Consider sequences $a : \mathbb{Z}^+ \rightarrow A$ on a set $A$. Define the relation $\sim$ over sequences by $a \sim b$ iff there are only finitely many indices $i$ at which $a_i \neq b_i$. Clearly $\sim$ is reflexive and symmetric.
It appears to me that it is also transitive! Suppose $a \sim b$, and $b \sim c$. Let $I$ be the set of indices $i$ such that $a_i \neq b_i$. Let $J$ be the set of indices at which $b_i \neq c_i$. Both $I$ and $J$ are finite, so $K = I \cup J$ is finite. Let $i \notin K$. Then $a_i = b_i$, and $b_i = c_i$, so $a_i = c_i$. So there are only finitely many points at which $a$ and $c$ differ (all of them are in $K$), so $a \sim c$.
This is extremely surprising to me, because it implies that $\sim$ is an equivalence relation. My difficulty is in understanding what equivalence classes this relation could possibly define.
Question 1: Is it really true that $\sim$ is an equivalence relation?
Question 2: Can anybody give me some kind of concrete description of what equivalence classes $\sim$ defines?
Thank you!
Yes, it’s an equivalence relation. Another way to describe is to say that $a\sim b$ iff there is an $n\in\Bbb N$ such that $a_k=b_k$ for all $k\ge n$. About the only way that I’ve ever found to visualize an equivalence class is to pick one member of it; if that member is $a$, then the class consists precisely of all sequences that agree with $a$ from some point on.
It turns out to be very important in studying the box topology on products of infinitely many topological spaces.