Show a set it open but not closed

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?

• If $\epsilon$ is arbitrary, then how is it possible that $\epsilon + |z_0| = 1$? Sep 13, 2015 at 7:44
• Why the downvote....? The reasoning is not perfect, I agree, but that's hardly a reason to downvote the question.
– saz
Sep 13, 2015 at 8:56

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.

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?

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.

• The reasoning is not correct. The OP wrote "let $\epsilon > 0$ be given. [...] Chooose $\epsilon = 1-|z_0|$."
– 5xum
Sep 13, 2015 at 7:54
• @5xum Good point. I did just see what I expected to see.:D I'll change it. Sep 13, 2015 at 7:58