Let the set
$D=\{(x, y)\in\mathbb{R}^2|0 < x^2 + y^2 \leq 4\}$
is this set closed?
I know that this set is closed if $\mathbb{R}^2\backslash D$ is open.
The set $\mathbb{R}^2\backslash D=\{0\}\cup \{(x, y)\in\mathbb{R}^2|x^2 + y^2 > 4\}$
The set $\{(x, y)\in\mathbb{R}^2|x^2 + y^2 > 4\}$ is already open, but the one point set $\{0\}$ is closed
I don't know if this union is already open. I guess that it's not open, nor closed; but i'm not so sure about it.
A set $S \subset \mathbb{R}^2$ is open in $\mathbb{R}^2$ iff for any point $x \in S$, there exists an $\varepsilon$-neighborhood of $x$ which is contained in $S$.
In our case, i.e. $S = \mathbb{R}^2 \setminus D$, any $\varepsilon$-neighborhood of $(0,0)$ is not contained in $\mathbb{R}^2 \setminus D$. This implies that $\mathbb{R}^2 \setminus D$ is not open in $\mathbb{R}^2$, and hence $D$ is not closed in $\mathbb{R}^2$.