Sets - Proving every set is a subset of itself

[skullers_ab]

Homework Statement

Prove that for every set S, S \subseteq S. Use 'proof by cases'.

Homework Equations

A \subseteq B iff {X: X \in A --> X \in 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 \subseteq S, all I can figure out, simply, is:

For every element x in set S, x is an element of S, therefore, S \subseteq S

I do not know how to express this proof in terms of 'cases'. Any help would be appreciated.

 

[HallsofIvy]

You are required to use "cases"? How strange.

Try this:
case 1: Suppose x\in S then ...

case 2: Suppose x\notin S then ...

 

[statdad]

Perhaps "cases" means to make a distinction between empty and non-empty sets.

 

[skullers_ab]

@HallsofIvy and @statdad: Thank you for the response.

Should it be something like this?

@HallsofIvy:

For S \subseteq S : \forall x(x \in S \rightarrow x \in S)

Case 1: Let x \in S, then x \in S. p\rightarrowp is true, therefore S \subseteq S

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


AND/OR


@statdad:

For S \subseteq S : \forall x(x \in S \rightarrow x \in S)

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

Case 2: Let S be a non-empty set, Let x \in S, then x \in S. p\rightarrowp is true, therefore S \subseteq S



I hope I interpreted the cases correctly. Please advise.

 

[skullers_ab]

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