I am having a hard time with the following problem:

In F(R), let f~g iff f(x)=g(x) for all x>c where c is some fixed real number.

I proved that it was a equivalence relation by the following:

  1. f~f __ f(x)=f(x) so that is fine.

  2. f~g implies g~f __ f(x)=g(x) and g(x)=f(x) f~g does imply g~f so that is fine.

  3. f~g, g~h implies f~h __ f(x)=g(x) and g(x)=h(x) therefore you can substitute h(x) for g(x) thus f(x)=g(x).

The part that I am having trouble with is describing the partition associated with this equivalence relation. I know that the partition is equivalent to the equivalence class but I am unsure about how to find that since I'm dealing with functions. My assumption is that the partition is the collection of functions of f(x) where x>c.


If $c$ is a fixed constant, independent of $f$ and $g$, then the equivalence classes are essentially the functions with domain $(c,\infty)$. (That is, two functions are equivalent if and only if their restrictions to $(c,\infty)$ are identical.)

On the other hand, if $c=c(f,g)$, then you still have an equivalence relation, but the partition is different. In this case, two functions $f$ and $g$ are equivalent when they are "eventually equal": $[f]$ is the set of functions that are eventually (for large enough $x$) equal to $f$.

  • $\begingroup$ Ahh, great point. I just assumed $c$ was uniform ... a bit careless on my part. $\endgroup$ – Doc Nov 19 '13 at 2:33

I would say that for any function $f\in F(\mathbb R)$, the equivlence class containing $f$ (denoted $[f]$) is adequately described by $$[f]=\lbrace{g\in F(\mathbb R)}\mid g-f\equiv 0 \mbox{ on } (c,\infty)\}.$$ There are other equivalent ways to describe $[f]$; that's just one way.

  • $\begingroup$ Im sorry but what does that triple bar mean? Is it = because that would make sense to me since f(x)=g(x) and therefore g-f=0 $\endgroup$ – Student Nov 18 '13 at 23:17
  • $\begingroup$ ahh, sorry. The triple bar means "identically zero". Thus saying $f-g \equiv 0$ on an interval $I$ is a (slightly) quicker way of saying $f(x)-g(x)=0$ for all $x\in I$. $\endgroup$ – Doc Nov 18 '13 at 23:26

Your proofs are missing some detail:

  • Reflexivity is obvious, as you noted.

  • Symmetry is almost obvious: if $f\sim g$, then $f(x)=g(x)$ for all $x>c$; then $g(x)=f(x)$ for all $x>c$; therefore $g\sim f$.

  • About transitivity, you miss a fact. Saying $f\sim g$ and $g\sim h$ means there exist $c$ and $d$ such that $f(x)=g(x)$ for all $x>c$ and $g(x)=h(x)$ for all $x>d$; however nothing allows you to assume $c=d$. But the argument can be easily fixed.

‘Describing’ the partition is a rather generic term.

  • $\begingroup$ How would i fix that argument for transitivity? I have been trying to figure out but I don't see it. $\endgroup$ – Student Nov 19 '13 at 0:05
  • $\begingroup$ You have either $c\ge d$ or $c\le d$; but all you need is that $f$ and $h$ agree in some interval $(r,\infty)$: can you see which one in the two cases? $\endgroup$ – egreg Nov 19 '13 at 0:07
  • $\begingroup$ I get that c≥d or c≤d. what do you mean about agreeing in some case. I want to say that c≥d but don't know exactly why. $\endgroup$ – Student Nov 19 '13 at 0:12
  • $\begingroup$ @Student Take the greatest of the two and call it $r$; can you say that $f(x)=h(x)$ for all $x>r$? $\endgroup$ – egreg Nov 19 '13 at 0:20
  • $\begingroup$ I believe you can. $\endgroup$ – Student Nov 19 '13 at 0:27

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