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• Mar 16th 2013, 03:01 PM
sebasvargasl
I don't understand this example. Which are the elements of R?
• Mar 16th 2013, 03:05 PM
sebasvargasl
Quote:

Originally Posted by Hartlw
The best I could come up with was an example:

A={1.2.3.4,5} R = <
R[A]=(1,2),(1,3),(1,4),(1,5),(2,3),(2,4),(2,5),(3,4), (3,5),(4,5)
B={3,4,5,6}
R[B]= (3,4),(3,5),(3,6),(4,5),(4,6),(5,6)
R[A]-R[B]= (1,2),(1,3),(1,4),(1,5),(2,3),(2,4),(2,5)
[A-B]={1,2}
R[A-B]=(1,2)

so R[A-B] is a subset of R[A]-R[B]

??
• Mar 18th 2013, 06:30 AM
Hartlw
Quote:

Originally Posted by sebasvargasl
??

Defininitions:

AXB is the set of all ordered pairs (a,b) with a ϵ A and b ϵ B.

A relation R from A to B is a subset of AXB.

If B =A then you have a relation from A to A: R[A], a subset of AXA.

Example: A = {1,2,3}
AXA = { (1,1), (1,2), (1,3), (2,1), (2,2), (2,3), (3,1), (3,2), (3,3)}
Let the relation R[A] be “less than,” <. Then
R[A] = {(1,2), (1,3), (2,3)}
Write R[A] for the relations “equal” and “greater than.”

Definition: A – B: All elements of A which are not also elements of B. (Draw a Venn diagram with B partially overlapping A and subtract B)
Examples:
If B = {2,3,4,5} and A as above,
A - B = {1}.

If A and B don’t have elements in common, A-B is empty.
• Mar 18th 2013, 08:31 AM
sebasvargasl
Your definition of R[A] is wrong. The elements of R[A] are not necessarily ordered pairs. In order to say things about R[A] first you have to define R. Again, R[A]={y: ∃x(x∈A∧xRy)}. When you state R[A] = {(1,2), (1,3), (2,3)} this implies that for some x (x,(1,3))∈R, therefore R is a ternary relation and binary at the same time, a contradiction.
• Mar 18th 2013, 09:47 AM
Hartlw
Quote:

Originally Posted by sebasvargasl
Your definition of R[A] is wrong. The elements of R[A] are not necessarily ordered pairs. In order to say things about R[A] first you have to define R. Again, R[A]={y: ∃x(x∈A∧xRy)}. When you state R[A] = {(1,2), (1,3), (2,3)} this implies that for some x (x,(1,3))∈R, therefore R is a ternary relation and binary at the same time, a contradiction.

I DID define R[A], by assuming that’s what you meant by your unintelligible definition of Post #3, and my conclusions from this definition are correct. If I misinterpreted your original definition, why wait till now before explaining it? If you weren’t sure and have been working your way into it, that’s understandable.

You are now proposing a new problem by giving, for the first time, your new, intelligible, definition of R[A], which is:

R[A] is the set of all y in xRy.

I note you don’t define xRy but I know what you mean. For a proper definition, you should define what you mean by xRy.

I am not a mind-reader. Why didn’t you just say in the first place what your definition of R[A] is instead of waiting till now?

The more judicious way of doing this would be a new post of the form: If I define R[A] as set of all y in xRy, then show …..
• Mar 18th 2013, 10:32 AM
Hartlw
Here is new example to go with your new definition of R[A] in post #19.

Def: R[A] is set of all y of xRy. R of xRy is “less than,” <.

A={1,2,3,4}
xRy={(1,2),(1,3),(1,4),(2,3),(2,4),(3,4)}
R[A]={2,3,4}

B={3,4,5}
xRy=(3,4),(3,5),(4,5)
R[B]={4,5}

R[A]-R[B]= {2,3}

[A-B]={1,2}
xRy={(1.2)}

R[A-B]={2}

so R[A-B] is a subset of R[A]-R[B]
• Mar 18th 2013, 11:00 AM
Hartlw
And here is new proof of OP to go with new definition of post#19

x,y ϵ [A-B] → x,y ϵ A and x,y ϵ’ B → (y ϵ xRy → y ϵ A and y ϵ’ B)

So R[A-B] is a subset of R[A]-R[B]
• Mar 18th 2013, 02:16 PM
sebasvargasl
Now you're using xRy to denote a set, which is wrong, xRy is another form to write (x,y)ϵR: a sentential function, not a set. How is "y belongs to R[A] if and only if there exists x belonging to A for which the ordered pair (x,y) belongs to R" unintelligible?. Once again your example is not correct for what you're doing is defining two different relations namely, {(1,2),(1,3),(1,4),(2,3),(2,4),(3,4)} and {(3,4),(3,5),(4,5)}. They have to be one and the same relation to serve as an example. To write R[A] you use the first one, but, when you write R[B] you use the second one, so you're not really doing R[A]-R[B] but, R[A]-R'[B]
• Mar 18th 2013, 02:27 PM
Plato
Quote:

Originally Posted by Hartlw
And here is new proof of OP to go with new definition of post#19
x,y ϵ [A-B] → x,y ϵ A and x,y ϵ’ B → (y ϵ xRy → y ϵ A and y ϵ’ B)
So R[A-B] is a subset of R[A]-R[B]

This is completely wrong.

Relpy #13 is the only clear and correct proof in the thread.

• Mar 18th 2013, 02:27 PM
sebasvargasl