Uncountability of Real numbers - fault in counter proof

Question

I came up with the below proof for the proposition “the set of real numbers is countable”. Can you please help me find the error in this proof.

Proof:

1. Let P be the set of primes. P is a subset of integers and hence it is countable. Also P is not finite.

2. Let real numbers be defined as $r = x.d_1 .d_2 .d_3 .\ldots. $ Let L be the set $\{x, d_1, d_2, d_3, \ldots\}. $ L is countable. [If L is not countable, L is a counter-example to the Continuum hypothesis as cardinality of real numbers is exponential to cardinality of L. ( Ref 1 )]

3. We could map the set L to set P as:

   $x => p_1$

   $d_1 => p_2$

   $d_2 => p_3$

   $\vdots$

4. Each real number can then be represented as: $p_1^x.p_2 ^{d_1}.p_3^{d_2} .\ldots. $

5. Thus real numbers are countable. (This technique is an extension of Ref 2 - Proposition: The Cartesian product of two countable sets A and B is countable.)

References:

1. https://en.wikipedia.org/wiki/Continuum_hypothesis
2. https://en.wikipedia.org/wiki/Countable_set#Some_technical_details

Answer 1

Let's patch up the technical error at the beginning of your argument by referring to the sequence $L = x,d_1,d_2,d_3,\ldots .$ There is then indeed a one-to-one correspondence between the sequence $L$ and the set of all primes.

Now let's consider the real number $\frac{10}9.$ In this example, $x = 1$ and $d_i = 1$ for all $i.$ Then you claim to represent $\frac{10}9$ by $$ p_1^x \cdot p_2^{d_1} \cdot p_3^{d_2} \cdot p_4^{d_3} \cdots = 2^1 \cdot 3^1 \cdot 5^1 \cdot 7^1 \cdot 11^1 \cdots = 2\cdot 3\cdot 5\cdot 7\cdot 11 \cdots.$$

But what is $2\cdot 3\cdot 5\cdot 7\cdot 11 \cdots$? If we take any finite product of primes, such as $2\cdot 3\cdot 5\cdot 7\cdot 11 \cdots 2017,$ we get a positive integer, but $2\cdot 3\cdot 5\cdot 7\cdot 11 \cdots$ is not a finite product. It's the product of all of the prime numbers. The "$\cdots$" part means we just keep multiplying and multiplying without end, and the partial products keep getting bigger and bigger. Consider any integer $n$; since every partial product is an integer and the partial products never stop getting bigger, eventually they get bigger than $n.$ Hence the product $2\cdot 3\cdot 5\cdot 7\cdot 11 \cdots$ cannot be an integer.

What you can show with this technique is the well-known fact that the set of terminating decimals is countable. Any terminating decimal has a last non-zero digit and therefore there is a last positive power of a prime in the product; all the factors after that are $1,$ and the partial products after that point are all the same. Hence for a terminating decimal $x.d_1d_2d_3\cdots,$ we find that $p_1^x \cdot p_2^{d_1} \cdot p_3^{d_2} \cdot p_4^{d_3} \cdots$ is an integer, giving us an injection from the set of terminating decimals to the set of integers.