Three questions regarding ZF set theory

Dale10101

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In “FUNDAMENTALS OF ZERMELO-FRAENKEL SET THEORY” by TONY LIAN

http://www.math.uchicago.edu/~may/VIGRE/VIGRE2011/REUPapers/Lian.pdf

The following definitions are given:

Definition of an “ordered pair” 3.1. An ordered pair (a, b) is defined to be (a, b) = { {a}, {a, b} }

Definition of a “binary relation” 3.2. A set R is a binary relation if all elements of R are ordered pairs. (i.e. for z element of R
there exists x and y such that z = (x, y).

Definition of a “Membership relation” 3.3. The membership relation on A is defined by
Ea = {(a, b) | a element of A, b element of B, and a element of b}

From elementary algebra an ordered pair might be (2,5) or (5,2) as points on the Cartesian plane and of course they are not equal.

Question #1:

In the definition of ordered pair given above (a, b) = { {a}, {a, b} } but since by definition the right hand side of the equation is an unordered set containing the elements {a} and {a,b} wouldn't it also be true by definition that:

(a, b) = { {a}, {a, b} } = { {a,b}, {a} }, but this is like saying (2,5) = (5,2) and that can’t be right.

I am thinking that the original definition assumes that one will understand that in (a,b) “a” refers to the singleton {a} and “b” refers to the pair {a,b} regardless of the order of the elements on the right side of the equation.

Can I get confirmation on that … or not.

Question #2:
The definition of “Membership” says that a is an element of b, which seems true enough for (2,5) but not for (5,2).

Is the order/relation being established here intended to be applied only to numbers on a number line where each succeeding number is a subset of a preceding number? Maybe that is the point? The “membership relation” is a special type of relation in which its definition obtains … maybe, or maybe I am going off the rails.

Question #3

Are there two fundamentally different types or categories of relationships, those that define a correspondence between elements of two different sets, A and B (even if B has the same elements as A but is a differnt set like the measurements defining the length and breadth of a rectangle) and a second type of relationship that pertains to establishing “order” within a set?

I ask this because I originally thought of a “function” as a “relation” that passes the “vertical test”, but then became confused about what this had to do with “equivalence relations”.

Thanks for any help, Dale
 
In “FUNDAMENTALS OF ZERMELO-FRAENKEL SET THEORY” by TONY LIAN

http://www.math.uchicago.edu/~may/VIGRE/VIGRE2011/REUPapers/Lian.pdf

The following definitions are given:

Definition of an “ordered pair” 3.1. An ordered pair (a, b) is defined to be (a, b) = { {a}, {a, b} }

Definition of a “binary relation” 3.2. A set R is a binary relation if all elements of R are ordered pairs. (i.e. for z element of R
there exists x and y such that z = (x, y).

Definition of a “Membership relation” 3.3. The membership relation on A is defined by
Ea = {(a, b) | a element of A, b element of B, and a element of b}

From elementary algebra an ordered pair might be (2,5) or (5,2) as points on the Cartesian plane and of course they are not equal.

Question #1:

In the definition of ordered pair given above (a, b) = { {a}, {a, b} } but since by definition the right hand side of the equation is an unordered set containing the elements {a} and {a,b} wouldn't it also be true by definition that:

(a, b) = { {a}, {a, b} } = { {a,b}, {a} }, but this is like saying (2,5) = (5,2) and that can’t be right.
Your arguement is wrong. (2, 5)= {{2}, {2, 5}}= {{2, 5}, {2}} while (5, 2)= {{5}, {2, 5}}= {{2, 5}, {5}}. Those are not the same thing. The point is not which set of "{a}, {a,b}" if first and which second (we are, after all, defining "order") but that {a} contains only one member and {a, b} contains two.

I am thinking that the original definition assumes that one will understand that in (a,b) “a” refers to the singleton {a} and “b” refers to the pair {a,b} regardless of the order of the elements on the right side of the equation.
No, b does not "refer to the pair {a,b}". The point is simply to be able to distinguish between "a" and "b".

Can I get confirmation on that … or not.

Question #2:
The definition of “Membership” says that a is an element of b, which seems true enough for (2,5) but not for (5,2).
Are you still talking about {{a}, {a,b}}? That does NOT say "a is an element of b". b us not necessarily as set.

Is the order/relation being established here intended to be applied only to numbers on a number line where each succeeding number is a subset of a preceding number? Maybe that is the point? The “membership relation” is a special type of relation in which its definition obtains … maybe, or maybe I am going off the rails.
We can define orders on many different things, not necessarily sets at all.

Question #3

Are there two fundamentally different types or categories of relationships, those that define a correspondence between elements of two different sets, A and B (even if B has the same elements as A but is a differnt set like the measurements defining the length and breadth of a rectangle) and a second type of relationship that pertains to establishing “order” within a set?

I ask this because I originally thought of a “function” as a “relation” that passes the “vertical test”, but then became confused about what this had to do with “equivalence relations”.

Thanks for any help, Dale
 
OK, good … I have pretty much everything wrong which explains why every new thing seems inconsistent.

So, starting anew.

So what you are saying"

“The point is not which set of "{a}, {a,b}" if first and which second (we are, after all, defining "order") but that {a} contains only one member and {a, b} contains two.”

... is that writing:

{ {dog}, {horse, donkey}}

I have used the ZF definition of an ordered pair to define the ordered pair (a,b)

a = {dog},
b = {horse, donkey}

also, extending this concept further, writing:

{ {dog}, {cat, mouse, lizard}, {horse, donkey}}

I have defined the ordered triplet (a,b,c)

a = {dog},
b = {horse, donkey}, and
c = {cat, mouse, lizard}

I am going to stop here because this might be wrong too so why try and build on it.

Thanks for your help.
 
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