philosophical logic: essay 2

7/30/2019 Philosophical Logic: Essay 2

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Phil. 427/627: Lecture 2 (January 31, 2013)

Chapter 1: Enumerability

Goal of Chapter 1:(i) To understand the concept of enumerability.

(ii) To understand the notion of partial function.

enumerable = countable

Idea: An enumerable set is one whose elements can be arranged in a sinlge list. Every memeberappears sooner or later, maybe more than once.

(Note: Every finite set is enumerable.)

Definition: A denumerable set is an infinte enumerable set.

Cardinal numbers

What do we mean by a set having 5 members? We will try to anwer this obvious questionby using a concept of function, since that will help our further discussions. A set A having five

members means that there is a one-to-one function from set {0, . . . , 4} onto set A. So, if a set Ahas n memebrs, then there is a bijection f : {0, . . . , n 1} A.

We can extend this modeling (of the notion of size) from finite sets to infinite sets, answeringthe following questions: How can we talk about the size of an infinte set? Is there one kind ofinfinity?

Definitions: Let A and B be sets.

1. We say that A and B have the same cardinality (write A = B) if and only if there is abijection f : A B.

2. We write A B if there is one-to-one function f : A B.

3. We say that A < B if A B and A = B.

(Note: Having the same cardinality is an equivalence relation.)

Definition: A set is finite if it has the same cardinality as set {0, . . . , n} for some natural numbern N. Otherwise, the set is infinite.

By the way, set N is the smallest(?) infinite set, we call its cardinality 0.

Definition:A set A is countable iff A 0.A set A is denumerable (or infinitely enumerable or countably infinite) iff A = 0.

A set A is uncountable iff A > 0.

(Examples) The following sets are denumerable sets:Set of negative integersSet of positive even numbersSet of positive rational numbers

(Note: The laws of arithmetic for infinite sets are different than they are for finite sets.)

After understing the notion of partial function, we will come back to enumerability again.

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Partial Function

We defined a function to be a set of ordered pairs with the following characteristic:for any x, there is one and only y such that x, y is in the set. That is,f : A B iff Dom(f) = A and Rng(f) B.

How about +(Mary, 3)?

Definition: f is defined at x (write f(x) ) iff x Dom(f). [i.e. yf(x) = y]f is undefined at x (write f(x) ) iff x Dom(f).

Definition: f is a total function on A iff f is a function and Dom(f) = A.f is a partial function on A iff f is a function, Dom(f) A.

Important: Partial functions arent really a special type of function, but offer a new and usefulway of considering old functions. Hence, it will be useful to have a way of talking about functionswithout saying whether they are total or partial.

Definition: f; A B(f semi-maps A into B) iff Dom(f) A and Rng(f) B.[I.e. A0 A such that f : A0 B]

Next, lets extend the notions of 1-1 and onto mappings to semi-mappings:

Definition: f; A 11 B iff A0 A such that f : A0 11 B

f; A onto B iff A0 A such that f : A0 onto B

Henceforth, the term function will mean total or partial. That is, when we talk about a functionfrom N to N, we will mean a semi-mapping.

With a notion of partial functions, lets continue our discussion of enumerability.

Redundancy in a list is okay, since we can always get rid of repetition.

How about gaps?For example, 0, - , 1, -, 2, -, 3, . . . (Call it a gappy list.)

How can we show that this is enumerable list? That is, how can we make a function from theset of natural numbers to this gappy list? This is where we will use a partial function.

f(0) = 0, f(1) , f(2) = 1, f(3) , . . .

Also,f(n) = n/2 if n is even

otherwise

The following function is the same as we did in the above, but lets remember that f doest nothave to be total on N.

Definition: f enumerates A iff f;N 11onto A

Definition: A is enumerable iff f such that f enumerates A.

[p.7: A is enumerable iff it is the range of some function of natural numbers.]Lets compare this definition with the above definition countable. That is,

A set A is countable iff A 0.That is, A is countable iff X N A = X.

Then,A = X iff f f : A 11onto X

iff f f : X 11onto Aiff f f ; N 11onto A

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Chapter 2: Diagonalization

Goal of Chapter 2:(i) To see that some sets are not enumerable, and(ii) to understand the technique involved in this kind of a proof.

Here is a time-honored theorem:Cantors theorem: For any set A, A < (A).

Proof: We need to show (i) A (A) and (ii) A = (A).Show (i): To find a one-to-one function f : A (A). Let f(x) = {x}. This function is 1-1.Show (ii): To show that there is no one-to-one onto function from A to (A). Suppose that

there is a bijection f : A (A). [Show that there is a contradiction.] Since f is an ontofunction, Rng(f) = (A).

Let B = {x | x A and x f(x)}. By the definition of this set, for every x B, x A. Thatis, B A. Accordingly, B (A). Since Rng(f) = (A), B Rng(f). Then, by the definitionof Rng(f), there is a such that f(a) = B where a A.

