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ฟิสิกส์และคณิตศาสตร์มหาวิทยาลัย => ปีสี่: Group Theory (2549) => Topic started by: เกียรติศักดิ์ on November 17, 2006, 12:34:45 AM



Title: Equivalence relation แบ่งสมาชิกออกเป็น distinct classes
Post by: เกียรติศักดิ์ on November 17, 2006, 12:34:45 AM
Equivalence relation แบ่งสมาชิกออกเป็น distinct classes.


Instead of the proof demonstrated in class*, can I also prove this statement by proving that

"If there are two set of equivalent elements, and one element from one set is equal to an element from the other set, then the two sets are considered the same class; that is, no different classes have the same element--the classes are distinct."?

* which prove that by proving that "If there are two classes, and one element from one class is equal to an element from the other class, then the two classes are equal.".

It is what I did in class and not sure whether it is a valid proof.


Title: Re: Equivalence relation แบ่งสมาชิกออกเป็น distinct classes
Post by: ปิยพงษ์ - Head Admin on November 17, 2006, 06:04:05 AM
...
* which prove that by proving that "If there are two classes, and one element from one class is equal to an element from the other class, then the two classes are equal.".
...

How do you prove that two classes are equal? Isn't it the same method that we did in the class?  :coolsmiley:


Title: Re: Equivalence relation แบ่งสมาชิกออกเป็น distinct classes
Post by: เกียรติศักดิ์ on November 17, 2006, 10:23:53 AM
...
* which prove that by proving that "If there are two classes, and one element from one class is equal to an element from the other class, then the two classes are equal.".
...

How do you prove that two classes are equal? Isn't it the same method that we did in the class?  :coolsmiley:

That's what we did in class on the blackboard (by starting with that claim; I gave it as a reference). I wonder whether it is correct to start with this alternative claim in the very beginning:
If there are two set of equivalent elements, and one element from one set is equal to an element from the other set, then the two sets are considered the same class; that is, no different classes have the same element--the classes are distinct."


Title: Re: Equivalence relation แบ่งสมาชิกออกเป็น distinct classes
Post by: เกียรติศักดิ์ on November 17, 2006, 11:05:33 AM
Here's how it went in my paper.

Equivalence relation แบ่งสมาชิกออกเป็น distinct classes.
Claim: If there are two sets of equivalent elements, and one element from one set is equal to an element from the other set, then the two sets are considered the same class; that is, no different classes have the same element--the classes are distinct.

Proof: Supposed that there are two sets of equivalent elements
 \{g_1, g_2, \cdots, g_r\} and  \{h_1, h_2, \cdots, h_s\} .

Let  g_i = h_j , then  g_i \equiv h_j (reflexivity).
But  g_i \equiv \forall g , so  h_j \equiv \forall g (transitivity),
and  h_j \equiv \forall h , so  g_i \equiv \forall h (transitivity).

Since each element of one set is equivalent to all elements of the other set,
the two sets can be merged into a single class.

Thus, two sets cannot be considered two classes if there exists at least one common element.
Therefore, two classes are distinct.