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Thread: constant subsequences

  1. #1
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    constant subsequences

    Let Si be a sequence in a set A.
    Prove that Si has a constant subsequence iff there exists a in A such that for all n is in the natural numbers, there exists j≥n such that Si=a

    Where I am confusing myself is with the j and the n. Is the n referring to the position of the subsequence in the original sequence?
    I understand that a sequence does not need to be constant in order have a constant subsequence.

    Where do i start?




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  2. #2
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    What about $\displaystyle 0,1,0,1,0,1,...$ there is no $\displaystyle n$ such that $\displaystyle j\geq n$ implies that $\displaystyle s_j$ is constant.
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  3. #3
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    PerfectHacker, it is meant "there exists some j", not "for all j"

    randallg, there should be "S_j=a" in the end. basically the problem says

    S_i has a constant subsequence iff there exists some a in A that for any huge number n, we can find some bigger j>n that S_j=a. In other words, there should be infinitely many of elements equal to a in our sequence.

    the proof is easy. $\displaystyle S_i$ has a constant subsequence means $\displaystyle S_{m_k}=a$, where $\displaystyle m_k$ is a mononically increasing sequence of integers going to infinity. Then for any big n, there'll be some $\displaystyle m_k$ bigger than n.

    The other direction: construct the subsequence inductively. take n=1, then you know there exist $\displaystyle j_1$ such that $\displaystyle S_{j_1}=a$. Then take $\displaystyle n=j_1+1$, and you know there exist some number $\displaystyle j_2>j_1$ that $\displaystyle S_{j_1}=a$. And so on -- you obtain $\displaystyle S_{j_k}=a$ for any k.
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