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According to wolfram mathworld (and wikipedia :), given a set $S$ and an equivalence relation $R$ on $S$, an equivalence class is a set $\{x\in S|xRa\}$ associated with an element $a\in S$.

However, many times I've seen other interpretations, so to say. For instance, here, the first answer says:

"It’s easily checked that $\sim$ is an equivalence relation on $U$ whose equivalence classes are pairwise disjoint open intervals in $R$".

ie. the elements of the equivalence class are intervals (well, pairs that are interpreted as intervals).

Is it just jargon, for the sake of pragmatism?

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hyperio
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  • The same equivalence class can be associated with more than one element. In general, the equivalence classes relative to some equivalence relation are always pairwise disjoint sets. – Carl Mummert Sep 26 '16 at 10:33
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    Not the elements of the equivalence class are said to be intervals, but the equivalence classes themselves are said to be intervals. – drhab Sep 26 '16 at 10:33
  • In the answer you refer to, the equivalence classes are not "pairs that are interpreted as intervals", they are intervals. And this is exactly the same as the notion of equivalence class you quoted fom Wolfram. – bof Sep 26 '16 at 10:34
  • The equivalence classes of equivalence on some set are subsets of that set. The link you quote says that the equivalence classes are subsets of points in $;U;$ which form disjoint intervals ...etc. – DonAntonio Sep 26 '16 at 10:35
  • @DonAntonio so, in the example, each equivalence class represents one interval? Also, an equivalence class should have an element of $S$ associated with it, right? ie. we could have an injective $f(x)=EquivClass(x)\subseteq S,\forall x\in S$ – hyperio Sep 26 '16 at 10:42
  • @hyperio Not exactly: in the example, each equivalence class is a set of points which forms an interval. That the set of points is an interval follows from the way the relation was defined, of course. Now, "equivalence class" actually means "equivalence class of this or that element", so yes: it is usually related to one element of the set on which the relation is defined, but observe that any given equiv. class can be related to maney, some times infinite, different elements of the set. – DonAntonio Sep 26 '16 at 10:50
  • @DonAntonio Very clear, thanks. So, in the answer I referred to, for $S=(a, b)$, I could pick any $x\in S$ and get the same equivalence class under $\sim$, ie. $S$ itself. Right? – hyperio Sep 29 '16 at 14:08

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