# Showing a set is neither open nor closed

Let $B = \{ z \in \mathbb{C} : 0 < |z| \leq 1 \}$. I claim $B$ is neither open nor closed. I know $B$ is open if for every $z \in B$, there is some $\epsilon > 0$ such that $D(z,\epsilon) \subseteq B$. The negation should be that $A$ is not open if there is some $z_0 \in B$ such that for all $\epsilon > 0$, $D(z, \epsilon) \cap B = \varnothing$.

I think to prove it is not open, it is better to use contradiction. Suppose $B$ is open. Let $z_0 = 1$ then it must be the case that for all $\epsilon > 0$ , $D(1,\epsilon ) \subseteq B$. Consider $w = 1 + \frac{\epsilon}{2}$. As $|1 - 1 - \frac{\epsilon}{2} | = \frac{\epsilon}{2} < \epsilon$ we have $w \in D(1, \epsilon)$. But, $w$ is not in $B$. Contradiction.

To show it is not closed, can show $B^c$ is not open. Say $B^c$ is open. choose $z_0 = 0 \in B^c$. For every $\epsilon>0$ we must have $D(0, \epsilon) \subset B^c$. But, $z = \frac{\epsilon }{2} \notin B^c$ (This is obvious, but how can I show it?)

IS this a correct approach?

• You have $0<|\epsilon/2|\le1$ for $\epsilon>0$ small enough. Sep 13, 2015 at 11:12

You negation of the criterium of openness is wrong though. It is enough that $D(z_0,\varepsilon) \not\subseteq B$ for some $z_0\in B$ and some $\varepsilon > 0$, i.e. it doesn't have to be all $\varepsilon > 0$ and $D(z_0,\varepsilon)$ can have a non-empty intersection with $B$, it just has to contains points outside of $B$. That is also all you prove.
$B$ is open if for each $z\in B$ there is some $\epsilon>0$ such that $D(z,\epsilon)\subset B$. So the negation of this is : $$\exists z_0\in B:\forall \epsilon>0 \,\,D(z_0,\epsilon)\cap B^c\neq \emptyset$$