1. CS 461 – Nov. 2 Sets Prepare for ATM finite vs. infinite Infinite sets
Countable
Uncountable
Prepare for ATM
Proving undecidable is similar to proving a set uncountable.
Please be sure you understand reasoning.

2. Finite/infinite
Finite set – means that its elements can be numbered. Formally: its elements can be put into a 1-1 correspondence with the sequence 1, 2, 3, … n, where n is some positive integer.
An infinite set is one that’s not finite. 
However, there are 2 kinds of infinity!

3. Countable
Countable set: An infinite set whose elements can be put into a 1-1 correspondence with the positive integers.
Examples we can show are countable:
Even numbers, all integers, rational numbers, ordered pairs.
Uncountable set = ∞ set that’s not countable. 
Examples we can show are uncountable:
Real numbers, functions of integers, infinite-length strings

4. Examples
The set of even numbers is countable.
The set of integers is countable. 2 4 6 8 10 12 … 1 3 5 1 -1 2 -2 3 -3 4 -4 … 5 6 7 8 9

5. Ordered pairs j = 1 j = 2 j = 3 j = 4 i = 1 1 2 4 7 … i = 2 3 5 86 9
i = 4 10
The number assigned to (i , j) is (i + j – 1)*(i + j – 2)/2 + i

6. Real numbers Real # Value 1 2.71828… 2 3.14159… 3 0.55555… 4 -1.23456…
Suppose real numbers were countable. The numbering might go something like this:
The problem is that we can create a value that has no place in the correspondence!
Real #
Value 1 … 2 … 3 … 4 … 5 …
X ? …

7. Infinite bit strings # Value 1 00000… 2 100000… 3 011010… 4 0010001… 5… … ?
11010…

8. Universal TM Let’s design “U” – the Universal TM:
Input consists of <M> and w:
<M> is the encoding of some TM
w is any (binary) string.
Assume: U is a decider (i.e. ATM is decidable.) U
No* means “does not accept.” Either enter reject state or loop forever.
<M>,w M
yes
yes w
no* no

9. ATM solution Start with U, the Universal Turing Machine
Suppose U decides ATM. Let’s build new TM D.
D takes in a Turing machine, and returns opposite of U’s answer. D
<M> U no
yes
<M>,<M> no
yes
If M accepts its own string rep’n, D rejects <M>.
If M doesn’t accept <M>, D accepts <M>.
What does D do with <D> as input?

10. For example <M1> <M2> <M3> <M4> … <D> M1
Yes No M2 M3 M4 D
Uh-oh
Contradiction  The TM D can’t exist  So U is not a decider.

11. In other words Let U = universal TM. Create 2nd TM called D.
Its input is a TM description <M> and a word <w>.
Determines if M accepts w.
Assume U halts for all inputs. (is a decider)
Create 2nd TM called D.
Its input is a TM description <M>.
Gives <M> to U as the TM to run as well as the input.
D returns the opposite of what U returns.
What happens when the input to D is <D>?
According to U, if D accepts <D>, U accepts, so D must reject!
According to U, if D rejects <D>, U rejects, so D must accept!
Both cases give a contradiction.
Thus, U is not a decider. ATM is undecidable.