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Lecture

Homework assignment 1(1).pdf

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Department
Philosophy
Course
PHL246H1
Professor
Franz Huber
Semester
Fall

Description
Homework assignment 1 (due Thursday, September 26) Intuitively, a set is a collection of objects (things, entities). For instance, the set T of people teaching this course is the collection of “objects” (or subjects, if you prefer) that is either a TA for PHL 246 or an instructor for PHL 246. We use curly brackets ‘{’ and ‘}’ to denote sets. One way to describe T is as follows: T = {Parisa, Seya, Franz}. A different way to describe the very same set T is as follows: T = {x: x is a TA for PHL 246 or x is an instructor for PHL 246}. The order in which the members are listed does not matter: T = {Parisa, Seya, Franz} = {Seya, Parisa, Franz}. Nor does it matter if a member is listed once or more than once: T = {Parisa, Seya, Franz} = {Seya, Parisa, Franz, Franz}. We use the symbol ‘∈’ to denote that an object is a member of a set, and the symbol ‘∉’ to denote that an object is not a member of a set: Franz ∈ T, but Rob Ford ∉ T. Every set C is identical to the set of all its members: for all sets C, C = {x: x ∈ C}. This will be useful below. We use the symbol ‘⊆’ to denote that all the members of one set are members of another set, and we call the former a subset of the latter (and the latter set of superset of the former): {Parisa, Seya} ⊆ T, which just means that the set of TAs for PHL 246 is a subset of the set of teachers for PHL 246. Two sets are identical just in case they have the same members. So for any sets B and C, B = C just in case B ⊆ C and C ⊆ B. Example: {Parisa, Seya, Franz} ⊆ T and T ⊆ {Parisa, Seya, Franz}, and so T = {Parisa, Seya, Franz}. There is exactly one set that has no members. This is the empty set, which is denoted by ‘∅’ or ‘{}’. The empty set is a subset of every set C, since all its members (namely none) are also members of any set C: for any set C, ∅ ⊆ C. Every set C is a subset of itself, since trivially all members of C are members of C: for any set C, C ⊆ C. If B and C are sets, we can form the intersection of B and C, (B∩C) (read: B intersection C), which is the set of objects that are members of B as well as of C: (B∩C) = {x: x ∈ B and x ∈ C}. Example: T∩{x: x is TA for PHL 246} = {Parisa, Seya}. If B and C are sets, we can form the union of B and C, (B∪C) (read: B union C), which is the set of objects that are members of either B or C or both: (B∪C) = {x: x ∈ B or x ∈ C}. Example: {Parisa, Seya}∪{Franz} = T. If B and C are sets, we can form the complement of B with respect to C, (C\B) (read: C complement/ without B), which is the set of objects that are members of C, but not of B: (C\B) = {x: x ∉ B and x ∈ C}. Example: T\{Franz} = {Parisa, Seya}. If C is a set, we can form the powerset of C, ℘(C), which is the set of all subsets of C: ℘(C) = {B: B ⊆ C}. For instance,℘(T) = {∅, {Parisa}, {Seya}, {Franz}, {Parisa, Seya}, {Parisa, Franz}, {Seya, Franz}, T}. I will often drop the parentheses ‘(’ and ‘)’ and write ‘B∩C’ instead of ‘(B∩C)’ etc. Consider an arbitrary non-empty set W, that is, an arbitrary set W that has at least one member and so is different from the empty set, ∅. Show that the following is true for all subsets B and C and D of W: (I) B = B∩B (1 point) I will illustrate how this works by showing that B = B∪B. B = {x: x ∈ B} = {x: x ∈ B or x ∈ B} = B∪B. The first equation holds, because B is identical to the set of all its members. The second equation holds, because
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