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Suppose that $Ω \in \mathcal{F}$ and that $\mathcal{F}$ is closed under the formation of the complements and finite disjoint unions. Show by an example that $\mathcal{F}$ need not be a field

Solution:

Hint: I think the solution is given as follows

Consider $ X=\{1,2,\ldots,2n\}$ for some $n \in \mathbb{N}$ and

$$\mathcal{F} := \{A \subseteq X; \sharp A \, \text{is even}\}, $$ where $\sharp A$ denotes the cardinality of the set $A$.

I am not completely sure what was the thought of solving process.

My Questions:

  1. First of all, is the only reason that $\mathcal{F}$ is not always a field is because of the word "disjoint"?

  2. Does this mean that the set $A \in \mathcal{F}$ could look like $A = \{ 1,2\}$ or $A = \{1,2,3,4\}$ etc.? I don't understand what the even condition secures and why $\mathcal{F}$ is closed under the formation of the complements and finite disjoint unions.

  3. To prove that $\mathcal{F}$ is not a field we say

    • $Ω \in \mathcal{F}$
    • $A \in \mathcal{F}$ and $A^c \in \mathcal{F}$

    The second is true because if $A = \{ 1,2\}$ then $\sharp A = 2$, so $A$ is of even cardinality. Then $A^c = \{ 3,4..,2n\}$ implies that $\sharp A = 2n-2$ which is also of even cardinality.

  4. $A \in \mathcal{F}$, $B \in \mathcal{F}$ so we must DISPROVE that $A \cup B \in \mathcal{F}$?? How can we do that?

Xander Henderson
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Pedros
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  • Suppose $A = { 1, 2}$ and $B = { 1, 3}$. Each of these has even cardinality, but their union has odd cardinality. – Xander Henderson Sep 29 '19 at 23:53
  • @XanderHenderson These are not diisjoint sets. – Marios Gretsas Sep 30 '19 at 00:00
  • I didn't claim that they were. You have a system which is closed with respect to the union of disjoint sets, but a $\sigma$-field (presuming that this is what you mean) is closed under countable unions (not necessarily disjoint). The goal is to show that closure under disjoint unions is not sufficient to show closure under unions more generally. – Xander Henderson Sep 30 '19 at 00:02
  • If that is not what you mean, you should define what you mean by a "field". – Xander Henderson Sep 30 '19 at 00:06
  • @XanderHenderson the O.P i believe that with Field he means an algebra of sets..In wikipedia,especially in probability theory a field is a pair (X,S) where X is the space and S is an algebra of sets. – Marios Gretsas Sep 30 '19 at 00:10
  • i believe this is related to https://math.stackexchange.com/questions/8173/what-is-an-example-of-a-lambda-system-that-is-not-a-sigma-algebra – Calvin Khor Sep 30 '19 at 00:17
  • @MariosGretsas Ah! Presuming that they mean an algebra (and not a $\sigma$-algebra), then it should still be closed under finite unions. A union of two sets is a finite union. – Xander Henderson Sep 30 '19 at 00:37
  • @XanderHenderson yes i know..the O.P wanted a family of sets that is closed under disjoint unions and complements,but is not an algebra. – Marios Gretsas Sep 30 '19 at 00:41
  • @MariosGretsas Cool. That's what I thought when I provided my example in the first comment above. Their continued confusion following that example led to my further comments. – Xander Henderson Sep 30 '19 at 00:43

1 Answers1

1

You're almost asking about what is known as a Dynkin system, or a $\lambda$-system; there, countable disjoint unions are allowed, but the situation is the same if the space is finite. I'm not sure if the version with finite disjoint unions has a name.

This answer to What is an example of a lambda-system that is not a sigma algebra? gives a different solution.

For your questions, some hints -

  1. Yes. You might want to read this question Difference between a Dynkin System and Sigma algebra, just pretend that there are only finite unions.

  2. Lets just pick say $n=4$ for concreteness ($|X|=8$). If two sets of $\mathcal F$ are disjoint, what is the size of their union? (in particular, odd or even?)

  3. If they are not disjoint, what are the possibilities for the sizes of their union? (can you make the cardinality odd?)

Calvin Khor
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