1.) Use the squeeze thm to prove
$f(x) = x \cos ( \frac{1}{x}) $ when $ x \ne 0 $ and
$f (x) = 0$ when $x = 0$
is continuous at $x= 0$.
2.) Write down a function that is continuous at exactly the points -1,1,5,42.
3.) Is there a function whose set of points of discontinuity is exactly the cantor set?
Also problems in the text:
4.3.1
4.3.3. $f(x) = ax +b $ is continuous
4.3.8 If f is continuous on $\mathbb{R}$ and $f(x) = 0$ for all $x \in \mathbb{Q}$
$f(x) = 0$ for all $x \in \mathbb{R}$
4.3.9. (a) Contraction mapping thm
I used the squeeze thm to prove #1. Does anyone know if we should use the definition of continuity as well?
I don't think so, because to use the squeeze theorem I believe you need pointwise continuity.
4.3.1 (a) Prove that $g(x)=^3\sqrt{x }$ is continuous at $c=0$
Proof: Let $\epsilon>0$ and choose $\delta=\epsilon^{3}$.
Since $ |g(x) − c| = |^3\sqrt x − 0| = |^{3}\sqrt x|$
then $|x-c|=|x-0|<\delta=\epsilon^{3}$ implies that $|^3\sqrt x|<\epsilon$.
Thus, g(x) is continuous at $c=0$.
Comments, suggestions, or continuation of the problem would be nice 
Prove $f$ below is continuous on $\mathbb{R}$.
Let $f : \mathbb{R} \rightarrow \mathbb{R}$ and assume there exists a $0< c <1$ with $$\left| f(x) - f(y) \right| \le c\left| x - y \right|$$ for all $x,y \in \mathbb{R}$.
Let $\varepsilon > 0$. Choose $\delta = \varepsilon$. Let $x,y \in \mathbb{R}$. Note when $\left| x - y \right| < \delta$ we have $$\left| f(x) - f(y) \right| \le c \left| x - y \right| < c\delta < \varepsilon$$ since $0< c<1$.
That seemed way too easy... comments?
Has anyone solved 2)
We have done it for discontinuity but I am having troubles figuring out how to do it for exact continuity.
Has anyone gotten #2?
When searching online I am finding functions of the form
$$f(x)=\left\{ \begin{array}{ccc}
(x+1)(x-1)(x-5)(x-42) & if &\in \mathbb{Q}\\
0 & if & x\notin \mathbb{Q}
\end{array} \right.$$
But I don't completely understand it. So because there is at least one irrational number between two rational numbers and at least one rational number between two irrational numbers then it holds true. I think.
Solution 3.
Let $C=\cup_{n=0} Cn$ with $C_n$ being the respective term in the cantor set.
Then the function
$$f(x)=\left\{ \begin{array}{ccc}
0 & if & x\in \mathbb{Q}/C\\
1 & if & x \in \mathbb{R}/\mathbb{Q} \cup C
\end{array}\right.$$
is continuous at all points of the cantor set.
@nklausen We can get some clue why this type of example works by looking at the graph below. The blue points represent a function on the rationals while the red points represent a function on the irrationals. The function is continuous only where these two graphs cross.
Why can you assume that $0 < c < 1$? @jmincey
I understand why you did it for this proof, but is it ok to assume that it is less than 1?
@ediazloa it is part of the initial assumptions of the problem. Problem 4.3.9 a (I think) from the book.
@jmincey I agree, isn't it because if $c > 1$ then the statement of c$\delta < \epsilon$ would not be true. Thus you can say $ 0 < c < 1.$ Right?
I think it's $C = \cap_{n=0}^\infty C_{n}$. The rest of it seems to be right to me though!
What functions are people using to prove this? My guess is $g(x) = \left| x \right|$ and $h(x) = - \left| x \right|$. Thus $h(x)\le f(x) \le g(x)$, right?
has anybody looked at this example mark gave...
$lim_{x \rightarrow a} (2f(x) - g(x)) = lim_{x \rightarrow a} 2f(x) - lim_{x \rightarrow a} g(x)$
any thoughts?? i tried approaching it just like a regular functional limit proof but got stuck about half way.
This is a general idea of what I was thinking
Letting $\lim_{x\rightarrow a}f(x)=L$ with $\epsilon >0$ and $\lim_{x\rightarrow a}g(x)=M$ with $\epsilon >0$
$$\left| (2f(x)-g(x))-(2L-M)\right|=\left|2f(x)-2L+M-g(x))\right|$$ using a triangle inequality
$$\leq2\left|f(x)-L\right|+\left|M-g(x)\right|\leq2\left|f(x)-L\right|+\left|g(x)-M\right|<\frac{\epsilon}{2}+\frac{\epsilon}{2}=\epsilon.$$
That looks really good. you chose $$|2f(x)-2L|<\epsilon/4$$ right?
I think so? In my head I wanted
$\frac{\epsilon}{2}>2\left|f(x)-L\right|$ so I believe it would be $\frac{\epsilon}{4}>\left|f(x)-L\right|$ but I could be totally off.