Proving the open disc in $\mathbb R^2$ is not closed 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 $\{\frac{1}{n}\}$
 A: 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?
A: You can prove that it is closed by proving that its complement is not open. Take $(1,0)\in D^\complement$. For each $r>0$, take $r'>0$ such that $r'<r$ and also that $r'<2$. Then $(1-r',0)\in D_r\bigl((1,0)\bigr)\cap D$. Therefore, $D_r\bigl((1,0)\bigr)\not\subset D^\complement$. This proves that $D^\complement$ is not an open set.
A: In this case the boundary of $D$ are all limit points of $D$, i.e. the set:
$$
\partial D = \{ (x,y) \in \mathbb{R}^2 \: | \: \sqrt{x^2 + y^2} = 1\}.
$$
Let's pick $(1,0) \in \partial D$. In order for $(1,0)$ to be a limit point of $D$, any neighborhood of $(1,0)$ must contain a point of $D$ different from $(1,0)$. Observe that $(1,0) \not \in D$, so we just need to show $N_\varepsilon (1,0) \cap D$ is nonempty for any $\varepsilon > 0$. Let $\varepsilon > 0$. Evidently, we have:
$$
\Big(1-\frac{\varepsilon}{2}, 0\Big) \in D, 
$$
as $\sqrt{(1-\varepsilon/2)^2 + 0} = 1-\varepsilon/2 < 1$. Moreover,
$$
\Big(1-\frac{\varepsilon}{2}, 0\Big) \in N_\varepsilon(1,0),
$$
since $\big|(1,0) - (1-\varepsilon/2, 0)\big| = \varepsilon/2 < \varepsilon$. Thus, the intersection $N_\varepsilon(1,0) \cap D$ in nonempty for any $\varepsilon > 0$, and so $(1,0)$ is a limit point of $D$ not contained in $D$. So $D$ cannot be a closed set.
