Results 1 to 3 of 3

Thread: Relations, part deux

  1. #1
    Senior Member oldguynewstudent's Avatar
    Joined
    Oct 2009
    From
    St. Louis Area
    Posts
    255

    Relations, part deux

    I am completely lost on the terminology on this one.

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

    Here [a$\displaystyle ]_\rho$ := {b $\displaystyle \epsilon$ X | a$\displaystyle \rho$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.
    Follow Math Help Forum on Facebook and Google+

  2. #2
    MHF Contributor

    Joined
    Aug 2006
    Posts
    21,800
    Thanks
    2829
    Awards
    1
    Quote Originally Posted by oldguynewstudent View Post
    I am completely lost on the terminology on this one.

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

    Here [a$\displaystyle ]_\rho$ := {b $\displaystyle \epsilon$ X | a$\displaystyle \rho$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 $\displaystyle X\ne\emptyset$.
    But to prove this we know that $\displaystyle \rho$ is reflexive.
    So $\displaystyle \left( {\forall x \in X} \right)\left[ {(x,x) \in \rho } \right]$ thus $\displaystyle \left( {\forall x \in X} \right)\left[ {x \in [x]_\rho } \right]$
    Follow Math Help Forum on Facebook and Google+

  3. #3
    Senior Member oldguynewstudent's Avatar
    Joined
    Oct 2009
    From
    St. Louis Area
    Posts
    255

    Thanks

    Quote Originally Posted by Plato View Post
    You are correct that we assume that $\displaystyle X\ne\emptyset$.
    But to prove this we know that $\displaystyle \rho$ is reflexive.
    So $\displaystyle \left( {\forall x \in X} \right)\left[ {(x,x) \in \rho } \right]$ thus $\displaystyle \left( {\forall x \in X} \right)\left[ {x \in [x]_\rho } \right]$
    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]$\displaystyle \rho$ 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.
    Follow Math Help Forum on Facebook and Google+

Similar Math Help Forum Discussions

  1. Replies: 1
    Last Post: Sep 19th 2011, 01:09 PM
  2. Replies: 25
    Last Post: Jan 15th 2010, 03:53 AM
  3. Sum of a part arithmatic and part geometric sequence
    Posted in the Pre-Calculus Forum
    Replies: 4
    Last Post: Oct 25th 2009, 10:53 PM
  4. evaluating limits. (part deux)
    Posted in the Pre-Calculus Forum
    Replies: 2
    Last Post: Oct 5th 2009, 06:00 PM
  5. Replies: 14
    Last Post: Feb 13th 2009, 04:27 AM

Search Tags


/mathhelpforum @mathhelpforum