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 .
i dont understand D part
For D, we get
if 1 is an element of A. Thus, it is true.
For D, we get , if 1 is an element of A. Hence, it is true.