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.