2
$\begingroup$

Can you help me to prove this claim :

$A$ is a subset of a normed vector space, closure of $A$ is closure of

$$B=\bigcap_{n=1}^\infty \left( A+{1\over n}B_1 (0)\right)$$

I tried to prove that closure of $A$ is a subset of closure of $B$ and vice versa. My trying was ineffective. Thank you.

$\endgroup$
  • $\begingroup$ What do you mean by $+$ at $A+(1/n)B_1(0)$? $\endgroup$ – Jimmy R. Dec 15 '15 at 14:51
  • 1
    $\begingroup$ @Stef usual additivity in vector spaces. $\endgroup$ – mohsen Dec 15 '15 at 14:54
1
$\begingroup$

Suppose that $x \in \def\cl{\operatorname{cl}}\cl A$, than for any $n \in \mathbf N$, we have that there is $x_n \in A$ such that $\|x-x_n\| < \frac 1n$, or $x \in A + \frac 1n B_1(0)$. That is $x \in \bigcap_n A + \frac 1nB_1(0)$.

On the other hand, if $x \in \bigcap_n A + \frac 1n B_1(0)$, for any $n$, there are $x_n \in A$, $y_n \in B_1(0)$, such that $x = x_n + \frac 1n y_n$. As $\|y_n \| \le 1$, we have $\frac 1n y_n \to 0$. Hence, $$ x_n = x - \frac 1n y_n \to x $$ As $x_n \in A$, we have $x \in \cl A$.

Altogether we have

$$ \cl A = \bigcap_n \left[ A + \frac 1n B_1(0) \right] $$

| cite | improve this answer | |
$\endgroup$
0
$\begingroup$

$\overline{A} \subset \overline{B}$:

Since $B$ is defined as the intersection of sets containing $A$, we have $A \subset B$ and hence $\overline{A} \subset \overline B$.

$\overline B \subset \overline A$:

We want to show that for all $b \in \overline B$, there is a sequence of points $a_n \in A$ such that $a_n \to b$. Since $b \in \overline B$, there exists a sequence of points $b_n \in B$ such that $b_n \to b$.

From the definition of $B$ as $$B = \bigcap\limits_{n=1}^\infty (A + \frac1n B_1(0))$$ we see that given a point $b_n \in B$ and an integer $n \in \mathbb{Z}$, there exists a point $a_n \in A$ such that $b_n \in a_b + B_1(0)$, and thus that $\|a_n - b_n\| < \frac1n$ (since $b_n$ is in a ball of radius $\frac1n$ around $a_n$). Therefore, $a_n$ converges, and converges to the same limit as $b_n$. So we have a sequence of points $a_n \in A$ such that $a_n \to b$, and thus $b \in \overline A$. $\square$

| cite | improve this answer | |
$\endgroup$
0
$\begingroup$

Since any ball centered at origin is symmetric, the definition of closure implies that

$$\begin{align} x\in\overline{A}\quad&\Longleftrightarrow\quad \left(x+\frac{1}{n}B_1(0)\right)\cap A\neq\varnothing,\quad\forall\ n\in\mathbb{N}\\\\ &\Longleftrightarrow \quad x\in A-\frac{1}{n}B_1(0),\quad\forall\ n\in\mathbb{N}\\\\ &\Longleftrightarrow \quad x\in A+\frac{1}{n}B_1(0),\quad\forall\ n\in\mathbb{N}\\\\ &\Longleftrightarrow \quad x\in \bigcap_{n=1}^{\infty} \left(A+\frac{1}{n}B_1(0)\right) \end{align}$$

So, $\overline{A}=B$ and thus $\overline{A}=\overline{B}$.

| cite | improve this answer | |
$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

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