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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.


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 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
up vote 7 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*}

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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? – Rudy the Reindeer Jul 25 '12 at 9:14
Oh, ok. If $U_0$ is a nbhood of zero then so is $-U_0$. – Rudy the Reindeer 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

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