '12 at 7:46
Yes, this is correct. –  t.b. Jul 26 '12 at 8:36
No problem :) Maybe it would be a good exercise for you to prove the following useful fact: If $U$ is a neighborhood of $0$ then there is a neigbhorhood of zero $V \subset U$ such that $V = - V$ and $V+V \subset U$. –  t.b. Jul 26 '12 at 8:51
The exercise is a general fact on topological abelian groups (if you solve it the solution of your question becomes a bit easier, but that's all). Of course not every neighborhood of $0$ is closed under taking inverses (consider $(-1,2)$ in $\mathbb{R}$, for example). Try to understand what happens when you endow the additive group $\mathbb{R}^2$ with the topology $U \times \mathbb{R}$ with $U \subset \mathbb{R}$ open. What is the closure of $0 \in \mathbb{R}^2$ with this topology? Given $U = (-1,3) \times \mathbb{R}$, how can you find a $0$-nbhd $V$ such that $V + V \subset U$ and $V = -V$? –  t.b. Jul 26 '12 at 10:01
All you say is correct (except the part about the metric which doesn't matter at all). In your solution to the exercise you found $M,N$ such that $M + N \subset U$. Take $\tilde{V} = M \cap N \cap U$, argue that $\tilde{V} + \tilde{V} \subset U$ and put $V = \tilde{V} \cap (-\tilde{V})$ and show that this $V$ does the job. –  t.b. Jul 26 '12 at 10:29

1 Answer 1

up vote 6 down vote accepted

$\def\cl{\mathop{\mathrm{cl}}}$ For $x \in G$, let $\mathcal U_x$ denote the set of all neighbourhoods of $x$. Then we have that $x - \mathcal U_0 = \mathcal U_x$ for each $x \in G$. It follows \begin{align*} x \in \cl\{0\} &\iff \forall U \in \mathcal U_x : U \cap \{0\} \ne \emptyset\\ &\iff \forall V \in \mathcal U_0: (x - V) \cap \{0\} \ne \emptyset\\ &\iff \forall V \in \mathcal U_0: 0 \in x - V\\ &\iff \forall V \in \mathcal U_0 : x \in V\\ &\iff x \in \bigcap \mathcal U_0. \end{align*}

share|improve this answer
Thank you but I'm not sure I understand: if $U_x$ is a nbhood of $x$ then $U_x - x = U_0$ is a nbhood of $0$. How do I get $x - U_0 = U_x$ from that? –  Matt N. Jul 25 '12 at 9:14
Oh, ok. If $U_0$ is a nbhood of zero then so is $-U_0$. –  Matt N. Jul 25 '12 at 9:15
Or, to put it another way: The map $y \mapsto y-x$ is a homeoorphism, so it maps nhoods of $x$ to nhoods of $0$. –  martini Jul 25 '12 at 9:17
The argument could be summarized as: $x$ being in all neighbourhoods of $0$ is equivalent to $0$ being in all neighbourhoods of $x$. –  joriki Jul 25 '12 at 9:35
@ClarkKent If $0 \in x - V$, then there is an $y \in V$ such that $0 = x - y$, that is $x = y \in V$ ... –  martini Jul 25 '12 at 9:47

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.