Open sets, countable unions of open rectangles

[infk]

Homework Statement

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 ([itex]\mathbb{R}^d[/itex], I suppose) is a countable union of open, disjoint rectangles if and only if Ω is itself a rectangle.

Homework Equations

N/A

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 [itex]\bigcup[/itex]U where U and V are open, non-empty and disjoint. Thus if we can write Ω = [itex]\bigcup^{\infty}_{k=1} R_k[/itex] where [itex]R_k[/itex] are open disjoint rectangles of which at least two are non-empty (lets say [itex]R_1[/itex] and [itex]R_2[/itex] ) we can then write Ω = [itex]R_1 \cup (\bigcup^{\infty}_{k=2} R_k) [/itex] and therefore Ω is not connected.
There is also another question; Show that an open disc in [itex]\mathbb{R}^2[/itex] 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 = [itex]\bigcup^{\infty}_{k=1} R_k[/itex] where [itex]R_k[/itex] 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?

 

[HallsofIvy]

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.