Set problem, new operation

[alpha2cen]

We know set calculation well.

A={1, 2, 3}

B={2, 3, 5}

A cup B ={1, 2, 3, 5}

A cap B ={ 2, 3}

, etc.

.

New problem

A={ red coat, red car, blue apple, blue dress}

B={white coat, blue car, red apple, blue dress}

A cap B = { blue dress} well known

.

New operation

A (red)cup B= { red coat, red car, red apple}

A (car)cup B= { red car, blue car}

A (car, apple)cup B={red car, blue apple, blue car, red apple}

How about this operation?

Edited January 27, 2011 by alpha2cen

[ajb]

Reminds me of forgetful functors. "Throwing away some of the structure".

 

So I don't think there is any real problems with the operations you suggest. The first two seem quite natural, you consider only the "colour" or "car " properties of the elements. (Sounds very categorical)

 

The last operation is just the intersection of the above?

 

Something like (In clear notation I hope)

 

[math]A \cup_{car , apple} B = (A \cup_{car} B )\cap A( \cup_{apple} B ) [/math]?

 

I am sure you could formulate this a lot neater than I suggest.

[alpha2cen]

Something like (In clear notation I hope)

 

[math]A \cup_{car , apple} B = (A \cup_{car} B )\cap A( \cup_{apple} B ) [/math]?

 

I am sure you could formulate this a lot neater than I suggest.

 

A (car, apple)cup B= (A (car)cup B ) cup ( A (apple)cup B)

={red car, blue apple, blue car, red apple}

A (a, b, c, ...)cup B = (A (a)cup B) cup (A (b)cup B) cup (A ( c)cup B) cup ...

Practically we use this operation very much.

Edited January 27, 2011 by alpha2cen

[ajb]

Great, so all you have to do is define things like

 

[math]A \cup_{car}B[/math],

 

carefully.

 

The "cap" should be "cup" in what I have written. My mistake. But anyway, without some extra condition these are simply as sets. Defining the cup and cap with a condition is all that is need, I expect.

[alpha2cen]

A={red car, blue car, red orange}

B={red car, blue car, blue dress}

 

A (red)cap B ={red car} more defined

A cap B={red car, blue car}

U- (A (red)cup B )= {blue car, blue dress}

 

I think we use this operation every day.

Edited January 27, 2011 by alpha2cen

[ajb]

I think we use this operation every day.

 

So, what would be the formal definitions here? Say we have an arbitrary, but finite number of "properties" or "conditions".

[alpha2cen]

It is more better the elements have their property words or common name, i.e., food, house, car, etc.. The properties are like this. Smell, color, age, hardness, darkness, feeling, etc.

Edited January 27, 2011 by alpha2cen

[ajb]

Maybe you want something like

 

[math]A\cup_{p}B = \{\left. a \in A \cup B\right| p(a) = \textnormal{True} \} [/math],

 

 

for some criterion [math]p[/math]. Then you would have to see if this is really what you are looking for.

[alpha2cen]

Maybe you want something like

 

[math]A\cup_{p}B = \{\left. a \in A \cup B\right| p(a) = \textnormal{True} \} [/math],

 

 

for some criterion [math]p[/math]. Then you would have to see if this is really what you are looking for.

 

p(a)=true has some problem. a={a1, a2,...}, p(a_i element {a})=true It has some different element property against generally used set. But how do we define number set or polygon set? This is not a number set, but a concept set.

Edited January 27, 2011 by alpha2cen

[ajb]

Say, it is an element of some other set (or class)? The class of all "red objects" or something similar.

[Xerxes]

Reminds me of forgetful functors. "Throwing away some of the structure".

Hmm, well, the structure that forgetful functors "forget" is algebraic, so unless our sets are algebraic, I'm not sure it applies. Two classic examples: the functor that sends the category of vector spaces to the category of abelian groups, "forgets" the algebraic operation of scalar multiplication by the field over which our vector spaces are defined.

 

Likewise, the forgetful functor that sends the category of general groups to their underlying sets "forgets" about the group binary operation

 

Although the OP was not very clear, even less so his/her follow-ups, I suspect the "new" operation being described is disjoint union of sets. So that is, say, we have a set of all colours, and a subset of all objects, then their disjoint union is the set {{red,car}, {red, dress}, {blue, car},....,{white, car},....,, {blue, moon},....}}

 

Contrast this to the the "ordinary" union of sets: here we do NOT forget the origin of each set element entering into the union. In a certain sense this is opposite to the forgetful functor!

[ajb]

Hmm, well, the structure that forgetful functors "forget" is algebraic, so unless our sets are algebraic, I'm not sure it applies.

 

Sure, but it did remind me of them.

 

I am not sure exactly what alpha2cen has in mind with this.

[alpha2cen]

We usually use cup , cap, etc. operations.

But, this operation is a cup against something or a cap against something.

There are many fruits, i.e., an apple, an orange, a melon, etc.

But they are all fruits set.

Airplane, car, ship , etc., are made of metal, so metal set.

