# Question on Topological vector space 1

I have numbered this question as (1) because I will be posting series of questions where I don't understand. I hope its allowed. I want to prove the following :

If $X$ is a topological vector space then : If $A\subset X$ then $\bar A = \cap(A+V)$, where $V$ runs through all nbd of $0$.

I tried like this : $x\in \bar A\subset A+V$ $\implies x\in A+V$ for all V. The other way , $x\in A+V$ $\implies$ $x\in A$ which means $x\in \bar A$.

Is my argument correct? If it fails what should I take care of? Thank you very much.

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The implication $x \in A + V \Rightarrow x \in A$ doesn't hold. –  J. J. Jun 14 '12 at 10:02
@J.J : How can i argue to give a reasonable answer ? –  Theorem Jun 14 '12 at 10:11

For the other direction it's helpful to remember that $x \in \bar{A}$ if and only if every neighbourhood $U$ of $x$ intersects $A$. So let $U$ be an arbitrary neighbourhood of $0$, whence $x + U$ is an arbitrary neighbourhood of $x$. If $x \in \bigcap (A + V)$, then in particular $x \in A - U$ so that we can write $x = a - u$ for some $a \in A$, $u \in U$. Thus $a \in x + U$ and we see that $x + U$ intersects $A$.

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Can you tell me why should $x\in A-U$ ? –  Theorem Jun 14 '12 at 10:31
@Theorem: $-U$ is a neighbourhood of $0$ and hence $A-U$ is included in the intersection. –  J. J. Jun 14 '12 at 10:47
I am getting confused with complements and addition :(. –  Theorem Jun 14 '12 at 10:52

We assume that $x\in A+V$ for all neighborhood $V$ of $0$. Let $W$ a neighborhood of $x$. We have to show that $W\cap A\neq\emptyset$. We can write $W=x+W'$, where $W'$ is a neighborhood of $0$. $x\in A+W'$ hence $x=a+w'$ where $a\in A$ and $w'\in W'$. We have $x-w'=a\in A$ and $x-w'\in W$ if we assume $W$ symmetric (without loss of generality, just substituting $W$ by $W\cap (-W)$.

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The main ingredients:

1. Show that $$\bigcap_{V\text{ a nbhd of }0}(A+V)=\bigcap_{V\text{ a symmetric}\atop \text{nbhd of }0}(A+V)\;.$$

2. Show that if $V$ is a symmetric nbhd of $0$, then $x\in A+V$ iff $A\cap(x+V)\ne\varnothing$.

3. Show that $x\in\operatorname{cl}A$ iff $A\cap(x+V)\ne\varnothing$ for every nbhd of $0$ iff $A\cap(x+V)\ne\varnothing$ for every symmetric nbhd of $0$.

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We need to recall the following:

Let $X$ be a topological space , $A\subseteq X$ and $x\in X.$ Then $x\in \overline{A}$ iff for any nbd $U$ of $x$, $U\cap A \ne \varnothing$. Also, if $X$ is a TVS with topology $\tau$, then it is known that $a+G$ and $tG$ belong to $\tau$ whenever $a\in X$, $G\in \tau$ and $t\ne 0$. We call this, the invariant property. We now prove that $$\overline{A} = \bigcap_{V\in \mathcal{V}}(A+V).$$ where $\mathcal{V}$ denotes the collection of all nbs of 0.

First, let $x\in \overline{A}$ and let $V\in \mathcal{V}$. Then $-V$ is nbd of 0. Hence, $x+(-V)=x-V$ is nbd of $x$. Thus, $(x-V)\cap A \ne \varnothing$. Choose $y\in x-V$ and $y\in A$. Then it follows that $x\in A+V$. Hence, $x\in A+V$ for every $V\in \mathcal{V}$. This implies that $x\in \displaystyle{\bigcap_{V\in \mathcal{V}}(A+V)}$.

Next, let $x\in \displaystyle{\bigcap_{V\in \mathcal{V}}(A+V)}$. We want to show that $x\in \overline{A}$. Let $U$ be any nbd of $x$. Then $-x+U$ is nbd of 0 and so is $-(x+U)=x-U$. Hence, $x-U\in \mathcal{V}$. Thus, $x\in A+(x-U)$. Hence, there exist $a\in A$ and $u\in U$ such that $x=a+(x-u)$. This implies that $a=u$. Hence, $U\cap A \ne \varnothing$. Thus, $x\in \overline{A}$. This completes the proof.

juniven

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