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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.

(Edit)

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 $g(N \times M) \subset V$.

Rudy the Reindeer
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    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 07:46
  • @t.b. No reason to apologize, no problem. Should I remove the first part of my question? Thank you for the link, it answers the first part of my question. – Rudy the Reindeer Jul 25 '12 at 07:56
  • @t.b. I added the answer to my first question to my question. Is it correct? – Rudy the Reindeer Jul 26 '12 at 08:31
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    Yes, this is correct. – t.b. Jul 26 '12 at 08:36
  • @t.b. Awesome! Thanks a lot!! : ) (For some reason your comment didn't ping me :,( ) – Rudy the Reindeer Jul 26 '12 at 08:47
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    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 08:51
  • @t.b. I'm sorry but I think I still don't understand it properly. We showed that $H$ is a subgroup, so in particular, closed with respect to taking additive inverses. But $H$ is also the intersection of all nbhoods of zero. So every nbhood of zero has also got to be closed w.r.t. taking additive inverses. But this means that if $v \in V$ then $-v$ is in $V$ and vice versa so $V = -V$. But then for every nbhood $N$ of zero we'd have $N = -N$. So also, $U = -U$ and of course $V = -V$ for all $V \subset U$. – Rudy the Reindeer Jul 26 '12 at 09:47
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    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
  • @t.b. Is the closure of ${(0,0)}$ in this topology ${0} \times \mathbb R$? Because if $(x,y)$ is in the closure I want every nbhood of $(x,y)$ to intersect with ${(0,0)}$. A nbhood of $(x,y)$ looks like $U \times \mathbb R$. $\mathbb R$ always intersects with ${0}$, ${0}$ only intersects with $U$ for all $U$ if $0$ is in all $U$ but then $x$ must equal $0$. – Rudy the Reindeer Jul 26 '12 at 10:14
  • @t.b. As for your second question: $V = (-\frac12, \frac12) \times \mathbb R$ seems to be such a nbhood. So in a metric space I want $V$ to be symmetric around the origin. But unfortunately my topological group is not necessarily metric so such a nbhood might not exist. – Rudy the Reindeer Jul 26 '12 at 10:19
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    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
  • @MattN., why "Since G×G has the product topology, N×M is a neighborhood of 0 if and only if N and M are neighborhoods of 0"? Are there some references about this fact? Thank you very much. – LJR Jun 15 '13 at 13:44
  • @LJR I can't think of any reference. Perhaps I should have written $(0,0)$ where I wrote $0$ to be less confusing. The open sets of a product of two topological spaces $T \times T'$ are precisely the sets of the form $O \times O'$ where $O$ and $O'$ are open in $T$ and $T'$ respectively. – Rudy the Reindeer Jun 16 '13 at 14:26
  • @Matt, thank you very much for your help. – LJR Jun 19 '13 at 12:10
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    @RudytheReindeer Just happened to see these comments because the question was linked elsewhere. No, there are usually open sets not of that form in the product (otherwise it would not be closed under unions). – Tobias Kildetoft Mar 06 '15 at 07:27
  • @TobiasKildetoft Thank you for your comment! – Rudy the Reindeer Mar 07 '15 at 06:44

1 Answers1

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$\def\cl{\mathop{\mathrm{cl}}}$ 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*}

martini
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