I am trying to explain why this proof is false. The easy way is to just assert that we have a valid proof that tells us that the cardinality of the irrational numbers is greater than the set of rationals and be done with it, but I want to be able to explain exactly where this proof is wrong.
Let $I$ be the set of irrational numbers, where the elements are $\{i_n: n \in R \land 0 < i_n < 1.0 \} $
Let $T_{i_n}$ be the set of all rational prefixes of $i_n$
For example if $i_{1.3721\dots} = 0.314159\dots$ then set $T_{i_{1.3721\dots}} = \{ 0.3, 0.31, 0.314, 0.3145, 0.314159, \dots\}$
Let $T = \bigcup T_{i_n}: n \in R$
Then $T$ will be a set of rational numbers
If $\exists q : q \in T_{i_k} \land q\notin \bigcup T_{i_n}: n \in (R \setminus k)$
Then $|I| \le |T|$ We are done
Else $|I| \not \le |T|\rightarrow \exists i_x \exists i_y: i_x \ne i_y \land T_{i_x} = T_{i_y}$ Contradiction
Two different irrational numbers can't have the same prefix set.
The proof is trying to say that every prefix set contributes at least one rational number to the union, and if that one prefix set were removed, then there would be some rational number missing from the union.
I would argue that in the prefix set given as an example, there are an infinite number of prefix sets that contribute 0.3 to the union, and there are an infinite number of sets that contributes 0.31, and the same for 0.314, and the same for all the elements in that prefix set. So you can remove any prefix set from the union and the union will not be any less. You can even remove an infinite number of irrationals and still have the same union. The only way to reduce the rational elements in the union, is by removing a segment from the union. For example if you remove an interval, like all the irrational numbers between 0.005 and 0.006 then the rational numbers from 0.005 to 0.006 will not be in the union.
Would this be one reason for why the proof fails and are there others?