proof of pseudoparadox in measure theory
Since this paradox depends crucially on the axiom of choice, we will place the application of this controversial axiom at the head of the proof rather than bury it deep within the bowels of the argument.
One can define an equivalence relation on by the condition that if and only if is rational. By the Archimedean property of the real line, for every there will exist a number such that . Therefore, by the axiom of choice, there will exist a choice function such that if and only if .
Next, we will show that is injective. Suppose that and . By what we already observed, , so is a non-negative rational number and . There are 3 possible cases: 1) In this case, implies that , which would imply that . 2) In this case, implies which, in turn, implies that , which is impossible if both and belong to . 3) In this case, implies that , which would imply that . The only remaining possibility is that , so implies that .
Next, we show that is surjective. Pick a number in . We need to find a number such that . If , we can choose . If , we can choose .
Having shown that is a bijection between and , we shall now complete the proof by examining the action of . As we already noted, is a rational number. Since the rational numbers are countable, we can arrange them in a series such that no number is counted twice. Define as
|Title||proof of pseudoparadox in measure theory|
|Date of creation||2013-03-22 14:38:43|
|Last modified on||2013-03-22 14:38:43|
|Last modified by||rspuzio (6075)|