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 ofwe 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.