Let R be an equivalence relation on a set S. Let E and F be two distinct equivalence classes of R. Prove that E and F = null set.

Do i show that itys transitive im a bit stuck. Help much appreciated thanks.

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- Dec 30th 2009, 02:33 AMadam_leedsDistinct equivalent classes the null set
Let R be an equivalence relation on a set S. Let E and F be two distinct equivalence classes of R. Prove that E and F = null set.

Do i show that itys transitive im a bit stuck. Help much appreciated thanks. - Dec 30th 2009, 02:43 AMSwlabr
Do you mean $\displaystyle E \cap F = \emptyset$?

Assume there exists an element in $\displaystyle E \cap F$, call it $\displaystyle x$. Then you can apply transitivity as every element of $\displaystyle E$ is equivalent to $\displaystyle x$, as is every element of $\displaystyle F$. $\displaystyle e \sim x$ and $\displaystyle x \sim f \Rightarrow e \sim f \Rightarrow E = F$, a contradiction. - Dec 30th 2009, 02:45 AMadam_leeds
- Jan 5th 2010, 10:21 AMadam_leeds
- Jan 5th 2010, 10:50 AMDefunkt
- Jan 5th 2010, 12:47 PMDrexel28
More of a forward-knowledge looking back approach (since you need to do this problem to prove what I'm about to say), but a relation $\displaystyle R$ on $\displaystyle S$ induces a partition $\displaystyle \pi$ of $\displaystyle S$ where the blocks are the equivalence classes. If that is a definition in your book the answer follows immediately.