# Open sets, countable unions of open rectangles

1. Oct 6, 2012

### infk

1. The problem statement, all variables and given/known data
So here is a "proof" from my measre theory class that I don't really understand. Be nice with me, this is the first time I am learning to "prove" things.
Show that a connected open set Ω ($\mathbb{R}^d$, I suppose) is a countable union of open, disjoint rectangles if and only if Ω is itself a rectangle.

2. Relevant equations
N/A

3. The attempt at a solution
Taught in class:
An open set Ω is connected if and only if it is impossble to write Ω = V $\bigcup$U where U and V are open, non-empty and disjoint. Thus if we can write Ω = $\bigcup^{\infty}_{k=1} R_k$ where $R_k$ are open disjoint rectangles of which at least two are non-empty (lets say $R_1$ and $R_2$ ) we can then write Ω = $R_1 \cup (\bigcup^{\infty}_{k=2} R_k)$ and therefore Ω is not connected.
There is also another question; Show that an open disc in $\mathbb{R}^2$ is not a countable union of open disjoint rectangles. To show this, the professor said that the previous result apllies since a disc is connected.

I don't understand:
Why does this prove the proposition?, Is the assumption that we can write Ω = $\bigcup^{\infty}_{k=1} R_k$ where $R_k$ are open disjoint rectangles of which at least two are non-empty, equvalent with saying that Ω is a rectangle? If it is, have we not then assumed that Ω is a rectangle and then shown that a rectangle is not connected?

2. Oct 6, 2012

### HallsofIvy

Staff Emeritus
First, if Ω is a rectangle, then we can write it as a "union of disjoint open rectangles" by taking the set of such rectangles to include only Ω itself.

To show the other way, that if Ω can be written as a "union of disjoint open rectangles then it is a rectangle", use proof by contradiction- if Ω is not a rectangle, then such a union cannot consist of a single rectangle. But the "lemma" you give shows that the union of two or more disjoint open rectangles is not connected, so cannot be a rectangle as all rectangles are connected. That gives a contradiction, proving the theorem.