I don't understand this example. Which are the elements of R?

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- Mar 16th 2013, 03:01 PMsebasvargaslRe: Proof about relations
I don't understand this example. Which are the elements of R?

- Mar 16th 2013, 03:05 PMsebasvargaslRe: Proof about relations
- Mar 18th 2013, 06:30 AMHartlwRe: Proof about relations
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 AMsebasvargaslRe: Proof about relations
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 AMHartlwRe: Proof about relations
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 AMHartlwRe: Proof about relations
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 AMHartlwRe: Proof about relations
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 PMsebasvargaslRe: Proof about relations
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 PMPlatoRe: Proof about relations
- Mar 18th 2013, 02:27 PMsebasvargaslRe: Proof about relations