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/sci/ - Science & Math


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9067111 No.9067111 [Reply] [Original]

I don't understand what this means, can someone help me plz? Maybe give an example? thx

>> No.9067138 [DELETED] 

oniii CHAAANNN

>> No.9067143

>>9067138
nyaaa

>> No.9067157

>>9067138
>>9067143
wrong neighbourhood /a/ fags

>> No.9067180

The intersection of unions is the same as the union of intersections

basically, it doesnt matter how you order your unions or intersections, as the operation is associative

kind of like how 1+1+1+1+1+1-1-1-1-1-1=1-1+1-1+1-1+1-1+1-1

>> No.9067266

>>9067111
Imagine you have a square array of sets.

Union the sets of each column to get a row of sets
then intersect the row of sets.
(left hand side)
=
(right hand side)
Intersect the sets of each row to get a column of sets
then union the column of sets.

Union distributes over Intersection and Intersection distributes over Union.

>> No.9067324
File: 29 KB, 633x758, tfw angry.png [View same] [iqdb] [saucenao] [google]
9067324

>>9067157
A-ANIME W-WEBS-SITE

>> No.9067366

>>9067180
>misleading and wrong

this is illustrating the distributive property between the two operations, not associativity of each (associativity is a property of a single operation, distributivity is a relation between two operations)

let I and J each have size 2, so that this indexes 4 sets which we'll just call A, B, C, and D numbered 11, 12, 21, and 22.

We first take the inner union ranging over J. So A_i1UA_i2. We then range over I's, intersecting. So (AUB)\cap (CUD). That's the LHS. The RHS first intersects over I's, so A_1j\cap A_2j, and we union over J's (A\cap C)U(B\cap D).

Pic says the two are equal.

>> No.9067371

>>9067366
you might be familiar with it in the form

a*(b+c)=a*b+a*c

swap out * for \cap and + for \cup. In lattices, the dual also holds and implies pic related.

>> No.9067373

>>9067371
ack sorry; i should say implies the finite version. the infinite distributivity law is stronger but does hold for set intersection and union.

>> No.9067455

>>9067266
This is not true in general though right?
For instance let A be an nxn matrix and let A_ij={i+j mod n}.

>> No.9067476

>>9067455
OP and that anon is clearly talking about arranging sets, not numerical entries. the operations are clearly set theoretic union and intersection.

these are not and have no reason to be related to matrix manipulations

>> No.9067496

>>9067476
In my question A_ij is a set though.

>> No.9067508

>>9067111
>I don't understand what this means, can someone help me plz? Maybe give an example? thx
sure!

(N denotes the intersection):

let A_11 = {1, 2, 3}, A_12 = {2,3,0}, A_21 = {1,3}, A_22 = {3,4,5}, I = {1,2}, J = {1,2}

then U_ j A_ij = A_i1 U A_i2
So N_i (U_ j A_ij) = N_i (A_i1 U A_i2) =
(A_11 U A_12) N (A_21 U A_22)

with A_11 U A_12 = {0,1,2,3} and A_21 U A_22 = {1,3,4,5} so in total we get
N_i (U_ j A_ij) = {1,3}

>> No.9067565

>>9067496
So what is A a matrix of and what does it have to do with the collection of sets?

The statement in OP's picture has nothing to do with algebraic structure other than the intersection and union of sets.

>> No.9067620

>>9067324
kill yourself wojak waifu feelfag

>> No.9067640

>>9067565
The matrix I gave was just a counterexample to OP's pic. I guess my main question is: what is the context of OP's pic is, and under what circumstances would such an identity apply?

>> No.9067802

>>9067111
>give an example
Say you have four rectangles in R^2: [0,2]X[0,2], [1,3]X[0,2], [0,2]X[4,6], and [1,3]X[4,6].
If you take ([0,2]X[0,2]∩[1,3]X[0,2]) U ([0,2]X[4,6]∩[1,3]X[4,6]) = [1,2]X[0,2] U [1,2]X[4,6] = [1,2]X([0,2]U[4,6]), that's the same as if you took ([0,2]X[0,2]U[0,2]X[4,6]) ∩ ([1,3]X[0,2]U[1,3]X[4,6]) = ([0,2]X([0,2]U[4,6]) ∩ ([1,3]X([0,2]U[4,6]) = [1,2]X([0,2]U[4,6])
The intersection will always give a figure with x ranging over [1,2] and the union will always give one with y ranging over [0,2] and [4,6], regardless of in which order you do the operations

>>9067455
>>9067640
>unions and intersections of matricies