# Thread: Construction of pairwise disjoint sets

1. ## [UPDATED PROBLEM] Construction of pairwise disjoint sets

I am given a problem asking me to identify pairwise disjoint sets that satisfy some property, unfortunately I do not know how I would construct them.

I'm given sets Y1, Y2, ... and I'm asked to identify pairwise disjoint sets Z1, Z2, ... that satisfy the following property.

$\bigcup_{i=1}^{N}Y_i = \bigcup_{i=1}^{N}Z_i$

I know that if Z1, Z2, ... are pairwise disjoint sets, then $\bigcup_{i=1}^{N}Z_i=\sum_{i=1}^{N}Z_i$

Substituting this into the property we get:

$\sum_{i=1}^{N}Z_i= \bigcup_{i=1}^{N}Y_i$

What should be the next step?

How does this revelation help construct the pairwise disjoint sets and how does it relate to the property?

EDIT: Rephrased question

2. ## Re: Construction of pairwise disjoint sets

Originally Posted by chaosier
I am given a problem asking me to identify pairwise disjoint sets that satisfy some property, unfortunately I do not know how I would construct them.
I'm given sets Y1, Y2, ... and I'm asked to identify pairwise disjoint sets Z1, Z2, ... that satisfy the property (UNION Zi for i = 1 to n) = (UNION Yi for i = 1 to n).

I know that if Z1, Z2, ... are pairwise disjoint sets, then (UNION Zi for i = 1 to n) = sum{from i = 1 to n}(Zi).

sum{from i = 1 to n}(Zi) = (UNION Yi for i = 1 to n).
What does sum{from i = 1 to n}(Zi) that mean?

Having work with set theory for years, I have never seem the expression sum{from i = 1 to n}(Zi)
In fact, I find the way you asked the question very confusing. I would ask you to repost the questions with a clear explication of the terms.

3. ## Re: Construction of pairwise disjoint sets

Sorry, I made up this notation since I don't know how to type with symbols on this forum.

sum{from i = 1 to N}(Zi) means the summation from i = 1 to N of Z_i, so Z1 + Z2 + Z3 + ... + ZN.

4. ## Re: Construction of pairwise disjoint sets

Originally Posted by chaosier
Sorry, I made up this notation since I don't know how to type with symbols on this forum.

sum{from i = 1 to N}(Zi) means the summation from i = 1 to N of Z_i, so Z1 + Z2 + Z3 + ... + ZN.
The point is if each of $Z_j$ is a set then $\sum_{i=1}^{N}Z_i$ has absolutely no meaning whatsoever.

The summation sign, $\sum$ is not usually used with sets. If you have a textbook that uses it, the you must tell us what it means.

5. ## Re: Construction of pairwise disjoint sets

Ok, I suppose that I have the wrong idea then. How would you propose I approach the problem? Looking at my book it seems that it's applied to events, and not sets.

6. ## Re: Construction of pairwise disjoint sets

Originally Posted by chaosier
Ok, I suppose that I have the wrong idea then. How would you propose I approach the problem? Looking at my book it seems that it's applied to events, and not sets.
I have tried to get you to simply state the problem exactly as it is written.
Is this part of a probability course?
Also note there is a difference is a disjoint collection and a pairwise disjoint collection. That distinction is really important.

Also any collection of sets can be written as pairwise disjoint collection with the same union.
But if the set are really events then that may be what you want.

7. ## Re: Construction of pairwise disjoint sets

Actually the problem is written as I have posted. It tells me to consider sets Y1, Y2, .... It then asks me to identify pairwise disjoint sets Z1, Z2, ... such that the property
I posted above is satisfied for every N that is an element of {1, 2, 3, ..., }

8. ## Re: Construction of pairwise disjoint sets

Originally Posted by chaosier
I'm given sets Y1, Y2, ... and I'm asked to identify pairwise disjoint sets Z1, Z2, ... that satisfy the following property.
$\bigcup_{i=1}^{N}Y_i = \bigcup_{i=1}^{N}Z_i$
If that is the real question, then let $Y=\bigcup_{i=1}^{N}Y_i$
Define $Z_1=Y_1$ and for $1 define $Z_J=Y_J\setminus \left[\bigcup_{i=1}^{J-1}Z_i\right]$.

I will let you show that works.

9. ## Re: Construction of pairwise disjoint sets

Thanks, I arrived at the same result after much struggling.