Philosophical Devices
Week 1 – Introduction to Naïve Set Theory

Definitions & Such

Set=_{df} any welldefined 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 uppercase letters, elements with lowercase 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 BA
Notation: we express subset using ‘’, and proper subset using ‘’

{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 nmembers has 2^{n} 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 ‘AB’, 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: AB=_{df }{  x ∈ A and y ∈ B}
We can extend the above story for all ordered ntuples, 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: ABC =_{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, xA iff x
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 selfmembered. Is there a set for this condition?
Suppose there is – call it R (for Russell). We can now ask, is RR or RR?
Let RR. By definition, R contains all and only those sets that are not members of themselves. This means that, if RR, then RR. So, RR. But, this is contradicts our initial assumption that RR. Therefore, it is not the case that RR.
Alternatively, let RR. By definition, R contains all and only those sets that are not members of themselves. This means that, if RR, then RR. So, by assumption, RR. But, this contradicts our initial assumption that RR. Therefore it is not the case that RR.
Problem: if it is not the case that RR, then RR, and if it is not the case that RR, then RR. Yet, given the reasoning above, we know that neither RR nor RR.
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 712}, {x: x is a child aged 1015}
(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}
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.
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