# Thread: Help with a continuity proof

1. ## 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; $\displaystyle \alpha$ if x=a; $\displaystyle \beta$ if x=b
Prove that $\displaystyle \alpha, \beta$ 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 $\displaystyle \alpha$ and the limit as x-->b of f from the left to find $\displaystyle \beta$, thus making g(x) continuous?

2. Originally Posted by paupsers
D is a nonempty bounded subset of R. Let a=inf D and b=sup D.
Let f-->R be a uniformly continuous function.
Let g(x)={f(x) if x in D; $\displaystyle \alpha$ if x=a; $\displaystyle \beta$ if x=b
Prove that $\displaystyle \alpha, \beta$ 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 $\displaystyle \alpha$ and the limit as x-->b of f from the left to find $\displaystyle \beta$, thus making g(x) continuous?
Here is a sketch of what is needed for this proof.

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

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

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

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

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

This should get you started.

3. Can anyone expand on this, not really sure how to do it

4. 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 $\displaystyle f(x)$ to be uniformly continous? Why do I need this hypothsis?

Things you need to show.

Here is the general idea

(1) Since $\displaystyle \{x_n\} \to b$ and $\displaystyle \{y_n\} \to b$ 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 $\displaystyle f(x)$ 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 $\displaystyle f(y_n)-b$ is small

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

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