# Sets - Proving every set is a subset of itself

1. Sep 7, 2009

### skullers_ab

1. The problem statement, all variables and given/known data

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

2. Relevant equations

A $$\subseteq$$ B iff {X: X $$\in$$ A --> X $$\in$$ B}

3. 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.

2. Sep 7, 2009

### 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 ...

3. Sep 7, 2009

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

4. Sep 7, 2009

### 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$$\rightarrow$$p 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

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$$\rightarrow$$p is true, therefore S $$\subseteq$$ S