1. ## Examples

Give an example or say why it is impossible:

1. A function which is discontinuous at every point of the set $\displaystyle \{\frac{1}{n}:n\in\mathbb{N}\}$ but continuous elsewhere on $\displaystyle \mathbb{R}$

I think it's possible since that set could be written as a countable union of closed sets, but I can't think of a particular example.

2. An infinite subset of [0,1] with no limit points.

I think this is impossible but I'm not sure about my reasoning - is this correct?: Any subset of [0,1] is bounded so by the Bolzano Weierstrass theorem any sequence in this subset must have a convergent subsequence, so it has a limit point.

3. A closed set whose supremum is not a limit point of this set.

not sure...

4.A power series that is absolutely convergent at only one point.

Probably something with the point being 0... not sure though

Thanks for any help

2. 1. Hint:
Spoiler:
$\displaystyle f(x)=\begin{cases}x&x\ne\frac1n\\[1mm]0&\mbox{otherwise}\end{cases}$

3. Hint:
Spoiler:
Any finite set that is bounded above.

4. Hint:
Spoiler:
$\displaystyle \sum_{n\,=\,1}^\infty\frac{(-1)^n}nx^n$

3. Thanks Abstractionist. Regarding your answer to question 4, is it always true that if $\displaystyle \sum a_n$ diverges, then $\displaystyle \sum a_nx^n$ also diverges, for any x besides 0? And if not, what makes your example special? Thanks again

4. Originally Posted by Aileys.
is it always true that if $\displaystyle \sum a_n$ diverges, then $\displaystyle \sum a_nx^n$ also diverges, for any x besides 0?
No. For example, $\displaystyle \sum_{n\,=\,1}^\infty\frac1n$ diverges but $\displaystyle \sum_{n\,=\,1}^\infty\frac{x^n}n$ converges for $\displaystyle -1\le x\le0.$ However, the convergence is only conditional, not absolute.

5. Ahhh OK so then is it always true that if $\displaystyle \sum a_n$ diverges then the only x for which $\displaystyle \sum a_nx^n$ converges absolutely is zero?

Also while you're here ( ) was my answer to #2 correct?