Diagonalization

Diagonalization Arguments: Overview

When do 2 sets have the same number of elements
Some examples:
- Positives and Negatives
- Positives and Naturals
- Positives and Integers
- Positives and Rationals
The examples lead up to our goal:
- Positives and Reals

Are these sets the same size: {a,b,c} and {x, y, z}
- How do you know?
- Both have 3 elements
And these sets:
- {a, b, c, d, e, f, g, h, i, j, k, l, m, n}
- {A, B, C, D, E, F, G, H, I, J, K, L, M, N}
How do you know?
- Did you count them?
- Perhaps, but not necessarily
Two sets have the same size if they
- Have the same number of elements (ie count them), or
- There is a one-to-one, onto correspondence between the elements
  - Written as 1:1,onto

How can we compare the sizes of infinite sets?
- Count them?
- Show that they are 1:1, onto
Let's look at some examples

Positives and Integers: could these sets have the same size?
What correspondence could we use?
- i = (p+1)/2 if p is odd
- i = -p/2 if p is even
- p = 2i-1 if i is positive
- p = -2i if i is negative
- Hmmm?
Thus, these sets are the same size!
- Hmmm?? Remember our definition of same size

Positives and Rationals: Are these sets the same size?
Is there a correspondence?
- Put solution on board ...
- Hmmm? We ignore some subtle points
Thus, these sets are the same size!

To Prove: The set of integers does not contain a largest element
Assume that the set of integers does contain a largest element
Let p be that largest element
Let q = p + 1
But q is an integer
- Both p and 1 are integers, and the set of integers is closed under addition
- Remember the concept of sets being closed under an operation
But, q is greater than p (Definition of greater, wave hands)
Thus q is greater than the largest element, a contradiction
Thus the assumption (there is a largest) was false
Conclusion: the set of integers does not contain a largest element

Positives and Reals (between 0.0 and 1.0):
Could these sets be the same size?
We will prove that they ARE NOT
- Proof by contradiction
- Assume they are the same size, and find a contradiction
- Assume they are the same size means assume that they are 1:1, onto
If they are 1:1, onto, then they can be listed 1, 2, 3, ...
This will lead to a contradiction

Assume they are 1:1, onto and thus a list of reals exists
- Every real must be on this list
Let Lⁱ be the list element at position i,
For some real r, let r_j be the digit at position j
Thus, Lⁱ_j is digit j of the real at position i
Now form real x as follows:
- Digit j of real x is 1 + L^j_j
- That is, digit j of x is 1 + digit j of the real at position j
Now, since the list contains all reals, then it must contain x
- Assume that x occurs at position k (ie x=L^k)
- Then L^k₁ is the first digit of x, and L^k₂ is the second digit of x
What is digit k of x????
- By definition of the list, it's L^k_k (since x is at position k)
- But by definition of x, it's 1 + L^k_k
- Thus, L^k_k = 1 + L^k_k!
This is a contradiction, and so the two sets are NOT 1:1, onto

Start with a list of all reals
Define real x: digit k of x is 1 + L^k_k
x is a real, and so it must appear in the list.
Assume it appears at position k (ie x = L^k)
What is digit k of x: L^k_k, of course
But, digit k of x is defined to be 1 + L^k_k
L^k_k = 1 + L^k_k is a contradiction, so conclude the list does not exist
Notice the two ways of looking at L^k_k
- If we look at x as member k of the list, we get digit k to be L^k_k
- If we look at x as defined, we get digit k to be 1 + L_k_k