is this response to cantors diagonal argument allowed?

Question

Take cantors set of points on a line [0,1] if we have a function that maps the points onto a list we will always find a number not on that list through the diagonal argument. Yet if we keep taking the midpoints of [0,1] to a list wont we eventually fill every point. it would look like this:

1. 1
2. 0
3. 0.5
4. 0.75
5. 0.25
6. 0.875 and so on...

I understand that its impossible for fractions to represent an irrational number yet with each repetition you are forming a new series that will converge towards every irrational number between [0,1]. Surely as the list tends to infinity i could argue that the fractions will converge to the irrational numbers? i apologize i don't have the mathematical rigor to formally construct the argument.

Answer 1

Given that you start with rational numbers, if you keep taking midpoints of earlier numbers, you will only get more rational numbers, since for any rational numbers $p$ and $q$, their midpoint $\frac{p+q}{2}$ is another rational number. So, you will never obtain any irrational number.

Now, you seem to realize that, and you say that we will get arbitrarily close to any irrational number, but the difference is essential: we are looking for a complete listing of all real numbers between $0$ and $1$, not just a bunch of numbers arbitrarily close to such numbers.

Answer 2

You did notice something relevant, but it does not contradict Cantor's argument. You have no way of knowing that that list covers all of the irrational numbers. However, you can modify your argument to provide another proof of the uncountability of the reals. Do you see how?