Class Notes (836,136)
Philosophy (1,521)
PHL246H1 (19)
Lecture

# Homework assignment 1(1).pdf

4 Pages
527 Views

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
More Less

Related notes for PHL246H1
Me

OR

Join OneClass

Access over 10 million pages of study
documents for 1.3 million courses.

Join to view

OR

By registering, I agree to the Terms and Privacy Policies
Just a few more details

So we can recommend you notes for your school.