Proving A ⊆ B using Set Theory

[glebovg]

How would I prove A \subseteq B \Leftrightarrow A \cap B^{c} = \emptyset ?

 

[glebovg]

This is what I have so far:

Suppose A \cap B^{c} = \emptyset. Let x \in A. We want to show x \in B. Since A \cap B^{c} = \emptyset and x \in A then x \notin B^{c}. Hence x \in B.

 

[PAllen]

Why not start by assuming either side of the two way implication, rather than assuming something completely different?

 

[glebovg]

It is supposed to be A \subseteq B \Leftrightarrow A \cap B^{c} = \emptyset.

 

[PAllen]

Ok, then what you've got is a good start. Now work it the other direction.

 

[glebovg]

So (\Rightarrow) is correct? First, I suppose A \subseteq B. Then I let x \in A and therefore x \in B by supposition, but then it does not lead me anywhere.

 

[PAllen]

Can something be in B and B complement?

 

[glebovg]

\emptyset ?

 

[glebovg]

Also, we know \emptyset \subseteq A \cap B^{c} and we want to show A \cap B^{c} \subseteq \emptyset, but I do not know how to do it.

 

[PAllen]

\emptyset ?

Can there be any element in B and B complement?

 

[glebovg]

No, that would be a contradiction.

 

[PAllen]

No, that would be a contradiction.

The rest should follow. Can't think of another hint that isn't the complete answer.

 

[glebovg]

Is this correct?

Suppose A \subseteq B. Let x \in A. Then x \in B \because A \subseteq B. A \cap B^{c} \Rightarrow x \in A and x \in B^{c} by definition of intersection. Since x \in B, A \cap B^{c} = \emptyset.

 

[PAllen]

Ok, but you can tighten up the the argument. If x in A, then x in B, then x not in B complement; then A intersect B complement is empty.