Edited January 28, 2011 by alpha2cen

[Xerxes]

this operation is a cup against something or a cap against something.

 

I'm afraid I don't recognize the qualifier "against" in this context

 

There are many fruits, i.e., an apple, an orange, a melon, etc.

But they are all fruits set.

Airplane, car, ship , etc., are made of metal, so metal set.

Hmm, how can I put this to you kindly?

 

Your understanding of set theory needs a little work. Look, but please pay close attention to the way I use notation, which is standard but arbitrary.

 

Define a set called FRUIT, and include as its elements all apples, all oranges, all pears etc (notice the plural here). Let's write [math]F=\{A,O,P,....\}[/math], so that the union [math]A\cup O\cup P \cup..... = F[/math].

 

So now we have the subsets {apples}, {oranges}, {pears}, so that any chosen apple, say, is an element in the subset {apples} of the set FRUIT. One writes [math] a \in A \subsetneq F[/math]. Notice that no apple can also be an orange (as far as I am aware) so the intersection is empty, i.e. [math]A \cap O = \O[/math]; this is called being "disjoint", so that the set FRUIT is the union of disjoint sets, what I called the "disjoint union". Of course one may always introduce a new constraint, say size, in which case the union need not be disjoint, since the intersection of subsets may not be empty.

Edited January 28, 2011 by Xerxes

[alpha2cen]

I'm afraid I don't recognize the qualifier "against" in this context

 

 

How about change like this, i.e., "element x union of A and B or element x intersection of A and B, and x is aij or bij."

A={a1, a2, ..., an}, B={b1, b2, ..., bm }

a1={a11, a12, ..., a1n1}, a2={a21, a22, ..., a2n2}, ..., an={an1, an2, ..., annn}

b1={b11, b12, ..., b1m1}, b2={b21, b22, ..., b2m2}, ..., bm={bm1, bm2, ..., bmmm}

 

Example

A={red car, blue car, red orange}

B={red car, blue car, blue dress}

A (red)cup B=?

A element

red car={red, car}, blue car={blue, car}, red orange={red, orange}

B element

red car={red, car}, blue car={blue, car}, blue dress={blue, dress}

 

A (red)cup B = {red car, red orange}

Edited January 29, 2011 by alpha2cen

[khaled]

Maybe you want something like

 

[math]A\cup_{p}B = \{\left. a \in A \cup B\right| p(a) = \textnormal{True} \} [/math],

 

 

for some criterion [math]p[/math]. Then you would have to see if this is really what you are looking for.

 

I think that's a good definition, for one constraint

 

Hmm, well, the structure that forgetful functors "forget" is algebraic, so unless our sets are algebraic, I'm not sure it applies. Two classic examples: the functor that sends the category of vector spaces to the category of abelian groups, "forgets" the algebraic operation of scalar multiplication by the field over which our vector spaces are defined.

 

Likewise, the forgetful functor that sends the category of general groups to their underlying sets "forgets" about the group binary operation

 

Although the OP was not very clear, even less so his/her follow-ups, I suspect the "new" operation being described is disjoint union of sets. So that is, say, we have a set of all colours, and a subset of all objects, then their disjoint union is the set {{red,car}, {red, dress}, {blue, car},....,{white, car},....,, {blue, moon},....}}

 

Contrast this to the the "ordinary" union of sets: here we do NOT forget the origin of each set element entering into the union. In a certain sense this is opposite to the forgetful functor!

 

I think the one big set would be complex to represent, but easier we would have:

 

Set of Objects = { car, dress, moon, ... }

 

Set of Colors = { white, black, red, green, ..., phi }

 

Set of Shapes = { cubic, rectangular, ..., phi }

 

Set of Sizes = { long, short, wide, huge, ..., phi }

 

Set of Measurements = { um, mm, cm, m, Km, Mm, Gm, ..., phi } .. and so on,

 

and you can form a definition using: Color,{Shape}+,Object,{[iNTEGER],Measurement,Size}+,...

 

where + means "1 or more recurrence" .. { .. }+ means "1 or more recurrence" of the sub-set

 

example: { "blue", { "spiral", "circular", "cylinder" }, "tube", { [40], "cm", "long" }, { [60], "mm", "wide" } }

 

Good luck

Edited January 29, 2011 by khaled

[Xerxes]

Sorry but I can make no sense of these last two posts. It might help to point out a coupla things;

 

1. The objects that enter into the set operations union and intersection MUST be sets, they cannot be elements (in topology we stretch the point slightly and informally refer to the sets in a topological space as "elements", but strictly this is wrong).

 

Likewise the result of these operations on sets is again a set, never an element. (This is called a "closed" operations, BTW)

 

2. As my tutor was fond of saying: "since notation is arbitrary, there is no good reason not to use the same as everyone else". Try it......

 

There is obviously some confusion about simple set theory here. If you would like a brief tutorial, just ask; I am sure some of us here can help out with that.

Edited January 29, 2011 by Xerxes