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

  1. Mar 15, 2017 #1
    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.

    Capture.PNG
    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. jcsd
  3. Mar 15, 2017 #2

    QuantumQuest

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    ##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##
     
  4. Mar 18, 2017 #3
    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.
     
  5. Mar 18, 2017 #4
    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##?
     
  6. Mar 18, 2017 #5

    QuantumQuest

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    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.
     
  7. Mar 18, 2017 #6
    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.
     
  8. Mar 18, 2017 #7

    QuantumQuest

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    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.
     
  9. Mar 18, 2017 #8
    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.
     
  10. Mar 18, 2017 #9

    QuantumQuest

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    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

     
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