Sets - Proving every set is a subset of itself

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In summary, to prove that for every set S, S \subseteq S, we use proof by cases. We consider two cases: when S is an empty set and when S is a non-empty set. In the first case, we show that the condition is true by vacuous proof, as \phi has no elements. In the second case, we show that for every element x in set S, x is also an element of S, thus fulfilling the definition of S being a subset of itself.
  • #1
skullers_ab
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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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  • #2
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 ...
 
  • #3
Perhaps "cases" means to make a distinction between empty and non-empty sets.
 
  • #4
@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.
 
  • #5
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
 

1. What is a subset of a set?

A subset of a set is a collection of elements that are all contained within the original set. This means that all the elements in the subset are also elements of the larger set.

2. How do you prove that a set is a subset of itself?

To prove that a set is a subset of itself, you must show that every element in the set is also contained within the set. This can be done by listing out all the elements in the set and showing that they are all present in the larger set.

3. Why is it important to prove that a set is a subset of itself?

Proving that a set is a subset of itself is important because it helps to establish the properties and relationships of different sets. It also allows us to make accurate statements about the elements contained within a set.

4. Can a set be a subset of itself more than once?

No, a set cannot be a subset of itself more than once. This is because a subset must have all of its elements contained within the larger set, and if the set is already a subset of itself, there are no new elements to add.

5. Is every set a subset of itself?

Yes, every set is a subset of itself. This is because a subset is defined as a collection of elements that are all contained within the original set, and by definition, all the elements in a set are also contained within that set.

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