TITLE: Finding a common subsequence given two sequences on the same elements
QUESTION [5 upvotes]: Let $A = \{x_1, x_2, x_3,\dots\}$.
Claim: If a sequence $y_k$ in $A$ contains inﬁnitely many distinct
  elements, then it has a subsequence $\{y_{k_m}\}$ that is a
  subsequence of the $x_n$ exhibited.
In short, two sequences constructed from the same set with infinitely many distinct elements
  must share some common subsequence.   (Is this rephrasing correct?)

Intuitively, this seems true, but how do I know for sure? Or is the claim false?

REPLY [4 votes]: It is true, and here is a constructive proof that such a subsequence exists:
Begin with the element $y_{k_1}=y_1$ which is equal to $x_{n_1}$ for some index $n_1$. There are $n_1-1$ elements that precede $x_{n_1}$. So, by the pigeonhole principle, one of the next $n_1$ elements  in the sequence $\{y_k\}$ must have an index in $\{x_n\}$ which is greater than $n_1$.  Call this element $x_{n_2}=y_{k_2}$.
This process can be repeated indefinitely to produce a subsequence of $\{y_k\}$ that is also a subsequence of $\{x_n\}$.  In other words, for a given $y_{k_i}=x_{n_i}$, we find $y_{k_{i+1}}$ by selecting a member of the next $n_i$ members of $\{y_k\}$ that is further along in the sequence $\{x_n\}$.