# Showing a Set is not Connected

1. Oct 21, 2016

### Bashyboy

1. The problem statement, all variables and given/known data
I am trying to show that $X = \{z : |z| \le 1 \} \cup \{z : |z-2|<1\} \subset \mathbb{C}$ is a connected set.

2. Relevant equations
Definition of connectedness that I am working with: A metric space $(X,d)$ is connected if the only subsets of $X$ that are both open and closed are $\emptyset$ and $X$. If $A \subseteq X$, then $X$ is connected if $(A,d)$ is connected

3. The attempt at a solution
I have been staring at this problem for quite some time and I don't really have much:

Let $C \subseteq X$ be nonempty and both closed and open. I want to show that $X \subseteq C$. Let $z \in X = C \cup (X-C)$. Then either $z \in C$, in which case we are done, or $z \in (X-C)$...I would like to show that the latter case implies a contradiction, but I am having difficulty.

Last edited: Oct 21, 2016
2. Oct 21, 2016

### Bashyboy

According to my book, they have to be disjoint open sets in order to conclude that $X$ is disconnected. The first set in the union is not open, if I am not mistaken.

3. Oct 21, 2016

### Staff: Mentor

You are correct -- both sets have to be open. I have deleted my earlier post.

4. Oct 21, 2016

### Staff: Mentor

Shouldn't this be " A metric space $(X,d)$ is connected if the only subsets of $X$ that are both open and closed are $\emptyset$ and $X$."?
X clearly has two subsets: {z : |z| ≤ 1} and {z | |z - 2| < 1}, but the first set is closed and the second set is open. Neither set is both open and closed.

I'm sure you already know this -- I'm just trying to clarify your statement above.

5. Oct 21, 2016

### Bashyboy

Whoops! I fixed it.

6. Oct 21, 2016

### Staff: Mentor

Your set C is clopen iff its boundary is empty. Can you use this to show that $z \in X - C$ is a contradiction?

7. Oct 21, 2016

### Staff: Mentor

The usual definition goes by separation. Two sets $A\, , \,B\subseteq X$ are called separated in $X$ if $\overline{A} \cap B = \emptyset$ and $A \cap \overline{B} = \emptyset$. Now $X$ is connected, if it cannot be written as union of two non-empty, separated sets.

So what you actually want to use is the equivalence of these two definitions: Your $X$ is connected, since the obvious choice of $A$ and $B$ are not separated (the second condition does not hold) and all other choices clearly neither.
???

If $A \subseteq X$, then $A$ is connected if $(A,d)$, i.e. $A$ with the induced topology from $X$ is connected.
However, one doesn't really need to use the detour of the induced topology.

Anyway, back to your proof. I cannot see, how the consideration of a single element $z$ could help, since we are talking about global properties.
Wouldn't it be more natural to consider $A \cap C$ and $B \cap C$ with the assumed clopen set $C$ and $A=\{z:|z|≤1\}$ and $B=\{z:|z−2|<1\}$? Are they clopen as well? And what does this mean for $X=A \cup B \;$?

8. Oct 23, 2016

### Bashyboy

I don't see the contradiction. Let $\partial C := \overline{C} \cap (\overline{X-C})$ be the boundary (this is how my book defines it). If $C$ is clopen, then $\overline{C} = C$ and $int ~C=C$, and therefore $\partial C = C \cap (\overline{X-C})$. Since $z \in X-C$, then $z \in \overline{X-C}$; and since $z \notin C = \overline{C}$, we can say $z \notin \partial C$. But can conclude neither that $\partial C$ is empty nor nonempty.

9. Oct 23, 2016

### Bashyboy

Hold on! Wouldn't that say no point $z$ in $X$ is in $\partial C$, and this would be the contradiction sought after?

What worries me is that the proof doesn't use any specific information about the metric space $X$ and the subsets $A$ and $B$.

10. Oct 23, 2016

### Staff: Mentor

Probably because one doesn't need neither a metric nor the induced topology. Both are only used to describe the sets $A$ and $B$ in this example. For the properties in question, they aren't necessary.