# 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.

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

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

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

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

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 $$b \in \overline A$$. $$\square$$

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 $\overline{A}=\overline{B}$.