Show a set it open but not closed Let $A = \{ z : |z| &lt; 1 \} $ where $z \in \mathbb{C}$. Show that $A$ is open but not closed.
Try:
Let $\epsilon &gt; 0$ be given. Pick any $z_0 \in A$ and choose $\epsilon = 1 - |z_0| $. We show $D(z_0,\epsilon) \subset A$. This will prove our set is open. Suppose $w \in D(z_0, \epsilon )$. this means that $|w-z_0| &lt; \epsilon$. We must show that $w \in A$. Notice:
$$ |w| =|w - z_0 + z_0| \leq |w-z_0| + |z_0| &lt; \epsilon + |z_0| &lt; 1 - |z_0| + |z_0| = 1$$
and we have our conclusion. Is this a correct solution? Also, I am stuck on trying to show it is not closed. Can someone help me?
First of all, your wording for the proof that $A$ is open is faulty. There are two things you need to fix:


You first say "let $\epsilon$ be given", and then say "Choose $\epsilon = 1-|z_0|$". The two sentences are in contradiction to each other.
You then show that $D(z_0,\epsilon) \subseteq A$ with that poorly defined $\epsilon$ and say "we have our conclusion". But the conclusion referenced here is only of the subsets, not that $A$ is open. You need to improve your wording. Preferably, to something like


  $A$ is open if, for every something, this is true. Let $x$ be any something. Then, [...], therefore, this is true for $x$, and because $x$ is arbitrary, it is true for every something.
No, there are some mistakes.

Let $\epsilon &gt; 0$ be given. Pick any $z_0 \in A$ and choose $\epsilon = 1 - |z_0| $. We show $D(z_0,\epsilon) \subset A$. This will prove our set is open. Suppose $w \in D(z_0, \epsilon )$. this means that $|w-z_0| &lt; \epsilon$. We must show that $w \in A$. Notice:
$$ |w| =|w - z_0 + z_0| \leq |w-z_0| + |z_0| &lt; \epsilon + |z_0| = 1 - |z_0| + |z_0| = 1$$

I has to be:

Let $z_0 \in A$ be given. Pick $\epsilon = 1 - |z_0|$. We show $D(z_0,\epsilon) \subset A$. This will prove our set is open. Suppose $w \in D(z_0, \epsilon )$. this means that $|w-z_0| &lt; \epsilon$. We must show that $w \in A$. Notice:
$$ |w| =|w - z_0 + z_0| \leq |w-z_0| + |z_0| &lt; \epsilon + |z_0| = 1 - |z_0| + |z_0| = 1$$

Why?
The (rather: one) definition is:
$A$ ist open $:\Leftrightarrow$
$$\forall z_0 \in A\exists  \epsilon &gt;0:D(z_0,\epsilon) \subset A$$
So for any given $z_0$ you have then to find some $\epsilon$.
Show that A is not closed.
Ony way is to show that $B:=A^C=\mathbb{C}\setminus A$ ist not open.
(The topological defintion of closed is that the complement is open.)
Suppose $B$ is open.
$z_0=1$ is in $B$.
There has so exist some $\epsilon &gt;0$ (w.l.o.g. $\epsilon &lt; 2$) such that $D(z_0,\epsilon) \subset B$.
Since $A \ni 1-\frac{\epsilon}{2} = z_0-\frac{\epsilon}{2} \in D(z_0,\epsilon) \subset A^C$ we have a contradiction.
Therefore $B$ is not open, so $A$ is not closed.