I'm trying to get use to calculating limit points in metric spaces, but am getting stuck.
Consider the following subset of $\mathbb{R}^{2}$
The set of all complex $z$ such that $|z| < 1$
I'm trying to show that this set is not closed.
To do this, I need to show that it doesn't contain all of its limit points. This means I need to show that there is some neighbourhood containing no point that is different to $p$ (say).
Let $D = \{(x,y) \in \mathbb{R}^{2}: \sqrt{x^{2} + y^{2}} < 1\}$
if I consider a neighbourhood $N_{r}(p) = \{q: d(p,q) < r\}$
I'm not really sure how to proceed from here. I know ultimately that the limit point will be the value $1$. I've only managed to do limit points of some very simple examples such as $\{\frac{1}{n}\}$
Best Answer
Hint: Consider the sequence of points $(a_n)$ in $\mathbb{R}^2$ given by $a_n=(1-1/n,0)$. Then each $a_n\in D$, and you can easily show that $a_n\to(1,0)\in\mathbb{R}^2$. However, this point is not contained in the open unit disk - how does this relate to $D$ being closed?