Thread: prove that the set of real numbers is not a countable set

Originally Posted by rosebud

1. Prove that R is equivalent to (0,1) and (0,1) is equivalent to [0,1]. Conclude that R is equivalent to R.

surely that is not what the problem says! That R is equivalent to [0,1] would follow from this but "R is equivalent to R" is trivial.

To prove that R is equivalent to (0,1), look at the function f(x)= . To prove that (0,1) is equivalent to [0,1], write the rational numbers in (0, 1) in an ordering (which is possible because the rational numbers are countable) and define f(x)= x if x is irrational, for n>2.

Now prove that [0,1] is uncountable. Consider an arbitrary function f: J--> [0,1] and prove that im f is not equal to [0,1]. Thus, there is no function from J onto [0,1], and so [0,1] is uncountable. Suppose that T is a function from J to [0,1].
2. Show that there are sequences {a_n}from infinity when n=1 and {b_n} from infinity when n=1 such that [a_1, b_1] is a subset [0,1], and for each n E J, [a_n+1 , b_n+1] is a subset [a_n, b_n] and T(n) is not an element of [a_n, b_n]
3. Show that {a_n} from infinity when n=1 converges; call the limit A.
4. Prove that A E [a_n, b_n] for each n E J. Conclude that A is not an element im T.

I have no idea what "from infinity" means here.

5. Finish the proof that [0,1] is uncountable.