# Trying to understand a proof about $\lim S$

1. Mar 15, 2017

### Bashyboy

1. The problem statement, all variables and given/known data
I am trying to understand the proof that $\lim S$ is a closed set in the metric space $M$, where $\lim S = \{ p \in M ~|~ p \mbox{ is a limit point of } S\}$.

Here is the definition of a limit point: $p$ is a limit point of $S$ if and only if there exists a sequence $(p_n)$ of points in $S$ such that $p_n \rightarrow p$. Attached is a snapshot from the book.

2. Relevant equations

3. The attempt at a solution

I don't understand why there exists $q_n = p_{n,k(n)}$ satisfying the conditions mentioned in the picture. What does $k(n)$ denote? A subsequence? The indices are causing me trouble. Why would only one of these sequences converge to $p$? It seems that all of them should. Clearly $p_{n,k} \rightarrow p_n$ means $d(p_n,p_{n,k}) < \frac{1}{k}$, but I don't see the connection to what is give in the picture. How does the replacement of $k$ with $k(n)$ factor into the proof? I tried showing there exists such a $q_n$, but was unsuccessful.

2. Mar 15, 2017

### QuantumQuest

$q_n$ is a sequence where $k$ depends on $n$. If you substitute some values e.g. $1,2,\cdots$ for $n$ you can see how the $q_n$ expression looks like.

Proof states that there exists $q_n$. In other words suffice it to find one such sequence. But follow carefully the dependence on indices.

$k \rightarrow\infty$ as stated in the proof. $k(n)$ is a sequence of values of $k$ as $n \rightarrow\infty$

3. Mar 18, 2017

### Bashyboy

Okay. I am still having trouble understanding this proof. Just above the author's application of the triangle inequality he writes, "Then, as $n \rightarrow \infty$ we have." This, along with the notation $k(n)$, suggests to me that the author is taking points from each sequence $p_{n,k}$, associated with each $p_n$, to form a sequence in $S$ that converges to $p$, thereby showing $p$ is also a limit point of $S$, in addition to $\lim S$. Does this sound right? If so, the next thing I am having trouble seeing is how they construct this sequence; the author merely asserts it exists. I could use a hint on showing such a sequence exists or constructing it.

4. Mar 18, 2017

### Bashyboy

It seems that $k=k(n)$ must be monotically increasing from $\mathbb{N}$ to $\mathbb{N}$, and in fact $k=k(n)=n$ would work. Does this seem right? That is, would $p_{n,n}$ converge to $p$?

5. Mar 18, 2017

### QuantumQuest

See carefully the wording in the claim of the proof: It states, "We claim that $p \in \lim{S}$". What does this imply for $\lim{S}$?

Look. $k \rightarrow \infty$, so the value of $k$ increases to $\infty$. Now, the crucial point is to see that author takes another sequence namely $q_{n}$ and the $n$th term of this sequence tends to the $n$th term of $p_{n}$ as $n \rightarrow \infty$. So, at the right hand side of the triangle inequality we have that $q_{n} \rightarrow p_{n}$ and $p_{n} \rightarrow p$ which was supposed in the beginning of the proof, reading the right hand side from right to left . So, that implies that $q_{n} \rightarrow p$ as stated in the proof and so the proof is complete.

6. Mar 18, 2017

### Bashyboy

Yes, I believe I understand this part. At this point I am trying to justify his presupposition, namely, that there does in fact existence such a sequence. He doesn't give a proof of this, but just asserts it.

7. Mar 18, 2017

### QuantumQuest

For $(p_{n,k}), k \in \mathbb N$ it is obvious that there is at least one such sequence as $p_{n}$ is a limit of $S$. Now, $(q_{n})$ is a subsequence of $(p_{n})$ where $k$ depends on $n$. So, because it is proved that $q_{n} \rightarrow p$, as also $p_{n} \rightarrow p$ , then $p \in \lim S$, so $\lim S$ is a closed set.

8. Mar 18, 2017

### Bashyboy

This is precisely my contention; it is obvious there is at least one such sequence. Strictly speaking, this needs to be proven.

I don't believe this is right; I don't believe $(q_n)$ is a subsequence of $(p_n)$. Each $(p_{n,k})$ is a sequence that converges $p_n$, not a subsequence of $(p_n)$. When you let $k=k(n)$, it seems you are building sequence $(q_n)$ from each $(p_{n,k})$, whose $n$-th term is $q_n = p_{n,k(n)}$; i.e., we pick a point from $p_{n,k}$ for each $k$ that depends upon $n$ in some way. I am not sure how else to say this.

9. Mar 18, 2017

### QuantumQuest

I really cannot see your point.

Yes my bad, I inadvertently did not write what I meant. What I mean is that $q_n$ are values constructed from $p_{n,k}$ by taking values for $k$ depended on $n$. We're interested in the distance between $p_n$ and $q_n$ as $n \rightarrow \infty$ and there, this distance tends to $0$ : $d(p_n,q_n) \lt \frac{1}{n} \rightarrow 0$. So, the rest is as stated in #5