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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.

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    $\begingroup$ When will $\frac{1}{3}$ appear in your list? $\endgroup$
    – Chas Brown
    Apr 13, 2017 at 23:25
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    $\begingroup$ What you're proving is that there exists a countable dense subset of $[0,1]$: your sequence contains subsequences that converge to any point in that interval. But note that that is much different from saying that $[0,1]$ is itself a countable set: your sequence does not contain every point in that interval. $\endgroup$ Apr 13, 2017 at 23:28
  • $\begingroup$ In a finite list it wont ever appear, yet if we keep repeating the midpoints we will keep ending up getting closer and closer to it. The argument however is that the list is infinite and therefore we should be able to say these convergences will tend to infinity and so will spit out the irrationals and other fractions missed. $\endgroup$ Apr 13, 2017 at 23:31
  • $\begingroup$ The crucial part of cantors diagonal argument is that we have numbers with infinite expansion (But the list also contains terminating expansions, which we can fill up with infinite many zeros). Then, an "infinite long" diagonal is taken and used to construct a number not being in the list. Your method will only produce terminating decimal expansions, so it is not only countable, but does not even produce all rational numbers. $\endgroup$
    – Peter
    Apr 13, 2017 at 23:32
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    $\begingroup$ @Matthew: That's what "dense" means -- everything in $[0,1]$ is a limit of a sequence of terms in your list. That does not mean they are actually present in the list: most elements of $[0,1]$ are not. $\endgroup$
    – user14972
    Apr 13, 2017 at 23:45

2 Answers 2

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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.

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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?

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  • $\begingroup$ Im stumped, maybe that each new midpoint causes 2 gaps between the points. As you repeat the process these gaps will get infinitesimally small yet there would be an infinite amount of them. i thought it would form an infinite amount of convergent series towards these gaps but perhaps these gaps really never go and show that points that will never be listed? $\endgroup$ Apr 14, 2017 at 0:14
  • $\begingroup$ @Mathew McLeod what I was reffering to is a proof by contradiction- you assume $x_n$ contains all reals, then see that the intersection of closed bounded intervals $I_n$ not containing $x_n$ has a non empty intersection by the nested interval therorem, call it $x$. Then x is not in $x_n$. This is very similar to your observation of splitting intervals. $\endgroup$ Apr 14, 2017 at 4:15

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