Let's solve the given problems step by step.
### (a) Write the cardinality of [tex]\( n(\overline{A \cup B \cup C}) \)[/tex].
To find [tex]\( n(\overline{A \cup B \cup C}) \)[/tex], we will use the principle of inclusion-exclusion:
1. Step 1: Calculate [tex]\( n(A \cup B \cup C) \)[/tex]:
[tex]\[
n(A \cup B \cup C) = n(A) + n(B) + n(C) - n(A \cap B) - n(A \cap C) - n(B \cap C) + n(A \cap B \cap C)
\][/tex]
2. Step 2: Calculate [tex]\( n(\overline{A \cup B \cup C}) \)[/tex]:
[tex]\[
n(\overline{A \cup B \cup C}) = n(U) - n(A \cup B \cup C)
\][/tex]
From the provided result, we know that:
[tex]\[
n(\overline{A \cup B \cup C}) = n(U) - [(n(A) + n(B) + n(C) - n(A \cap B) - n(A \cap C) - n(B \cap C) + n(A \cap B \cap C))]
\][/tex]
### (b) Find the value of [tex]\( n(C - B) \)[/tex].
To find [tex]\( n(C - B) \)[/tex], we simply subtract the cardinality of the intersection of B and C from the cardinality of C:
[tex]\[
n(C - B) = n(C) - n(B \cap C)
\][/tex]
### Prove that [tex]\( n(A) = n(A \cap B) + n(A \cap C) + n_0(A) \)[/tex].
This proof involves understanding how the elements of set A can be partitioned:
- [tex]\( n(A \cap B) \)[/tex] represents the number of elements that are in both A and B.
- [tex]\( n(A \cap C) \)[/tex] represents the number of elements that are in both A and C.
- [tex]\( n_0(A) \)[/tex] represents the number of elements that are only in A, not in B or C.
Given these, we can write the cardinality of A as:
[tex]\[
n(A) = n(A \cap B) + n(A \cap C) + n_0(A)
\][/tex]
This indicates that the set A is partitioned into disjoint subsets: elements common with B, elements common with C, and elements unique to A.
### When [tex]\( n(A) = n(B) \)[/tex], is [tex]\( A = B \)[/tex]?
No, [tex]\( n(A) = n(B) \)[/tex] does not necessarily imply [tex]\( A = B \)[/tex].
1. Cardinality and Equality:
- Two sets A and B having the same cardinality means they have the same number of elements.
- However, they may have different elements despite having the same number of them.
Conclusion:
- Two sets A and B can have different elements even if the number of elements is the same, i.e., [tex]\( n(A) = n(B) \)[/tex] does not imply [tex]\( A = B \)[/tex].
These detailed steps form the complete solution to the given problems.
