Homework Help: Set Algebra

1. Jan 23, 2012

glebovg

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

Last edited: Jan 23, 2012
2. Jan 23, 2012

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 $x \in B$.

3. Jan 23, 2012

PAllen

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

4. Jan 23, 2012

glebovg

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

5. Jan 23, 2012

PAllen

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

6. Jan 23, 2012

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.

7. Jan 23, 2012

PAllen

Can something be in B and B complement?

8. Jan 23, 2012

glebovg

$\emptyset$ ?

9. Jan 23, 2012

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.

10. Jan 23, 2012

PAllen

Can there be any element in B and B complement?

11. Jan 23, 2012

glebovg

No, that would be a contradiction.

12. Jan 23, 2012

PAllen

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

13. Jan 24, 2012

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 $x \in B$, $A \cap B^{c} = \emptyset$.

14. Jan 24, 2012

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.