Closure of a subset of normed vector space 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.
 A: 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

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

A: $\overline{A} \subset \overline{B}$:
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 $b \in \overline A$.
$\square$
A: 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 $\overline{A}=\overline{B}$.
