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

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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](http://math.stackexchange.com/q/13368/) which only addresses the first part of the question), so I retract my vote for closure.2012-07-25
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    @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.2012-07-25
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    @t.b. I added the answer to my first question to my question. Is it correct?2012-07-26
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    Yes, this is correct.2012-07-26
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    @t.b. Awesome! Thanks a lot!! : ) (For some reason your comment didn't ping me :,( )2012-07-26
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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 $V+V \subset U$.2012-07-26
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    @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 $V \subset U$.2012-07-26
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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 $V = -V$?2012-07-26
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    @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 $0$.2012-07-26
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    @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.2012-07-26
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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.2012-07-26
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    @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.2013-06-15
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    @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.2013-06-16
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    @Matt, thank you very much for your help.2013-06-19
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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).2015-03-06
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    @TobiasKildetoft Thank you for your comment!2015-03-07

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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*}

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    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?2012-07-25
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    Oh, ok. If $U_0$ is a nbhood of zero then so is $-U_0$.2012-07-25
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    Or, to put it another way: The map $y \mapsto y-x$ is a homeoorphism, so it maps nhoods of $x$ to nhoods of $0$.2012-07-25
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    The argument could be summarized as: $x$ being in all neighbourhoods of $0$ is equivalent to $0$ being in all neighbourhoods of $x$.2012-07-25
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    How do I go from $0 \in x - V$ to $x \in V$?2012-07-25
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    @ClarkKent If $0 \in x - V$, then there is an $y \in V$ such that $0 = x - y$, that is $x = y \in V$ ...2012-07-25
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    @martini Thank you so much! I understand it now!2012-07-25