1. ## 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?

2. What about $0,1,0,1,0,1,...$ there is no $n$ such that $j\geq n$ implies that $s_j$ is constant.

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

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