# Closed set proof

1. Sep 5, 2015

1. The problem statement, all variables and given/known data

Show that the set of limit points of the set $A \subseteq \mathbb{R}$ given by $L$ is a closed set.

2. Relevant theorems

Definition: A closed set F (subset of R) is such that it contains its limit points.

Definition: A limit point x of a set A (subset of R) is such that the intersection Of every epsilon neighborhood with A excluding x is not empty.

Theorem: A number x is a limit point if and only if some cauchy sequence in A is convergent to x while each term in that cauchy sequence is not equal to x.

Theorem: A set F is closed if and only if every cauchy sequence in F has a limit that is also an element of F.

3. The attempt at a solution

Proof: Since $L$ is the set of limit points $A \subseteq \mathbb{R}$, then if $(a_n)$ is a cauchy sequence in $A$ then $\lim a_n \in L$. Suppose $x$ is a limit point of $L$, then we can form a cauchy sequence satisfying $l_n \neq x$ for all $n$, and $\lim l_n = x$. However, each term in the sequence $(l_n)$ is a limit reached by some cauchy sequence in $A$.

I will try to construct a cauchy sequence in A that converges to x hence showin that $x \in L$ and $L$ is a closed set.

We start by choosing $(a_n)$ such that;
$$|a_n - l_n| < \frac{1}{n}$$
For this definition of $(a_n)$, given any $\epsilon > 0$, we choose $\frac{1}{N_1} < \frac{\epsilon}{2}$ such that for all $n \geq N_1$;
$$|a_n - l_n| < \frac{\epsilon}{2}$$
Since $(l_n)$ converges to $x$, then, for all $n \geq N_2$;
$$|l_n - x| < \frac{\epsilon}{2}$$
Choosing $N = \max{(N_1,N_2)}$;
$$|a_n - x| = |(a_n - l_n) - (l_n - x)| \leq |a_n - l_n| + |l_n - x| < \epsilon$$
Hence, this contruction of $(a_n)$ converges to $x$.

Q.E.D.

Is this correct?

Last edited: Sep 5, 2015
2. Sep 5, 2015

### PeroK

I think you have the essense of the proof, but it's not very clear. For example, the first statement in your proof is:

"Since $L$ is the set of limit points $A \subseteq \mathbb{R}$, then if $(a_n)$ is a cauchy sequence in $A$ such that $\lim a_n \in L$."

This doesn't make sense. Also, you would benefit from writing what you are trying to prove. You've left the "relevent equations" section blank, but actually your definition of a limit point and a closed set are relevant here.

And, at the end, you omit stating the significance of $(a_n)$ converging to $x$.

I would take what you have and seriously tidy things up. For example, I would start:

"To show that $L$ is closed, we will show that $L$ contains its limit points. Let $x$ be a limit point of $L$ ..."

3. Sep 5, 2015

My sincerest apologies

"Since $L$ is the set of limit points $A \subseteq \mathbb{R}$, then if $(a_n)$ is a cauchy sequence in $A$ THEN $\lim a_n \in L$."

i will update the relevant theorems sections in a second, and polish some parts of the proof. Thanks for the feedback.

Last edited: Sep 5, 2015
4. Sep 9, 2015

I read that the only way to improve in proof writing is to let someone rip apart the proofs you write. Please, any suggestions regarding this proof are most welcome :)

5. Sep 9, 2015

### andrewkirk

A couple of suggestions, as you requested Ahmad:
This can be expressed more clearly and succinctly as:
A number x is a limit point of a set A if and only if there is a Cauchy sequence in A-{x} whose limit is x.
This can be expressed more clearly and succinctly as:

Suppose $x$ is a limit point of $L$, then we can form a cauchy sequence of points in L-{x} with limit $x$.

6. Sep 9, 2015

Thank you very much for the suggestions!

Also, could u provide insight into the validity of the proof. I am specially concerned about the part where i construct the cauchy sequence, since i feel the construction is very artificial.

7. Sep 9, 2015

### andrewkirk

Are you allowed to use the axiom of choice Ahmad? You have used it at least twice, once in choosing the sequence $(l_k)$ and once in choosing $(a_k)$.

I'm pretty sure the proof can be done without using the axiom of choice, but generally such proofs are longer and require more care than proofs that use it.

Also, the following is not valid without some justification being provided.
A justification is available, but requires a few steps and should not be omitted.
in any case, I suspect there's a better way of putting this, that is easier to justify. But first we need to know whether you're allowed to use Choice.

8. Sep 10, 2015

I am not familiar with the axiom of choice, could you please give me a reference on the axiom of choice.

I had as my goal the proof that L is closed. I assumed that x is arbitrary limit point of L, which allows me to say that there exists a sequence in L-{x} such that it converges to x. This allowed me to say that (ln) exists. Now i wanted to show that there exists a sequence in A such that it converges to x as well, this would imply that x is in L be definition of L. Since i am trying to prove a "there exists" statement i assumed that i had the right to show that a certain construction works.

