Suppose U is the power set of the set S={1,2,3,4,5,6}. For any T∈U, let ∣T∣ denote the number of elements in T and T′ denote the complement of T. For any T,R∈U, let T∖R be the set of all elements in T which are not in R. Which one of the following is true?
Correct answer
(D) (D) ∀ X ∈ U ∀ Y ∈ U (X Y = Y' X')
Solution
Given: S={1,2,3,4,5,6}, so ∣S∣=6. U is the power set of S. T′ is the complement of T with respect to S, so ∣T′∣=∣S∣−∣T∣=6−∣T∣. T∖R means elements in T but not in R.Let's evaluate each option:(A) ∀X∈U(∣X∣=∣X′∣)
This statement claims that for every set X in the power set U, its cardinality is equal to the cardinality of its complement.
If ∣X∣=∣X′∣, then ∣X∣=6−∣X∣, which implies 2∣X∣=6, so ∣X∣=3.
This statement would only be true if every set X∈U had exactly 3 elements. This is false. For example, if X={1}∈U, then ∣X∣=1 and ∣X′∣=5. Clearly, 1=5.
Therefore, statement (A) is false.
(B) ∃X∈U∃Y∈U(∣X∣=5,∣Y∣=5 and X∩Y=∅)
This statement claims there exist two sets X,Y in U such that both have 5 elements and they are disjoint.
If ∣X∣=5 and ∣Y∣=5, and X∩Y=∅, then the union of X and Y would have ∣X∪Y∣=∣X∣+∣Y∣=5+5=10 elements.
However, X and Y are subsets of S, so X∪Y must also be a subset of S. This means ∣X∪Y∣≤∣S∣=6.
Since 10≤6 is false, it is impossible for two disjoint sets to each have 5 elements from a universal set of 6 elements.
Therefore, statement (B) is false.
(C) ∀X∈U∀Y∈U(∣X∣=2,∣Y∣=3 and X∖Y=∅)
This statement claims that for any sets X,Y in U such that ∣X∣=2 and ∣Y∣=3, it must be that X∖Y=∅.
X∖Y=∅ means that all elements of X are also in Y, i.e., X⊆Y.
So the statement claims: for all X,Y with ∣X∣=2 and ∣Y∣=3, it must be that X⊆Y.
This is false. Consider S={1,2,3,4,5,6}. Let X={1,2} and Y={3,4,5}. Here, ∣X∣=2 and ∣Y∣=3. However, X is not a subset of Y, so X∖Y={1,2}=∅.
Therefore, statement (C) is false.
(D) ∀X∈U∀Y∈U(X∖Y=Y′∖X′)
Let's analyze the left side: X∖Y is the set of elements in X that are not in Y. This can be written as X∩Y′.
Let's analyze the right side: Y′∖X′ is the set of elements in Y′ that are not in X′. This can be written as Y′∩(X′)′.
By the double complement rule, (X′)′=X.
So, Y′∖X′=Y′∩X.
Since set intersection is commutative, Y′∩X=X∩Y′.
Thus, X∖Y=X∩Y′ and Y′∖X′=X∩Y′. Both sides are equal.
This is a fundamental set identity (also known as the definition of set difference in terms of intersection and complement, applied to both sides).
