19. Continuity Continued - Density

19.1 Definition:

Let be a sequence in a Metric Space $\QTR{Large}{(M,d)}$ . We say the sequence converges to $\QTR{Large}{x\in }$ $\QTR{Large}{M}$ , written . If for any 0 there exists an n such that for n.

19.2 Theorem:

Given Metric Spaces $\QTR{Large}{(M,d)}$ , and a function , $\QTR{Large}{f}$ is continuous a point $\QTR{Large}{x\in }$ $\QTR{Large}{M}$ if and only if for every ,

Note that this implies that $\QTR{Large}{f}$ is is continuous at every point if and only if for every convergent sequence

Proof:

Suppose $\QTR{Large}{f}$ is continuous a point $\QTR{Large}{x\in }$ $\QTR{Large}{M}$ , then given any 0 there is a 0 such that

Since the sequence converges, for there exists an n $_{\delta }$ such that for n. Hence, also, for n

Conversly, suppose $\QTR{Large}{f}$ is not continuous at $\QTR{Large}{x}$ , then there exits an 0. Such that for

any 0 we have

For every

choose

such that and .

restated,

Use the Countable Axiom of Choice to choose

Clearly but

19.3 Definition:

Given a Metric Space $\QTR{Large}{(M,d)}$ , and a subset Define

and

is called the closure of $\QTR{Large}{A}$ .

19.4 Lemma:

$\$ For any $\QTR{Large}{A}$ , is open.
That is, the closure of the closure of $\QTR{Large}{A}$ is the closure of $\QTR{Large}{A}$ .

Proof:(To be turned in. Due April 6)

Solution:

1. Suppose that a given $\QTR{Large}{x}$ for any 0 we have ,

then we could choose a sequnce in $\QTR{Large}{A}$ that converged to $\QTR{Large}{x}$ ,then

On the other had suppose that for any 0 we have

Let then we can find 0 and a sequence such that

, and hence

2. By definition Next suppose , where . For each

$\QTR{Large}{i}$ choose such that . One checks that

19.5 Definition:

We call a subset dense if .

Example:

is dense in .

19.6 Theorem:

Given Metric Spaces $\QTR{Large}{(M,d)}$ , , continuous functions, , and a dense subset , if then

Proof:

Suppose andis such that.

then

19.7 Example:

Let be continuous and suppose $\QTR{Large}{f(r)=}$ 0 for all , then $\QTR{Large}{f(r)=}$ 0 for all
is said to be additive if for all If $\QTR{Large}{f}$ is also continuous then

Proof of 2: (To be turned in. Due April 6)

Hint:

Prove it first for $\QTR{Large}{r\ =p}$ and with

Next for

Solution:

Given A simple induction argument shows that

MATH

Hence We can now apply 1. since the function

is zero on

19.8 Bonus Question:

Given , we call a function bounded near $\QTR{Large}{r}$ . if there exists

0 such that .

Note that this does not say that for any $\QTR{Large}{b}$ there exists a .

Prove that if is additive, and there exists some such that $\QTR{Large}{f}$ is bounded near $\QTR{Large}{r}$ , then $\QTR{Large}{f}$ is continous everywhere and hence

Solution:

If $\QTR{Large}{f}$ is bounded near one point then by linerarity it is bounded everywhere.
If $\QTR{Large}{f}$ is bounded near $\QTR{Large}{r}$ then it is continuous at $\QTR{Large}{r}$ .

Suppose there exists 0 such that

. Suppose . Let 2

Then and hence

$\QTR{Large}{|f(}$ 222

and by induction

Finally, given 0 choose n such that