There is a justification for the choice of (an) which i believe i mentioned in passing. We know that there exists a sequence in A-{ln} that converges to ln, so since there is such sequence for each n, i chose an element from each such sequence that is within 1/n from ln.

There is a much easier proof given in my book, but i self study real analysis and try to guess the reasoning and if possible work it out on my own.

Last edited: Sep 10, 2015
9. Sep 10, 2015

### andrewkirk

The wikipedia article on the axiom of choice is a pretty good start. Basically the axiom asserts that if you have an infinite collection of non-empty sets you can choose a set that has exactly one element from each. Most theorems in topology and analysis can be proved without it. But a surprising number of ordinary-looking theorems do require it. It's generally preferred not to use it if possible because if we accept it then some really weird consequences follow.

My statement that you used it twice was incorrect. You didn't use it when you chose the sequence $(l_k)$, because that is only choosing one sequence from a set of sequences converging on x, which we know to be non-empty because x is a limit point. Hence it's making only one choice. But you used the axiom when you defined the sequence $(a_k)$ because for every $k\in \mathbb{N}$ you are choosing one value of $a_k$ from among the infinite number of possibilities, and an infinite number of such choices has to be made - one for each $k$.

10. Sep 11, 2015

Hmmm my book "understanding analysis" followed the same argument in many occassions so i just assumed that i had the right to use it as well. So i will post-pone my understanding here until i get a good grip on the axiom of choice.

Thank you :)

11. Sep 11, 2015

### andrewkirk

I think that the axiom of choice may be needed to prove it by using a sequence that converges to x, as you've tried to do, because one has to define that sequence, which may require making an infinite number of choices. Hence I don't think sequences are a good way to approach this proof.

I think it would be easier to prove by using your above definition of limit point 'A limit point x of a set A (subset of R) is such that the intersection Of every epsilon neighborhood with A excluding x is not empty.'

If we assume that x, a limit point of L, is not in L, then it's not a limit point of A, so what does the definition of limit point tell us about how 'far away' x must be from A? Can that lead us to a contradiction with the assumption that it's a limit point of L?

12. Sep 11, 2015

### andrewkirk

I know how you feel. My topology book - the venerable 'Topology' by Munkres - explains the Axiom of Choice and why it's important, all very nice and clearly, but then proceeds to freely assume it in all sorts of proofs where it doesn't need to, and without even making it clear when he's doing it*. That annoys the heck out of me because I consider it very important to know whether a theorem requires Choice to prove it, given how contentious the axiom is.

*it's a good exercise to go through a proof and try and work out whether it has been used. I often get that wrong, as I did when I thought that your $(l_n)$ sequence used it.

13. Sep 11, 2015

I browsed through the wiki article to get an essence of the axiom of choice. This particular segment seems to me relevant here:

"In many cases such a selection can be made without invoking the axiom of choice; this is in particular the case if the number of bins is finite, or if a selection rule is available: a distinguishing property that happens to hold for exactly one object in each bin. "

Since i have prescribed a rule for the selection of (an) does that imply that i am not invoking the axiom of choice? Or is it not as simple as wiki makes it seem to be, and i need to do yet more reading to come to a concrete comclusion? Thanks.

14. Sep 11, 2015

### andrewkirk

No, because you have not given a procedure for identifying a unique $a_n$. You've just observed that every $l_n$ has a bunch of sequences that converge to it, and you arbitrarily choose one of those sequences from which to pluck $a_n$. And you have to do that an infinite number of times. I think there's a much quicker root to the proof by not using sequences.

15. Sep 11, 2015

There is a much easier way in my book. I will read about the axiom of choice in depth and then revisit this thread. Thank you :)

16. Sep 11, 2015

### andrewkirk

If you don't use sequences, you don't have to worry about the axiom of choice for this problem. It can be done in three lines.

If $x\notin L$ then $\exists\epsilon>0$ such that $(x-\epsilon,x+\epsilon)\subseteq \sim A$.

But $x$ is a limit point of $L$. Can we find a point in $L$ close enough to $x$ that it must be some minimum distance from any point in $A$? Can we get a contradiction from that point?

17. Sep 12, 2015

### verty

If $L$ is defined by $L = \{x \in \mathbb{R}: \phi(x)\}$, one can prove $\phi(x)$ without needing $(a_n)$ to exist.

1. WLOG, for each $n$ let $s_n$ be an arbitrary sequence of $A$ converging to $l_n$.
2. WLOG, for each $n$ let $a_n$ be an arbitrary element of $s_n$ so assigned, such that $|a_n - l_n| < {\epsilon \over 2}$.
3. The $a_n$'s so assigned are such that $\phi(x)$ is satisfied, but no generality was lost, therefore $\phi(x)$.

I confess to not being an expert at rigour but I personally didn't see a problem with the given sequences proof.