Is ∼ an Equivalence Relation on the Power Set of a Finite Set?

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Homework Help Overview

The problem involves defining a relation on the power set of a finite set and determining whether this relation is an equivalence relation. The original poster presents a finite set S and its power set, 2^{S}, and defines a relation ∼ based on the cardinality of subsets.

Discussion Character

  • Conceptual clarification, Assumption checking

Approaches and Questions Raised

  • Participants discuss the properties required for a relation to be an equivalence relation, including reflexivity, symmetry, and transitivity. There is an attempt to clarify the definitions and the notation used for the power set.

Discussion Status

Some participants provide guidance on verifying the properties of the relation ∼, while others express confusion regarding the total number of subsets in the power set and the specific elements within each equivalence class. Multiple interpretations of the definitions and notation are being explored.

Contextual Notes

There is mention of the original poster's misunderstanding regarding the notation for the power set and the number of subsets, as well as the need to clarify the definitions of the equivalence relation in the context of the problem.

kathrynag
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Homework Statement



Let S be a finite set and denote by 2^{S} = {A|A ⊆ S} the set of all subsets of S. Define
a relation ∼ on 2^{S} by A ∼ B if and only if A and B have the same number of elements.
(a) Show that ∼ is an equivalence relation on 2^{S}.
(b) Let S = {1, 2, 3, 4}. List the (sixteen) elements of 2^{S} and explicitly list the
elements in each equivalence class determined by ∼.

Homework Equations





The Attempt at a Solution


I started by determining what an equivalence relation is:
i. (a,a) is in ~
ii. For all (a,b) in S, if (a,b) is in ~, then (b,a) is in ~
iii. For all a,b,c in S, if (a,b) is in ~ and (b,c) is in ~, then (a,c) is in ~.
I have trouble using the definitions.
I tried doing something like 2^{a}=2^{a}
 
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Just follow the definitions, the relation defined was: "have the same number of elements"

So in order to check this: (a,a) is in ~
you just need to check: do 'a' and 'a' have the same number of elements

etc.(also, you meant: ii. For all a and b in 2^{S}, if (a,b) is in ~, then (b,a) is in ~... as the relation was defined on 2^{S}, not on S)
 
Last edited:
a and a have the same number of elements since a has the same number of elements as itself.
If a and b have the same number of elements, b and a have the same number of elements.
If a and b have the same number of elements and a and c have the same number of elements, then since a has the same number of element of b and c, b must have the same number of elements as c.

Ok for part b)
elements are 2, 4, 8, 16, but only 2, 4 are in S. I don't see how to get 16 elements.
 
No, for a set, S, the notation "2^S" means "the collection of all subsets of S". That notation is used because if set S contains n elements then it has 2^4 subsets. {1, 2, 3, 4} has 4 members so it has 2^4= 16 subsets.

The only subset that has 0 members is the empty set.

Their are 4 subsets that contain 1 member: {1}, {2}, {3}, {4}.

There are 6 subsets that contain 2 members, 4 that contain 3 members, and 1 that contains 4 members.

In general, if a set contains n members, there are the binomial coefficient,
\begin{pmatrix}n \\ m\end{pmatrix}
subsets that contain m members.
 

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