Philosophical Devices Week 1 – Introduction to Naïve Set Theory

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  Set=_df any well-defined collection of ‘objects’
  
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  Element=_df the objects in a set
  
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  Membership=_df the relation elements bear to the sets they are in
  

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  x  A – the element x is a member of set A
  
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  x  A – the element x is not a member of set A
  
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  A ^- x – the set A has as a member the element x
  
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  A ^- x – the set A doesn’t have as a member the element x
  

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  {1, 2, 3}, {Schnieder, Schramme, Recki, Puster, Gahde}, {Socrates, {Socrates}}
  

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  {x| x is a integer 1 x  3}, {x: x is a professor}, {x| (x=Socrates) or (x={Socrates}) }
  

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  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

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  {Socrates, Plato}  {Cicero} = {Socrates, Plato, Cicero}
  
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  {Socrates, Plato}  {Plato}= Plato
  
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  {Socrates}  {Plato} = 
  

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  {Socrates}  {Socrates, Plato}
  
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  {Socrates}  {Socrates, Plato}
  
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  {Socrates}  {Socrates}
  
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    {Schnieder, Schramme, Recki, Puster, Gahde}
  
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  Socrates  {Socrates}
  

{Gähde, Schnieder, {Schramme, Recki}}

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  Schnieder: a member, not a subset
  
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  {Schramme, Recki}: a member& a subset
  
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  { Gähde }: not member, is a subset
  
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  Recki: neither member nor subset
  

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  {Socrates, Plato} = {{Socrates, Plato}, {Socrates}, {Plato}, }
  
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  {} = {}
  
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  {Hesperus, Phosphorus} =
  

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  A – B =
  
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  A – {Socrates} =
  
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  {Socrates} – A =
  

Complement: the compliment of a set A, written A’, is the set consisting of everything not in A

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  Using A from the previous example, A’=
  
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  {{Socrates} – A}’ =
  

Definition: AB=_df { | x ∈ A and y ∈ B}

(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

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  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.

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(Note: these are identical to the exercises included at the end of Chapter 1)

(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}
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?

Which of the following items are (a) members (b) subsets or (c) neither?