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

• Bashyboy
In summary: This is precisely my contention; it is obvious there is at least one such sequence. Strictly speaking, this needs to be proven.Okay, I am not sure if I have understood it completely. So, I am going to give you some hints and you try to understand it. 1) The sequence ##(p_{n})## converges to ##p## because ##p## is a limit point of ##S##.2) This means that for each ##\epsilon > 0## there is an integer ##N## such that ##n > N## implies ##d(p_{n},p) < \epsilon##.3) Now, in the proof author constructs a sequence ##(q_{n})## such
Bashyboy

## Homework Statement

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.

## 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.

vr0nvr0n
Bashyboy said:
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?

##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.

Bashyboy said:
Why would only one of these sequences converge to ##p##? It seems that all of them should.

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

Bashyboy said:
How does the replacement of ##k## with ##k(n)## factor into the proof?

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

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.

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##?

Bashyboy said:
thereby showing ##p## is also a limit point of ##S##, in addition to ##\lim S##.

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}##?

Bashyboy said:
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##?

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.

QuantumQuest said:
k→∞k→∞k \rightarrow \infty, so the value of kkk increases to ∞∞\infty. Now, the crucial point is to see that author takes another sequence namely qnqnq_{n} and the nnnth term of this sequence tends to the nnnth term of pnpnp_{n} as n→∞n→∞n \rightarrow \infty. So, at the right hand side of the triangle inequality we have that qn→pnqn→pnq_{n} \rightarrow p_{n} and pn→ppn→pp_{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 qn→pqn→pq_{n} \rightarrow p as stated in the proof and so the proof is complete.

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.

Bashyboy said:
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.

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.

QuantumQuest said:
For (pn,k),k∈N(pn,k),k∈N(p_{n,k}), k \in \mathbb N it is obvious that there is at least one such sequence as pnpnp_{n} is a limit of SSS.

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

QuantumQuest said:
Now, (qn)(qn)(q_{n}) is a subsequence of (pn)(pn)(p_{n})

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.

Bashyboy said:
Strictly speaking, this needs to be proven

I really cannot see your point.

Bashyboy said:
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.

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

QuantumQuest said:
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 ##p \in \lim S## and the proof is complete.

## What is a proof?

A proof is a logical argument or series of steps that demonstrate the validity of a mathematical statement or theorem. It is used to verify the truth or falsehood of a statement and provide a deeper understanding of a concept.

## Why is understanding a proof important?

Understanding a proof is important because it allows us to fully comprehend the logic and reasoning behind a mathematical statement or theorem. It also helps us to apply the concept to other problems and build upon our understanding of the subject.

## How can I understand a proof about ##\lim S##?

To understand a proof about ##\lim S##, it is important to have a solid foundation in calculus and real analysis. You should also be familiar with the definitions and properties of limits, sequences, and series. It may also help to break down the proof into smaller steps and work through each one carefully.

## What are some common mistakes when trying to understand a proof about ##\lim S##?

Some common mistakes when trying to understand a proof about ##\lim S## include not having a strong understanding of the prerequisite concepts, not paying attention to the assumptions and implications of the proof, and not checking each step for accuracy and logic. It is important to take your time and carefully analyze each step of the proof.

## What resources can I use to help me understand a proof about ##\lim S##?

There are various resources available to help you understand a proof about ##\lim S##, such as textbooks, online lectures, and practice problems. You may also benefit from seeking help from a tutor, professor, or study group. It is important to actively engage with the material and ask questions when necessary.

• Calculus and Beyond Homework Help
Replies
5
Views
1K
• Calculus and Beyond Homework Help
Replies
5
Views
1K
• Calculus and Beyond Homework Help
Replies
3
Views
1K
• Calculus and Beyond Homework Help
Replies
3
Views
1K
• Calculus and Beyond Homework Help
Replies
4
Views
2K
• Calculus and Beyond Homework Help
Replies
1
Views
715
• Calculus and Beyond Homework Help
Replies
5
Views
2K
• Calculus
Replies
10
Views
2K
• Calculus and Beyond Homework Help
Replies
4
Views
631
• Calculus and Beyond Homework Help
Replies
6
Views
6K