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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!

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4 Answers 4

up vote 3 down vote accepted

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.

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That alternative definition helps a lot, Brian! Thanks for the answer! –  Nick Thomas Feb 12 '13 at 7:05
    
@Nick: You’re welcome! –  Brian M. Scott Feb 12 '13 at 7:06
    
Do you think the weak direct sum could visualize the OP this relation? thanks. –  Babak S. Feb 12 '13 at 7:12
    
@Babak: Maybe: its elements are exactly one equivalence class in the full product. –  Brian M. Scott Feb 12 '13 at 7:14

This is a very small hint in the light of Brian's complete one. I think the weak direct sum $\sum A_k$ when, for example, $\{A_k\}$ is a family of groups indexed by a set $K$, can help you to find out what is happening in the relation. Try to Google it for details.

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Nice hint...+ 1 ;-) –  amWhy Feb 12 '13 at 15:55

And as to what that equivalence class might be/look like?

Well, that depends on the nature of set A.

If A={0,1} then your function can be interpreted as producing a binary Real number corresponding to each function a, such as .01110011.. or .1101110... etc. In this case, the equivalence classes are Real numbers which vary by a finite sum of negative powers of two. So x and x + 1/4 + 1/32 would be in the same equivalence class. A different set A and a different interpretation of the sequence as a Real might produce (for example) equivalence classes of Reals which differ by a rational, such that x, x + 1/7, x + 23/788 etc.

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It is an equivalence relation. There's nice Brian answer, but it is short on examples, so here is something to gain more intuition:

  • Try to imagine what is the class of abstraction of sequence of zeros $(\ldots, 0, 0, 0, \ldots)$.

  • Another example would be a convergent sum $\sum_{i \in \mathbb{Z}} a_i < \infty$, what can you tell about other sequences in its class of abstraction?

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