Ball in 2D plane as a countable union of rectangles

[Chasing_Time]

Hi all, I'm getting stuck on this problem.

Homework Statement

I am asked to show that that the open ball in the plane {|x|} < 1} can be written as a countable union of rectangles [a_1, a_2] x [b_1,b_2], but the closed ball in the plane {|x| <= 1} cannot be written as a countable union of rectangles.

The Attempt at a Solution

I believe it has something to do with the fact that the rationals are countable but the reals are not, and extending this relationship to 2-D. For a given rectangle (a1, a2) X (b1, b2), to get the open ball one could keep halving the distances |a2 - a1| and |b2 - b1| for each rectangle, and since the operation is halving the set the result is still countable, and further, the product will be countable as well as the union of all such rectangles. But to get the closed ball must invoke irrational numbers- I believe this has something to do with the fact that the ball is convex, but I'm not sure how to incorporate that either. At least a starting point would be very helpful. Thanks so much.

 

[Dick]

How about the set of ALL rectangles R with a1, b1, a2 and b2 rational and R is contained in O={|x|<1}. That's countable, right? Can you show it covers O? To go the other way, take a point x such that |x|=1 and, say, x is in the first quadrant. Can you show that if x is in a rectangle contained in C={|x|<=1} then x is the upper right corner of the rectangle? How many points are there on the first quadrant of the of the circle |x|=1? Can you show it's uncountable?

 

[Chasing_Time]

Yes- this is all coming together now. I'm just unable to show that the set R can cover O = {|x| < 1}.

 

[Dick]

Yes- this is all coming together now. I'm just unable to show that the set R can cover O = {|x| < 1}.

Pick a point y in O. Can't you show pretty easily that there is a rational rectangle that contains y but doesn't touch the boundary circle C={|x|=1}? The distance from y to C is positive.