# Open set

1. Oct 7, 2012

### stauros

Prove if the following set is open

$$\displaystyle{A=\begin{cases} \vec{x}=(x_1,...,x_n)\in\mathbb{R}^n:x_n>0\\ \end{cases}}$$ .

I have written the following proof and please correct me if i am wrong

Let : $$\displaystyle{\vec{x}\in A}$$

Then we have : $$\displaystyle{\vec{x}=(x_1,...,x_n)}$$ with $$\displaystyle{x_n>0}$$

Choose $$\epsilon$$ such that $$\displaystyle{0<\epsilon<x_n}$$ and then $$\displaystyle{B(\vec{x},\epsilon)\subseteq A}$$

This happens because if $$\displaystyle{\vec{y}=(y_1,...,y_n)\in B(\vec{x},\epsilon)}$$ then $$\displaystyle{||\vec{y}-\vec{x}||<\epsilon}$$

and $$\displaystyle{y_i\in\left(x_i-\epsilon,x_i+\epsilon\right)}$$

Then we have $$\displaystyle{y_n\in\left(x_n-\epsilon,x_n+\epsilon\right)}$$ and thus $$\displaystyle{y_n>0}$$[/quote]

2. Oct 7, 2012

### HallsofIvy

Staff Emeritus
That makes no sense because there is no one number labeled "$x_n$". What you mean to say is that $\epsilon< min(x_n)$.

Can you prove this? That is, after all the whole point of the exercise! In particular, what is the definition of $||\vec{y}-\vec{x}||$?

3. Oct 7, 2012

### stauros

Yes you right,how about the inequality: $$\displaystyle{0<\epsilon<x_k, \forall 1\le k \le n}$$.

But i think the center point of the problem is that:

$|x_{i}-y_{i}|\leq ||x_{i}-y_{i}||<\epsilon$ using the Euclidian norm

Last edited: Oct 7, 2012