# Distinct equivalent classes the null set

• Dec 30th 2009, 02:33 AM
Distinct 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 AM
Swlabr
Quote:

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.

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 AM
Quote:

Originally Posted by Swlabr
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.

Yep thats what i meant thanks.
• Jan 5th 2010, 10:21 AM
Quote:

Originally Posted by Swlabr
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.

acually im a bit confused how is this a contradiction, it doesnt show its in the null set, just that e is in f?
• Jan 5th 2010, 10:50 AM
Defunkt
Quote:

You assumed that there is some element in $\displaystyle E\cap F$. As Swlabr showed, this implies $\displaystyle E=F$, however you took E,F to be distinct equivalence classes, ie. $\displaystyle E \neq F$ - so this is a contradiction.
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.