Philosophical Devices
Week 1 – Introduction to Naïve Set Theory
--------------------------------------------------------------------------------------------------------DEFINITIONS & SUCH
- Set=df any well-defined collection of ‘objects’
- Element=df the objects in a set
- Membership=df the relation elements bear to the sets they are in
Notation. Typically, sets are denoted with upper-case letters, elements with lower-case letters
– e.g. ‘A, B, C…’ for sets, ‘a, b, c…’ for elements. Membership is standardly expressed using
‘’.




x  A – the element x is a member of set A
x  A – the element x is not a member of set A
A - x – the set A has as a member the element x
A - x – the set A doesn’t have as a member the element x
Extensive specification: name all the members of a particular set
 {1, 2, 3}, {Schnieder, Schramme, Recki, Puster, Gahde}, {Socrates, {Socrates}}
Intensive specification: specify the ‘feature’ all the members share
 {x| x is a integer 1 x  3}, {x: x is a professor}, {x| (x=Socrates) or (x={Socrates})
}
Axiom of Extensionality: two sets are identical iff they have all and only the same elements
 For all sets A & B: A=B iff (x)( x  A iff x  B)
Two special sets:
 - The Empty set; the set that has nothing as a member
 - The Universal set; the set that has everything as a member
Is there more than one empty set? How about more than one Universal set?
--------------------------------------------------------------------------------------------------------FURTHER RELATIONS & OPERATIONS
Union: The union of two sets is the set which contains everything that belongs to either or
both. Similar to logical ‘or’ operator.
Intersection: the intersection of two sets is the set which contains everything that belongs to
both. Similar to logical ‘and’ operator.
Notation: we express union using ‘’, and intersection using ‘’
 {Socrates, Plato}  {Cicero} = {Socrates, Plato, Cicero}
 {Socrates, Plato}  {Plato}= Plato
 {Socrates}  {Plato} = 
Subset: B is a subset of A iff all the members of B are members of A.
Proper Subset: B is a proper subset of A iff all the member of B are members of A and BA
Notation: we express subset using ‘’, and proper subset using ‘’
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




{Socrates}  {Socrates, Plato}
{Socrates}  {Socrates, Plato}
{Socrates}  {Socrates}
  {Schnieder, Schramme, Recki, Puster, Gahde}
Socrates  {Socrates}
{Gähde, Schnieder, {Schramme, Recki}}
- Schnieder: a member, not a subset
- {Schramme, Recki}: a member& a subset
- { Gähde }: not member, is a subset
- Recki: neither member nor subset
Power Set: the set of all a set’s subsets; the power set of a set with n-members has 2n
members
 {Socrates, Plato} = {{Socrates, Plato}, {Socrates}, {Plato}, }
 {} = {}
 {Hesperus, Phosphorus} =
Difference: The difference of two sets consists of the elements of the first set that do not
belong to the second set. Thus, for two sets A & B, the difference of A and B is the set
consisting of those elements of A that are not in B, standardly written as ‘A – B’ (note: A – B
is sometimes also called the relative complement of B relative to A)
A – B =def {x| x ∈ A and x ∉ B}
Let A = {Socrates, Schnieder, Schramme} & B = {Schnieder, Cicero, Schramme}
 A–B=
 A – {Socrates} =
 {Socrates} – A =
Complement: the compliment of a set A, written A’, is the set consisting of everything not in
A
A’=df {x| x ∉ A}
 Using A from the previous example, A’=
 {{Socrates} – A}’ =
--------------------------------------------------------------------------------------------------------AN ASIDE ON ORDERING, CARTESIAN PRODUCTS, RELATIONS & FUNCTIONS
Ordered pair: written ‘<a,b>’, where a is considered the first element and b is the second
element of the pair. Importantly, <a,b> < b,a >, unlike with standard sets!
Definition: <a,b> =df {{a}, {a,b}}
Cartesian product: the Cartesian product of A and B, written ‘AB’, is the set consisting of
all ordered pairs which take an element of A as the first member of the pair and an element of
B as the second.
Definition: AB=df {<x,y> | x ∈ A and y ∈ B}
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We can extend the above story for all ordered n-tuples, where n is any natural number. For
example, ordered triples are usually defined as: <a,b,c> =df <<a,b>,c>. And for three sets A,
B and C the Cartesian product can be defined as: ABC =df ((A  B)  C)
Using these tools, it is possible to formalize talk about relations – e.g. ‘part of’, ‘sister of’, ‘in
same room as’, etc.
Similarly for functions – A relation F from A to B is a function from A to B iff:
(i) Each element in the domain of F is paired with just one element in the range, i.e.,
<a,b>F and <a,c>F iff b=c
(ii) The domain of F is equal to A
--------------------------------------------------------------------------------------------------------RUSSELL RUINS EVERYTHING – THE AXIOM OF COMPREHENSION
Comprehension: For any condition , there exists a set A such that for all x, xA iff x
- There is a set of all professors, all cars, all students, all students in this room, etc.
Take the set of all sets with just one member. Does it have itself as a member?
How about the set of all set with more than one member?
Let  = is not self-membered. Is there a set for this condition?
Suppose there is – call it R (for Russell). We can now ask, is RR or RR?
Let RR. By definition, R contains all and only those sets that are not members of
themselves. This means that, if RR, then RR. So, RR. But, this is contradicts our initial
assumption that RR. Therefore, it is not the case that RR.
Alternatively, let RR. By definition, R contains all and only those sets that are not members
of themselves. This means that, if RR, then RR. So, by assumption, RR. But, this
contradicts our initial assumption that RR. Therefore it is not the case that RR.
Problem: if it is not the case that RR, then RR, and if it is not the case that RR, then
RR. Yet, given the reasoning above, we know that neither RR nor RR.
The only solution is to go back and find an assumption we relied upon to reject.
--------------------------------------------------------------------------------------------------------EXERCISES
(Note: these are identical to the exercises included at the end of Chapter 1)
1. What is the union of the following pairs of sets?
(a) {Abe, Bertha}, {Bertha, Carl}
(b) {2, 5, 7, 11, 13}, {1, 5, 11, 13}
(c) {x: x is a child aged 7-12}, {x: x is a child aged 10-15}
(d) {France, Germany, Italy}, {Germany, Italy}
(e) {France, Germany, Italy}, {India, China}
(f) {x: lives in Germany}, {x: x lives in Europe}
(g) {x: x lives in China}, {x: lives in Europe}
(h) {x: x weighs more than 10 kilos}, {x: x weighs more than 7 kilos}
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2. What is the intersection of each of the above pairs of sets?
3. List all the subsets of the following sets.
(a) {Abe, Bertha}
(b) {7, 8, 9}
4. Give the power sets of the following sets.
(a) {1, 7}
(b) {London, Manchester, Birmingham}
5. Consider the set {1, 2, 3, 7, 8}
Which of the following items are (a) members (b) subsets or (c) neither?
2, {7,8}, {2,3}, {}, 3, {1, 2, 3, {7,8}}
6. Consider the set {1, 2, 3, {7,8}, {2,3}}
Which of the following items are (a) members (b) subsets or (c) neither?
2, {7,8}, {2,3}, {}, 3, {1, 2, 3, {7,8}}
7. (A): ‘This sentence is false.’ Show carefully that the supposition that (A) has a truth value
leads to a contradiction.