You have an ink stamp that is so amazingly precise that, when inked and pressed down on the plane, it makes every circle whose radius is an irrational number (centered at the center of the stamp) black.
Is it possible to use the stamp three times and make every point in the plane black?
If it is possible, where would you center the three stamps?
(In reply to
re(2):my proof by Brian Wainscott)
Brian, the assertion is not an immediate result of anything.
You wrote:
" Given a random real number in [0,1] it MIGHT be rational, but the odds are it won't be."
Once again, the odds are not the issue. Existence is the issue.
You wrote:
"So not only does it have a non-empty complement from which D can be chosen, amost every point in the plane is in the complement."
Your assertion (that "Any point D chosen from the complement of S(A,B) will give you three points (A,B,D) where you can stamp and cover the plane.") is not an immediate result of the claim.
"Almost every..." doesn't cut it.
To, hopefully clear up your confusion, here's what I am expecting from your proof:
You must demonstrate that, in the non-empty complement, there exists a point that, if stamped, will ensure that the set of non-black points (are not only countable, but) equals the NULL SET.
Equalling the NULL set means, of course, there does not exist even one non-black point.
re(3):my proof
