I know how to prove when the two equivalence classes are not disjoint, i.e. $[a]=[b]$. I see that the proof works for proving a equivalence class is disjoint, but I don't get it. Can someone explain it to me?
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Suppose we have two equivalence classes $[a]\not=[b]$. We'll show they are disjoint. Suppose $x\in[a]\cap[b]$. Then, $x\sim a$ and $x\sim b$. By symmetry, $a\sim x$, and hence, by transitivity, $a\sim b$. Therefore, $[a]=[b]$, which is a contradiction.
ervx
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So it is the same proof that I was referring to, but with the added contradiction. – Henry Lee May 17 '16 at 14:50
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Pretty much. You can basically prove both in one shot. Either $[a]\cap [b]=\emptyset$, or there is some $x\in[a]\cap[b]$, in which case, we get that $[a]=[b]$. – ervx May 17 '16 at 14:51
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It's not really a proof by contradiction. This is a direct proof that if two classes intersect, they are identical. – tomasz May 17 '16 at 14:52