3 ms·

What axiom do you have to add, are you talking about the axiom of infinity, defining the set of natural numbers? Because you do need that, but without that woul

What axiom do you have to add, are you talking about the axiom of infinity, defining the set of natural numbers? Because you do need that, but without that wouldn't even be able to define uncountability. After that, you can define the reals using Dedekind cuts or Cauchy sequences, which was known at the time (I believe Cantor actually worked on the Cauchy sequence construction). But there's an even simpler uncountable set: the power set of the natural numbers. Super easy to define, and the diagonalization argument falls right out.

GoblinSlayer 6y ago
To construct R you need to repeat Dedekind cut uncountable number of times, i.e. you need to iterate over all real numbers, which would make them countable.
- openasocket 6y ago
  You don't need any sort of "repetition" of Dedekind cuts to construct the real numbers, it's actually fairly straightforward. The set of all Dedekind cuts is a subset of the power set of rational numbers (i.e. each Dedekind cut is a set of rational numbers) satisfying the following properties: for each Dedekind cut A 1. A is not the empty set 2. A is not the set of all rational numbers 3. A is closed downwards, meaning if x is in A and y < x, then y is in A 4. A has no greatest element, meaning for all x in A, there is a y in A such that y > x. And each Dedekind cut is in one-to-one correspondence with a real number. You can define the usual arithmetic operations on them, show that every rational number has a corresponding Dedekind cut (for any rational q, we have {x in Q | y < q } as the corresponding cut). I haven't seen a proof of the uncountability of Dedekind cuts using the diagonalization argument, but you can prove there is a one-to-one correspondence between Dedekind cuts and Cauchy sequences of rational numbers, which is another construction of the reals. And there is a fairly straightforward proof that those are uncountable using diagonalization. But as you can see, you don't need any sort of repetition to do the construction, it's just a set of sets of rational numbers satisfying a few simple properties. For more information, check out https://en.wikipedia.org/wiki/Construction_of_the_real_numbers https://en.wikipedia.org/wiki/Construction_of_the_real_numbe...
  - GoblinSlayer 6y ago
    So you just assume that the set exists.
    - openasocket 6y ago
      Well I assume the set of natural numbers exist, by the axiom of infinity. From there you can construct the set of rationals. And you can construct the power set of the set of rationals, by the axiom of the power set. And the set of all Dedekind cuts is a subset of the power set of rationals satisfying those properties listed above, which we can construct by the axiom schema of separation. All of this is Zermelo-Frankel set theory, don't even need the axiom of choice.

GoblinSlayer 6y ago

To construct R you need to repeat Dedekind cut uncountable number of times, i.e. you need to iterate over all real numbers, which would make them countable.

openasocket 6y ago
You don't need any sort of "repetition" of Dedekind cuts to construct the real numbers, it's actually fairly straightforward. The set of all Dedekind cuts is a subset of the power set of rational numbers (i.e. each Dedekind cut is a set of rational numbers) satisfying the following properties: for each Dedekind cut A 1. A is not the empty set 2. A is not the set of all rational numbers 3. A is closed downwards, meaning if x is in A and y < x, then y is in A 4. A has no greatest element, meaning for all x in A, there is a y in A such that y > x. And each Dedekind cut is in one-to-one correspondence with a real number. You can define the usual arithmetic operations on them, show that every rational number has a corresponding Dedekind cut (for any rational q, we have {x in Q | y < q } as the corresponding cut). I haven't seen a proof of the uncountability of Dedekind cuts using the diagonalization argument, but you can prove there is a one-to-one correspondence between Dedekind cuts and Cauchy sequences of rational numbers, which is another construction of the reals. And there is a fairly straightforward proof that those are uncountable using diagonalization. But as you can see, you don't need any sort of repetition to do the construction, it's just a set of sets of rational numbers satisfying a few simple properties. For more information, check out https://en.wikipedia.org/wiki/Construction_of_the_real_numbers https://en.wikipedia.org/wiki/Construction_of_the_real_numbe...
- GoblinSlayer 6y ago
  So you just assume that the set exists.
  - openasocket 6y ago
    Well I assume the set of natural numbers exist, by the axiom of infinity. From there you can construct the set of rationals. And you can construct the power set of the set of rationals, by the axiom of the power set. And the set of all Dedekind cuts is a subset of the power set of rationals satisfying those properties listed above, which we can construct by the axiom schema of separation. All of this is Zermelo-Frankel set theory, don't even need the axiom of choice.