Sets - Proving every set is a subset of itself

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Homework Help Overview

The discussion revolves around proving that every set S is a subset of itself, denoted as S ⊆ S. The original poster is tasked with using a 'proof by cases' approach to demonstrate this property in set theory.

Discussion Character

  • Exploratory, Conceptual clarification, Problem interpretation

Approaches and Questions Raised

  • The original poster attempts to understand how to express the proof using cases, noting that for every element x in set S, x is also in S, which suggests S ⊆ S. Some participants propose distinguishing between empty and non-empty sets as a way to structure the proof.

Discussion Status

Participants are exploring different interpretations of the requirement to use 'cases.' Some have suggested specific cases involving elements belonging to the set and the implications of the set being empty or non-empty. Guidance has been offered regarding the structure of the proof, but no consensus has been reached on the final form.

Contextual Notes

The original poster is required to use a 'proof by cases' method, which has led to discussions about the implications of different types of sets (empty vs. non-empty) in the proof process.

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Homework Statement



Prove that for every set S, S [tex]\subseteq[/tex] S. Use 'proof by cases'.


Homework Equations



A [tex]\subseteq[/tex] B iff {X: X [tex]\in[/tex] A --> X [tex]\in[/tex] B}

The Attempt at a Solution



I know that A is a subset of B if every element of A is also an element of B. In the case of S [tex]\subseteq[/tex] S, all I can figure out, simply, is:

For every element x in set S, x is an element of S, therefore, S [tex]\subseteq[/tex] S

I do not know how to express this proof in terms of 'cases'. Any help would be appreciated.
 
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You are required to use "cases"? How strange.

Try this:
case 1: Suppose [itex]x\in S[/itex] then ...

case 2: Suppose [itex]x\notin S[/itex] then ...
 
Perhaps "cases" means to make a distinction between empty and non-empty sets.
 
@HallsofIvy and @statdad: Thank you for the response.

Should it be something like this?

@HallsofIvy:

For S [tex]\subseteq[/tex] S : [tex]\forall[/tex] x(x [tex]\in[/tex] S [tex]\rightarrow[/tex] x [tex]\in[/tex] S)

Case 1: Let x [tex]\in[/tex] S, then x [tex]\in[/tex] S. p[tex]\rightarrow[/tex]p is true, therefore S [tex]\subseteq[/tex] S

Case 2: Let x [tex]\notin[/tex] S, then p is false. Since the antecedent is false in a conditional statement, the condition is true by vacuous proof. Therefore S [tex]\subseteq[/tex] S.


AND/OR


@statdad:

For S [tex]\subseteq[/tex] S : [tex]\forall[/tex] x(x [tex]\in[/tex] S [tex]\rightarrow[/tex] x [tex]\in[/tex] S)

Case 1: Let S be an empty set, then S = [tex]\phi[/tex]. Let x [tex]\in[/tex] S. For S [tex]\subseteq[/tex] S : [tex]\forall[/tex] x(x [tex]\in[/tex] [tex]\phi[/tex] [tex]\rightarrow[/tex] x [tex]\in[/tex] S). Since [tex]\phi[/tex] has no elements, the first statement is false and thus the whole condition is true by vacuous proof. Therefore S [tex]\subseteq[/tex] S

Case 2: Let S be a non-empty set, Let x [tex]\in[/tex] S, then x [tex]\in[/tex] S. p[tex]\rightarrow[/tex]p is true, therefore S [tex]\subseteq[/tex] S



I hope I interpreted the cases correctly. Please advise.
 
Upon discussion with the lecturer (apparently I was wrong, earlier, to think that lecturers are not supposed to help with assignments), he mentioned the same thing as statdad: use the two cases of S being an empty and a non-empty set.

Thank you everyone. PF and its helping members are great.

Cheers
 

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