Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

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!

share|cite|improve this question
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.

share|cite|improve this answer
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.

share|cite|improve this answer
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.

share|cite|improve this answer

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?

share|cite|improve this answer

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.