# Show that open rectangle is open set in metric space $(\mathbb{R}^n, d_2)$

For given real numbers $a_i \lt b_i$, $i=1,...,n$ prove that open rectangle $$S=\langle a_1,b_1 \rangle \times...\times \langle a_n, b_n \rangle$$

is open set in $(\mathbb{R}^n, d_2)$ where $d_2$ is standard Euclidean metric.

We usually prove this by finding minimal radius of open ball such that whole ball is inside of $S$, and then take random point from ball and prove that it is in $S$. My partial solution, haven't finalized it:

Let $x_0=(x_1^0,...x_n^0) \in S.$ We define $r:=\min_{1\le i\le n}{(x_i^0-a_i,b_i-x_i^0)}.$ $x_0$ is in $S$ so $r \gt 0$. Lets show that $K(x_0,r) \subset S$. Let $x=(x_1,...,x_n) \in K(x_0,r)$. I have to show that $x\in S$.

$K(x,r)$ is open ball around x with radius of $r$.

I am stuck here.

• Think about what means "$x\in K(x_0, r)$". – Rodrigo Dias Nov 1 '16 at 10:18
• Well, by that I know that $d(x,x_0) \lt r$ – Rush ThaMan Nov 1 '16 at 10:24

## 1 Answer

This is the correct way!

$$x=(x_1,...,x_n)\in K(x_0, r) \Rightarrow$$ $$d_2(x,x_0)<r=\min_{1\leq i\leq n} \{x_i^0-a_i, b_i-x_i^0\}$$

We know that $$\max_{1\leq i\leq n} \{|x_i-x_i^0|\} \leq \sqrt{\sum_{i=1}^n (x_i-x_i^0)^2}=d_2(x,x_0)\text{ (why?)}$$ Then $$|x_i-x_i^0|<r\;\; \forall i\in \{1,...,n\}$$ and therefore $x\in S$.