i.e., if

Let be a subset of . We say

If , then the constant function is continuous since for all , and all complex sequences ,

Notice that and (Real part and imaginary part) are functions from
to
. In theorem 7.39 we showed if is any complex sequence and
, then

and

Hence and are continuous functions on .

Proof: Let
and let be any sequence in
such that ;
i.e., is a null sequence. By the reverse triangle inequality,

and

so we have

and hence

It follows by the comparison theorem that is a null sequence; i.e., . Hence is continuous.

Since
, the comparison theorem shows that

i.e., is continuous.

then is not continuous at , since

but

Notice that to show that a function is

Proof: Let be a sequence in domain such that . Then
for all and
for all , and by continuity of
and at , it follows that

By the sum theorem for sequences,

Hence is continuous at . The proofs of continuity for and are similar.

Proof: First we show is continuous at . Let be a sequence in
such that ; i.e., such that is a null sequence.
Then by the root theorem for null sequences (Theorem 7.19),
is a null sequence; i.e.,
, so is
continuous at .

Next we show that is continuous at . By the formula for a finite
geometric series (3.72 ), we have for all

Let be a sequence in . Then

so

Hence is continuous at .

Finally we show that is continuous at arbitrary . Let , and let be a sequence n . Then

Thus is continuous at .

Then

and

If
and
are defined by

and

then

and

Proof: Let be a sequence in such that . Then for all , we have and . By continuity of at , , and by continuity of at , .