CSCE 411 Lecture 34

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Lecture Slides

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Homework 9 posted on course web page (read through all textbook NPC problems carefully)

Undecidability

Models of Computation

Clearly and unambiguously specify how computation takes place.

  • Turing machines
  • Random Access machines (more realistic, but with infinite memory)
  • ...

They are all equivalent!

Church-Turing Thesis

Anything we reasonably think of as an algorithm can be computed by a Turing Machine

Set Theory Concepts

If A and B are sets, then the set of all functions from A to B is denoted by BA.

If A is a set, then 𝒫(A) denotes the power set (i.e. 𝒫(A) is the set of all subsets of A)

Cardinality

Two sets A and B are said to have the same cardinality iff there exists a bijective function from A onto B. (See CSCE 222 Lecture 11 and MATH 302 Lecture 8#Function)

How do set theorists count?

  • 0={}
  • 1={0}={{}}
  • 2={0,1}={{},{{}}}
  • Keep including all previously created sets as elements of the next set

Power set cardinality: |𝒫(X)|=|2X| since the bijection is given by the characteristic function:

  1. ∅↦f(a)=f(b)=0
  2. {a}↦f(a)=1,f(b)=0
  3. {b}↦f(a)=0,f(b)=1
  4. {a,b}↦f(a)=f(b)=1

We say |A|≤|B| if there is an injective function from A to B

We say |A|<|B| if there is no biective function from A to B

Cantor's Theorem

Let S be any set. Then |S|<|𝒫(S)|

Proof

Since function i from S to 𝒫(S) given by i(s)={s} is injective, we have |S|≤|𝒫(S)|.

Claim. There does not exist any function f from S to P(S) that is surjective.

T={s∈S:s∉f(s)} is not contained in f(S).

An element s∈S is either contained in T or not:

  • s∈T⟹s∉f(s) by definition of T, so T≠f(s).
  • s∉T⟹s∈f(s) by definition of T, so T≠f(s).

Therefore

f

is not surjective. This proves the claim.

Q.E.D.

Countable Sets

Let ℕ be the set of natural numbers.

A set X is called countable iff there exists a surjective function from ℕ onto X.

Thus finite sets are countable, ℕ is countable, but the set of real numbers is not countable.

An Uncountable Set

The set ℕℕ={f∣f:ℕ→ℕ} is not countable.

Proof. We have |ℕ|<|𝒫(ℕ)| by Cantor's theorem.

Since |𝒫(ℕ)|=|2ℕ| and 2ℕ is a subset of ℕℕ we can conclude that

|ℕ|<|𝒫(ℕ)|=|2ℕ|≤|ℕℕ

The cardinality of computer programs is countable (|ℕ|), but the cardinality of problems is not (|ℕℕ|)