Thread: 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.

Originally Posted by cgiulz

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 .

However, will always have a subsequence with a limit point.

Proof: Let . Because there are only finitely many points, is compact, and thus is a sequence in a compact set. So it follows from the Bolzano-Weierstrass Theorem that has a convergent subsequence (and therefore a limit point in that subsequence).

Originally Posted by redsoxfan325

However, will always have a subsequence with a limit point.

Proof: Let . Because there are only finitely many points, is compact, and thus is a sequence in a compact set. So it follows from the Bolzano-Weierstrass Theorem that 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 , there exist an infinite] number of points other than p itself whose distance to p is less than . 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 . A finite set does NOT have a cluster point and so no finite valued sequence has a cluster point.

Originally Posted by HallsofIvy

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 , there exist an infinite] number of points other than p itself whose distance to p is less than . 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 . A finite set does NOT have a cluster point and so no finite valued sequence has a cluster point.

OK. I was treating as a distinct point from , even if they both had the same value . I was thinking that if , then for all greater than some , and was the infinite set (even though all members of that set have the same value). It was a definition mix-up on my part.