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8.2 Continuity

8.6   Definition (Continuous) Let be a complex function and let . We say is continuous at if

i.e., if

Let be a subset of . We say is continuous on if is continuous at for all . We say is continuous if is continuous on ; i.e., if is continuous at every point at which it is defined.

8.7   Examples. If for all , then is continuous. In this case for every sequence so the condition for continuity at is

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 .

8.8   Theorem. If and are functions from to defined by

then and are continuous.

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.

8.9   Example. If

then is not continuous at , since

but

Notice that to show that a function is not continuous at a point in its domain, it is sufficient to find one sequence in such that and either converges to a limit different from or diverges.

8.10   Theorem (Sum and Product theorems.) Let be complex functions, and let . If and are continuous at , then , , and are continuous at .

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.

8.11   Theorem (Quotient theorem.) Let be complex functions and let . If and are continuous at , then is continuous at .

8.12   Exercise. Prove the quotient theorem. Recall that

8.13   Theorem (Continuity of roots.) Let and let for all . Then is continuous.

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

 (8.14)

If we replace by in (8.14), we get , i.e.,

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 .

8.15   Definition (Composition of functions.) Let be sets, and let , be functions. We define a function by the rules:

8.16   Examples. Let , be defined by

Then

and

If and are defined by

and

then

and

8.17   Theorem (Compositions of continuous functions.) Let be complex functions. If is continuous at , and is continuous at , then is continuous at .

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

8.18   Example. If for all , then is continuous (i.e., is continuous at for all .)

8.19   Exercise. A Let be defined by for all . Is continuous?

Next: 8.3 Limits Up: 8. Continuity Previous: 8.1 Compositions with Sequences   Index