Regarding Cantor's Diagonal Argument

[ryan]

Then the axiom of Archimedes uses a proof by contradiction based on the above definition. Assume there is a least upper bound for N that we will call b; so b = l.u.b N.

Least upper bounds are not guaranteed for all sets, in general you must argue that b exists, not simply assume it.

I don't know what you mean here because we are trying to show that b can't exist. Why would we argue for its existence?

A set can be bounded yet still not have a least upper bound. Showing that b cannot exist does not mean that N is unbounded unless you complete the argument.

Also, we generally want to specify which set the least upper bound comes from. N? R?

Yeah, I should have included that b E R.

Then you can complete the above argument using the least upper bound property of R. Of course, it is still possibly circular and certainly overkill to use properties of the reals to prove properties of the naturals...

b = or > k, for any k E N.

b - 1 < k, for k = b (this line is my own interpretation)

b < k + 1

But we defined N to be that if k E N, then k + 1 E N.

Therefor b is not the l.u.b of N.

So it all goes back to what we defined the set of N to be.

... hmmm ...

The standard proof that I think you're trying for is:

Since b is the least upper bound, nothing less than b is an upper bound, so there exists a k such that k > b - 1. That means that for that k, k + 1 > b. But b is an upper bound for N and k + 1 is in N, so k + 1 > b ≥ k + 1, which is a contradiction.

But what about my interpretation? Is there anything wrong with it?

The main issue with what you wrote is that because you are finding the least upper bound in R, you need an argument that b is a natural number in order to choose k = b.

Ohhhhh yeah, okay thanks beero, you are a good teacher.