Relations, part deux

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December 5th 2009, 11:42 AM
oldguynewstudent

Relations, part deux

I am completely lost on the terminology on this one.

Let $\rho$ be an equivalence relation on a non-empty set X.
Let a $\epsilon$ X. Show that [a $]_\rho$ $\not=$ $\emptyset$

Here [a $]_\rho$ := {b $\epsilon$ X | a $\rho$ b}

Is this supposed to be b=a? What is this set X?

The only thing I get is that X is non-empty so it must contain at least one element b. Then the equivalence relation has to be non-empty by definition of equivalence.
December 5th 2009, 11:52 AM
Plato

Quote:

Originally Posted by oldguynewstudent

I am completely lost on the terminology on this one.

Let $\rho$ be an equivalence relation on a non-empty set X.
Let a $\epsilon$ X. Show that [a $]_\rho$ $\not=$ $\emptyset$

Here [a $]_\rho$ := {b $\epsilon$ X | a $\rho$ b}

Is this supposed to be b=a? What is this set X?

The only thing I get is that X is non-empty so it must contain at least one element b. Then the equivalence relation has to be non-empty by definition of equivalence.

You are correct that we assume that $X\ne\emptyset$ .
But to prove this we know that $\rho$ is reflexive.
So $\left( {\forall x \in X} \right)\left[ {(x,x) \in \rho } \right]$ thus $\left( {\forall x \in X} \right)\left[ {x \in [x]_\rho } \right]$
December 5th 2009, 01:41 PM
oldguynewstudent

Thanks

Quote:

Originally Posted by Plato

You are correct that we assume that $X\ne\emptyset$ .
But to prove this we know that $\rho$ is reflexive.
So $\left( {\forall x \in X} \right)\left[ {(x,x) \in \rho } \right]$ thus $\left( {\forall x \in X} \right)\left[ {x \in [x]_\rho } \right]$

I just found out that we skipped four sections in the textbook. I had not heard of an equivalence relation before and did not know that it is reflexive, symmetric and transitive. I also did not understand that [a] $\rho$ was an equivalence class. So I have some more reading to do. I see the answer you supplied is extremely helpful.

Wish I had you for my professor instead of my current one!

Merry ChrisKwanuka.