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Math Help - Cluster point of a finite sequence

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    Cluster point of a finite sequence

    The terms of a sequence X_n take on only finitely many values a_1,...,a_k.

    That is, for every n, X_n = a_i for some i (the index i depending on n). Prove X_n has a cluster point.
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    Super Member redsoxfan325's Avatar
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    Quote Originally Posted by cgiulz View Post
    The terms of a sequence X_n take on only finitely many values a_1,...,a_k.

    That is, for every n, X_n = a_i for some i (the index i depending on n). Prove X_n has a cluster point.
    This seems false. Consider the sequence X_n=(-1)^n.
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    Super Member redsoxfan325's Avatar
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    However, X_n will always have a subsequence with a limit point.

    Proof: Let A=\{a_1,...,a_k\}. Because there are only finitely many points, A is compact, and thus X_n is a sequence in a compact set. So it follows from the Bolzano-Weierstrass Theorem that X_n has a convergent subsequence (and therefore a limit point in that subsequence).
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    Quote Originally Posted by redsoxfan325 View Post
    However, X_n will always have a subsequence with a limit point.

    Proof: Let A=\{a_1,...,a_k\}. Because there are only finitely many points, A is compact, and thus X_n is a sequence in a compact set. So it follows from the Bolzano-Weierstrass Theorem that X_n has a convergent subsequence (and therefore a limit point in that subsequence).
    Yes, but that is irrelevant to the question. Any such convergent subsequence is "eventually constant". The definition of "cluster point" is that p is a cluster point if, given any \epsilon> 0, there exist an infinite] number of points other than p itself whose distance to p is less than \epsilon. If the sequence has only a finite number of points then the set of distances a given point p and points in the sequence, other than p itself, is finite and has a non-zero minimum. Let [itex]\epsilon[/itex] be that minimum distance. Given any p, there exist no other point whose distance is less than \epsilon. A finite set does NOT have a cluster point and so no finite valued sequence has a cluster point.
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  5. #5
    Super Member redsoxfan325's Avatar
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    Quote Originally Posted by HallsofIvy View Post
    Yes, but that is irrelevant to the question. Any such convergent subsequence is "eventually constant". The definition of "cluster point" is that p is a cluster point if, given any \epsilon> 0, there exist an infinite] number of points other than p itself whose distance to p is less than \epsilon. If the sequence has only a finite number of points then the set of distances a given point p and points in the sequence, other than p itself, is finite and has a non-zero minimum. Let [itex]\epsilon[/itex] be that minimum distance. Given any p, there exist no other point whose distance is less than \epsilon. A finite set does NOT have a cluster point and so no finite valued sequence has a cluster point.
    OK. I was treating x_n as a distinct point from x_m, even if they both had the same value a_i. I was thinking that if x_n\to x, then for all n greater than some N, d(x_n,x)=0<\epsilon and \{x_N,x_{N+1},...\} was the infinite set (even though all members of that set have the same value). It was a definition mix-up on my part.
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