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(5): I think it's... by David Shin)
Let me explain why the apparent paradox occurs in my previous example:
"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", ...."
The fallacy here is the mistake of infinite induction. It is true that if you can get infinitely many uncountable many binary strings, you should be able to get uncountably many by restricting the first bit to be a 0. And same for the second bit, and same for the third bit...
However, you cannot extend this argument to infinity. Let me give the classic example of why this argument doesn't work:
"I shall prove that the set of all positive integers is finite:
1. Let S(k) denote the set {1,2,...,k}.
2. Clearly, S(1) is finite.
3. If S(k) is finite, then S(k+1) is also finite.
4. By induction, then, the set of all positive integers is finite."
So Tristan, in your "only left compartments" proof of this problem, you can use your argument to claim that if uncountably many figure eights can be placed in the plane, then, for any one figure eight, you can assume that no other figure eights are placed in its right compartment. But you cannot make such a claim for all figure eights.
re(6): I think it's... - Elaboration
