# Limit Points

1. Mar 11, 2008

### rbzima

This is review from class the other day that I managed to miss because of illness and I was wondering if someone could explain how to go about solving these problems:

#1
Let

$$B = \left\{ \frac{(-1)^nn}{n+1}:n = 1,2,3,...\right\}$$

Find the limit points of B
Is B a closed set?
Is B an open set?
Does B contain any isolated points?
Find $$\overline{B}$$.

#2
Let $$a \in A$$. Prove that a is an isolated point of A if and only if there exists an $$\epsilon$$ neighborhood $$V_\epsilon(a)$$ such that $$V_\epsilon(a) \bigcap A = \left\{a\right\}$$.

#3 - This is a proof the class worked on:
A set $$F \subseteq \texttt{R}$$ is closed if and only if every Cauchy sequence contained in F has a limit that is also an element of F.

This is all that was talked about according to the professor, and there were some other examples shown of the first problem. I suppose that since I wasn't there in class, I have less of an edge than those who were, so I'd love to see how and more precisely why things work the way they do. Help would be greatly appreciated!

2. Mar 11, 2008

### grmnsplx

well I don't wanna do you're homework for you so I'll give you some hints/

#1
obviously we want to use the definition of an open/closed set. for every point b in B, does there exist a neighbourhood contained in B. (answer - yes. now you have to show it)
As for isolated points, is there a neighbourhood (or an open ball) that around any point b in B that does not contain any other points in B? (no) This is almost never the case. A good example though is the discrete set which is entirely made up of isolated points.

sorry i forgot what$$\overline{B}$$ means. Cover?

#2 easy. proof by contradiction.

if I recall there are two main points. The first seems mundane, but is important
The following is NT a proof:
i. Assume that F is closed. if {f_j} is a cauchy sequence in R, then it must have a limit f because R is complete. we then show that s is in R. if f is not in F it means that f is not the limit point of f_j.
ii. Assume every cauchy sequence in F has a limit in F. we prove by contradiction. if F were not closed then the complement (R\F) would not be open and then we show that f_j would converge to some point in the closed set R\F. Here lies the contradiction. Hence F must be closed.
You should easily be able to find a proof on the web somewhere. google ["is closed if and only if every cauchy sequence" proof]

RBZIMA, what class is this for? I want to be helpful without giving away too much. this sounds like my second anal course which was in our 3rd year. Our intro to anal course was in our second year and didn't involve open sets and the like, just epsilon/delta mostly.

i can give more detail if neccessary

3. Mar 11, 2008

### rbzima

This is for a general Real Analysis course at my college. We are a relatively small college that doesn't offer many variations of Analysis, just the basics.

Thanks for the help though. I feel a little behind because I missed a day last week, but we'll see how things go.

4. Mar 11, 2008

### d_leet

How did you decide B is open? What definition of open are you using?

Assuming the usual topology on the real numbers and considering B, B is certainly not open in the topology because every neighbourhood of a point in the reals must contain an irrational number, but all the numbers in B are rational, so no neighborhood of a point in B can be contained in B unless we are considering a topology different from the usual on on the reals.

rbizma: What is the definition of closed you are using? If it is that a closed set contains all of its contact points then is B closed or not?

5. Mar 12, 2008

### grmnsplx

d_leet
you are exactly right. clearly I was a bit too quick to answer
B would be open in Q (correct?) but not in R

B is not open in R because I can always find points between any two points in B which are not in B. Namely, irrationals

Now, the question remains "is B closed?" That is to say, "does it contain all of it's limit points. RBZIMA, I think you can answer that.