Connected Sets and Their Interiors: Baby Rudin Exercise 2.20 Example

[Mogarrr]

Homework Statement

This is from Baby Rudin Exercise 2.20- Are closures and interiors of connected sets always connected? (Look at subsets of \mathbb{R}^2).

Homework Equations

The interior is the set of all interior points for a set E that is a subset of a metric space X.

A subset Y of a metric space X is disconnected if and only if there are open subsets U, V of X satisfying:
1) U \cap Y \neq \emptyset, V \cap Y \neq \emptyset
2) Y \subset U \cap V
3) U \cap V \cap Y = \emptyset

The Attempt at a Solution

It's best if I draw a picture, but I was basically thinking of a dumbbell in \mathbb{R}^2, that is two closed circles with a line segment connecting them.

I can prove that the interior of this shape is disconnected, however I am having some trouble showing that the original set, the dumbbell is connected.

This is homework for a class where the teacher takes points off (or fails to give points, you know you're perspective of the half empty/ half full glass) for not having justification.

So how to justify this set, the dumbbell, is connected. I've thought of using proof by contradiction, but I'm not sure where the contradiction will come, and I'm running low on time. Any help would be much appreciated.

 

[RUber]

I you can show that the ends of the dumbbell are both connected sets, then the line between them is a connection. If you can connect all subsets of a set without leaving the set, then the set is connected.

A subset Y of a metric space X is disconnected if and only if there are open subsets U, V of X satisfying:
1) U \cap Y \neq \emptyset, V \cap Y \neq \emptyset
2) Y \subset U \cap V
3) U \cap V \cap Y = \emptyset
.

I am not sure that these are consistent requirements.

 

[Mogarrr]

Then, I'd have to show the closed circle in \mathbb{R}^2 is connected. But I'm not really sure how to do that?

I suppose a proof by contradiction, assuming the closed circle is disconnected... (that's pretty much the only technique I got when it comes to connected sets).

I suppose if any 2 of the three statements for disconnected sets is assumed to be true, then I'd get a contradiction when using the last statement.

 

[RUber]

You can define then to be connected. Your goal is to show that the interiors are not connected.
Go back to the definition of connected.
You have connected sets A and B that are disjoint and disconnected from each other. Connect them with a line.
Now they are connected.

 

[gopher_p]

You can define then to be connected. Your goal is to show that the interiors are not connected.
Go back to the definition of connected.
You have connected sets A and B that are disjoint and disconnected from each other. Connect them with a line.
Now they are connected.

I think the difficulty that Mogarrr is having is working within the confines of the definition he/she has been given (which, by the way, should read $Y\subset U\cup V$ in the second condition). If you could demonstrate how the intuitive notion of "connecting" a disconnected pair of connected sets with a line actually gives a connected set per the definition, that might be more helpful.

 

[RUber]

Thanks gopher p,
If all you have to work with is the definition of disconnected, then I recommend approaching it like this:
It should be easy to show (or simply by assumption) that your two subsets (the ends of the barbell), A and B are disconnected. Let L be the line connecting them.
$L \cap A =\{p_a\}$ where $p_a$ is the point where L meets A.
$L \cap B=\{p_b\}$ where $p_b$ is the point where L meets B.
Now apply the definition of disconnected to show that $Y=A \cup B \cup L$ is not disconnected...and therefore connected.
Note that the fact that the subsets have to be open prevents you from dividing up the subsets cleanly.

 

[Mogarrr]

Thanks guys. Actually, I switched the example to 2 closed balls that just touch each either in \mathbb{R}^2 and to show this was connected, I just looked at a line that intersects the disconnections, and then showed how there can't be a disconnection there.