Let and . Prove or disprove the following.

- .
- .
- .
- .
- .
- .

* True. *

* Proof. * Since 1 is the only element in , and , we have that every element of is contained in . Hence, . Further, since , but is not in , we see that . Therefore, . ∎
* True. *

* Proof. * Again, since , we have by part (a). ∎
* True. *

* Proof. * By the definition of we have that . Since , we have . ∎
* True. *

* Proof. * By definition ; hence, .∎
* False. *

Since 1 is not a set, we cannot have .
* False. *

Since 1 is still not a set, we cannot have .

*Related*