What about there is no such that implies that is constant.
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?
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. has a constant subsequence means , where is a mononically increasing sequence of integers going to infinity. Then for any big n, there'll be some bigger than n.
The other direction: construct the subsequence inductively. take n=1, then you know there exist such that . Then take , and you know there exist some number that . And so on -- you obtain for any k.