Is a B? If so, a f(a) by the definition ofB. That is, a B, since f(a) = B. Contradiction.

Is a B? If so, a f(a) since f(a) = B. Then, by the definition of B, a B. Contradiction.2

We can illustrate the same point in a slighly different(?) way, which the book does.

The set of all sets of natural numbers (i.e. the set of the subsets of N) is not enumerable. It istoo big(!). In the proof for this proposition, we will learn a very useful technique: For a given anylist L of sets of natural numbers, we will construct a set D(L) of natural numbers which doesnot show up in the list L. What if we add D(L) to the list? Then, for this new list L, we cancome up with a set D(L) which does not occur in the list L.

Important Method: Diagnolization (How to construct D(L), given the list L.)

Given the list L of the sets of natural numbers,S1, S2, S3, . . .(Note: An infinite set is included in the list as well.)

We will show this list is always missing an element, that is, we cannot enumerate them. Wesuggest the following element as the candidate:

For each n, n D(L) iff n Sn. (Note: D(L) is infinite.)

(Example) {1}, E, {1,2}, E {2}, {1,3}, . . .S1, S2, S3, S4, S5, . . .

Then, D(L) = {3, 5, . . . } (since 1 S1, 2 S2, 3 S3, 4 S4, 5 S5, . . . )

Show that D(L) is not in the list L.

Proof: Suppose by reductio that D(L) appears on the list. Say, Sm = D(L).

Then, by the definition of D(L), we know that m D(L) iff m Sm.Since Sm and D

(L) are the same set, that m D(L) iff m D(L).This is a plain contradiction!

Lets look at this important diagonalization method from a slightly different point of view:

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Characteristic function

For a given set A, the characteristic function CA is a function that outputs 1 if its argument is inthe set A and 0 if it is not. That is,

CA(x) = 1 if x A

0 if x A.

We can set up the characteristic function CSn for each set of natural numbers, Sn, in the followingway:

CSn(x) = 1 if x Sn0 if x Sn.

(Note: In B & J, the characteristic functions sn is the same as CSn in the above.)

Then, we can draw the following array for the given list L: (Figure 2-1 on p.12)

1 2 3 4 . . .CS1 CS1(1) CS1(2) CS1(3) CS1(4) . . .CS2 CS2(1) CS2(2) CS2(3) CS2(4) . . .CS3 CS3(1) CS3(2) CS3(3) CS3(4) . . .

CS4 CS4(1) CS4(2) CS4(3) CS4(4) . . ....

Lets go back to our example to see how the array looks like:

(Example) {1}, E, {1,2}, E {2}, {1,3}, . . .S1, S2, S3, S4, S5, . . .

1 2 3 4 . . .CS1 1 0 0 0 . . .CS2 0 1 0 1 . . .CS3 1 1 0 0 . . .CS4 0 0 0 1 . . .CS5 1 0 1 0 . . ....

(Note: From this array, we can tell what the original list is. For example, by looking at thefirst row, we know that S1 = {1}, . . . , by the fifth row, S5 = {1, 3}. etc.

Lets go back to our array, and draw a diagonal sequence like the following:

d = CS1(1), CS2(2), CS3(3), CS4(4), . . .

This diagonal sequence might be on the list. So, we need one more trick to make sure that wecreate a sequence which does not appear on the list. We know that each element of sequence d iseither 1 or 0. So, we reverse each element of d by changing 1 to 0 and 0 to 1. So,

d = 1 CS1(1), 1 CS2(2), 1 CS3(3), 1 CS4(4), . . .

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Back to our example:

d = 1, 1, 0, 1, . . .d = 0, 0, 1, 0, . . .

We claim that d does not appear on the list L.

Proof : Suppose (by reductio) it does. Lets say that d is the mth row in the array. Mth rowlooks like this:

CSm CSm(1) CSm(2) CSm(3) CSm(4) . . . . . . CSm(m) . . . . . .

Since d = 1 CS1(1), 1 CS2(2), 1 CS3(3), 1 CS4(4), . . . , 1 CSm(m), . . . , we get the followingequations:

CSm(1) = 1 CS1(1),CSm(2) = 1 CS2(2),CSm(3) = 1 CS3(3),

CSm(4) = 1 CS4(4),...

CSm(m) = 1 CSm(m),...

That is,

1 2 3 4 . . . m . . .

CS1 1 0 0 0 . . .CS2 0 1 0 1 . . .

CS3 1 1 0 0 . . .CS4 0 0 0 1 . . .CS5 1 0 1 0 . . .

......

......

......

CSm 1 CS1(1) 1 CS2(2) 1 CS3(3) 1 CS4(4) . . . 1 CSm(m)...

But, CSm(m) = 1 CSm(m) is a plain contradiction, which leads to 0=1! So, sequence d cannot

appear on the list. 2

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philosophical logic: essay 2

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