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Let $\Omega={1,2,3,4,5}$ be a sample space.

Consider the collection of subsets of $\Omega$

F=$\{{\emptyset, \{1,2,4\},\{1,3,4,5\},\{1,4\},\{2\},\{2,3,5\},\{3,5\},\Omega\}}$.

A. Show tht F is a field.

it must satisfy the following criterias

(i) $\emptyset \in F$

(ii) If $A \in F$ then $A^c \in F(A^c= \Omega\backslash A)$

(iii) If$ A_1,A_2 \in F$, then $A_1 \cup A_2 \in F$

(i) We can clearly see that $\emptyset \in F$ then $\checkmark$.

(ii) This would be $A^c= \Omega \backslash A$ which would be $A^c= \emptyset$ which $\emptyset \in F$ then $\checkmark$?

(iii) We can see that $\{1,2,4\} \cup \{1,3,4,5\} \forall$ values in F.

Am I doing it right?

3 Answers 3

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  • (i) is correct
  • (ii) means that if $A$ is any set in $F$, then $A^c$ is in $F$ as well.
  • (iii) means if $A,B$ are any sets in $F$, then $A \cup B \in F$.
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    So are my answers technically right? – 2017-02-15
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    @guestuser46 only for (i), redo (ii) and (iii) for **all** possible choices. – 2017-02-15
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    for (ii) and example would be $(\{1,2,4\})^c= \{3,5\}$ which $\{3,5\} \in F$? – 2017-02-15
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    @guestuser46 example will not do. you need to go through *all* possible choices, but you are on the right track with that one. – 2017-02-15
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    Yeah that was just a part, so essentially all possible outcomes then? – 2017-02-15
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    @guestuser46 indeed – 2017-02-15
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    Thank you so much for making it clear – 2017-02-15
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I think because of how explicitly $F$ is defined, then you might have to check everything by hand.

For i, that is fine, it's clear by the definition of $F$.

For ii, just go set by set. As in, consider each element of $F$. What is the complement of $∅$? What is the complement of $\{1,2,4\}$? etc.

For iii, similar to ii. Consider each combination and show that the union is indeed still in $F$.

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    Ah okay, so since it's just a finite amount of numbers I should be able to set up all combinations then, thanks – 2017-02-15
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(i) is correct.

For (ii) you need to show the property holds for any set in $F$, not just the empty set. For instance, since $\{1,2,4\}\in F,$ $\{1,2,4\}^c = \{3,5\}$ must be in $F.$ This is true in this case. You must check that it holds true for all the sets in $F.$

For (iii) I don't know what you mean. I guess you're saying that $\{1,2,4\}\cup \{1,3,4,5\} = \Omega$ ? What you need to show is that for any two sets in $F$, their union is also in $F.$ For instance, since $\{1,4\}$ and $\{2\}$ are both in $F,$ then their union $\{1,2,4\}$ must be in $F$. This is true in this case and you must check the other possibilities to show that there are no counterexamples.