# Help with a continuity proof

• Dec 6th 2009, 09:05 AM
paupsers
Help with a continuity proof
D is a nonempty bounded subset of R. Let a=inf D and b=sup D.
Let f:D-->R be a uniformly continuous function.
Let g(x)={f(x) if x in D; $\alpha$ if x=a; $\beta$ if x=b
Prove that $\alpha, \beta$ exist such that g(x) is continuous.

Well, g(x) is continuous on (a,b) because f(x) is continuous, by hypothesis.
Can I take the limit as x-->a of f from the right to find $\alpha$ and the limit as x-->b of f from the left to find $\beta$, thus making g(x) continuous?
• Dec 6th 2009, 09:42 AM
TheEmptySet
Quote:

Originally Posted by paupsers
D is a nonempty bounded subset of R. Let a=inf D and b=sup D.
Let f:D-->R be a uniformly continuous function.
Let g(x)={f(x) if x in D; $\alpha$ if x=a; $\beta$ if x=b
Prove that $\alpha, \beta$ exist such that g(x) is continuous.

Well, g(x) is continuous on (a,b) because f(x) is continuous, by hypothesis.
Can I take the limit as x-->a of f from the right to find $\alpha$ and the limit as x-->b of f from the left to find $\beta$, thus making g(x) continuous?

Here is a sketch of what is needed for this proof.

As you have thought let
$\{ x_n\} \in (a,b) ;\{ x_n\} \to b \text{ as } n \to \infty$

Since $\{ x_n\}$ is Cauchy $f(\{ x_n\})$ is Cauchy (why?)

Define $g(b)=\lim_{n \to \infty}f(\{ x_n\})$

So now there is a canidate for the limit, but you must show that it does not matter what sequence you use to approximate it!

Now let $\{ y_n\} \in (a,b)$ you need to show that

$\lim_{n \to \infty}f(\{ y_n\}) = \lim_{n \to \infty}f(\{ n_n\})$

This should get you started.
• Dec 8th 2009, 02:53 PM
Kurt14
Can anyone expand on this, not really sure how to do it
• Dec 9th 2009, 07:49 AM
TheEmptySet
Quote:

Originally Posted by Kurt14
Can anyone expand on this, not really sure how to do it

What part are you stuck on? What have you tried?

Here a few more hints: Please show what you have done, and where you are stuck.

First question: What does it mean for $f(x)$ to be uniformly continous? Why do I need this hypothsis?

Things you need to show.

Here is the general idea

(1) Since $\{x_n\} \to b$ and $\{y_n\} \to b$ the two sequences must be getting close to each other. (You need to justify this)
(2)Since the two sequences are close to each other and $f(x)$ the immage of the sequences must be getting close to each other, and hence close to b. (This also needs to be justified.)
Use the above to show that $f(y_n)-b$ is small

For (1) why can't this be the case the open interval is $(0,2\pi)$ and $x_n=\frac{2}{\pi(1+4n)}$ $y_n=\frac{2}{\pi(3+4n)}$ These two sequences get as close to each other as we want but
For (2) let $f(x)=\sin\left( \frac{1}{x}\right)$ This immage of the seqences in never less than one. Why isn't this a counter example?

i.e why can I not extend $f(x)$ to be continous at 0?