Show a set it open but not closed

Question

Let $A = \{ z : |z| < 1 \} $ where $z \in \mathbb{C}$. Show that $A$ is open but not closed.

Try:

Let $\epsilon > 0$ 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 $|w-z_0| < \epsilon$. We must show that $w \in A$. Notice:

$$ |w| =|w - z_0 + z_0| \leq |w-z_0| + |z_0| < \epsilon + |z_0| < 1 - |z_0| + |z_0| = 1$$

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?

Answer 1

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.

Answer 2

Some remarks concerning your proof:

1. Explain at the beginning of the proof what you are trying to prove and why it suffices to show this in order to prove the statement. So, instead of writing

   We show $D(z_0,\epsilon) \subset A$. This will prove our set is open.

   it is much more better to write

   By definition, a set $A$ is open iff for any $z_0 \in A$ there exists $\epsilon>0$ such that $D(z_0,\epsilon) \subset A$. We show that for any $z_0 \in A$ the inclusion $D(z_0,\epsilon) \subset A$ holds for $\epsilon := 1-|z_0|$.

2. As pointed out by several others, the two sentences

   Let $\epsilon>0$ be given. Pick any $z_0 \in A$ and choose $\epsilon = 1-|z_0|$.

   contradict each other. If you had started with the definition of a set being open (as suggested in 1.), then this mistake wouldn't have happened.

3. The idea of your proof is correct.

Depending on your definition of a closed set, there are several ways to prove that $A$ is not closed:

1. Consider the sequence $z_n := 1-\frac{1}{n} \in A$. Is $z_n$ convergent and, if so, is the limit an element of $A$? What does this say about the closedness of $A$?
2. Consider the set $B := A^c = \{z; |z| \geq 1\}$. Then $A$ is closed if and only if $B$ is open. Consider $z_0 := 1$. Does there exist $\epsilon>0$ such that $D(z_0,\epsilon) \subseteq B$? Is $B$ open?

Answer 3

No, there are some mistakes.

Let $\epsilon > 0$ 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 $|w-z_0| < \epsilon$. We must show that $w \in A$. Notice:

$$ |w| =|w - z_0 + z_0| \leq |w-z_0| + |z_0| < \epsilon + |z_0| = 1 - |z_0| + |z_0| = 1$$

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 $|w-z_0| < \epsilon$. We must show that $w \in A$. Notice:

$$ |w| =|w - z_0 + z_0| \leq |w-z_0| + |z_0| < \epsilon + |z_0| = 1 - |z_0| + |z_0| = 1$$

Why?
The (rather: one) definition is:
$A$ ist open $:\Leftrightarrow$ $$\forall z_0 \in A\exists \epsilon >0:D(z_0,\epsilon) \subset A$$ 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 $\epsilon >0$ (w.l.o.g. $\epsilon < 2$) 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.