Suppose that I drew an infinite number of disjoint closed curves in the plane (such as circles, squares, etc.). Suppose that I then tell you that there is one curve for each positive real number.
You would not have too much trouble believing my assertions at this point. For example, I could have drawn all circles with center at the origin. They are all disjoint, and for each positive real number x, there is a corresponding circle - namely, the circle of radius x.
But suppose that I also tell you that all the curves I drew were figure eights. Can you believe my assertions now?
(A figure eight is a curve in the plane obtained from the basic "8" shape by any combination of translation, rotation, expansion, or shrinking.)
(In reply to
re(4): I think it's... by Tristan)
Sorry to keep poking at your proof...
You say, "Let's assume that we can enclose uncountably many figure 8s in a single largest one. If this is true, then we should be able to fit just as many even when we limit the area used to only the left compartments."
Very well. I shall use this same technique to show that the set of all binary strings (finite or infinite) is countable:
Start with the empty string. We add either 0 (left) or 1 (right) to it. From there, we add either 0 (left) or 1 (right) again, ad infinitum. Let's assume that there are uncountably many binary strings. If this is true, then we should be able to get just as many by only using "left compartments" (i.e., 0's) . But the number of binary strings with only 0's is clearly countable: "0", "00", "000", "0000", ....
Now, I was being pedantic here - the point is that we know that the set of all binary strings is in fact uncountable, so there must be something wrong with your argument.
re(5): I think it's...
