Statement to prove:

If x and y are in R (Real Number) and x<y, then the interval (x,y) contains countably many rational numbers and uncountably many irrational numbers.

The hint suggests me to use the following two theorems:

1. Q is dense in R. That means:

(i) if x and y are in R and x < y, there is an r in Q with x<r<y.

(ii) if x in R and epsilon >0, there is an r in Q with |x-r| < epsilon.

2. The interval (0,1) in R is uncountable.

My work so far:

1. First of all, note that since the interval (0,1) is a subset of R and it is uncountable by Theorem 2, then it implies that the whole real numbers system R is uncountable.

2. Now, lets consider in interval (x,y) such that x and y are in R and x<y.

3. Since (x,y) is a subset of R, then it is uncountable.

4. By Theorem 1(i), it implies that there is an r1 in Q such that x<r1<y.

Now, lets consider the interval (x,r1), which is a subset of (x,y). By reapplying Theorem 1(i), we know that there exists an r2 in Q such that x<r2<r1. Continue this process.

.....This is what I have so far. Up to this point, I am not sure how to conclude that there are countably many rational numbers by continuing this process.

Any help is greatly appreciated. Thanks again for reading my problem and your time