Showing a Set is not Connected

[Bashyboy]

Homework Statement

I am trying to show that ##X = \{z : |z| \le 1 \} \cup \{z : |z-2|<1\} \subset \mathbb{C}## is a connected set.

Homework 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

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.

 

[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.

 

[Mark44]

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.

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

 

[Mark44]

Homework Statement

I am trying to show that ##X = \{z : |z| \le 1 \} \cup \{z : |z-2|<1\} \subset \mathbb{C}## is a connected set.

Homework Equations

Definition of connectedness that I am working with: A metric space ##(X,d)## is connected if the only subsets of ##X## are ##\emptyset## and ##X##. If ##A \subseteq X##, then ##X## is connected if ##(A,d)## is connected

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.

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.

 

[Bashyboy]

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.

Whoops! I fixed it.

 

[Mark44]

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

 

[fresh_42]

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##.

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 ##X## is connected if ##(A,d)## is connected.

?

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 \;##?

 

[Bashyboy]

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

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.

 

[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##.

 

[fresh_42]

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

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.