GATE CSE 2015 Set 3 | Question: 23

Question

Suppose $U$ is the power set of the set $S = \{1, 2, 3, 4, 5, 6\}$. For any $T \in U$, let $|T|$ denote the number of elements in $T$ and $T'$ denote the complement of $T$. For any $T, R \in U \text{ let } T \backslash R$ be the set of all elements in $T$ which are not in $R$. Which one of the following is true?

• A. $\forall X \in U, (|X| = |X'|)$
• B. $\exists X \in U, \exists Y \in U, (|X| = 5, |Y|=5$ and $X \cap Y = \phi)$
• C. $\forall X \in U, \forall Y \in U, (|X| = 2, |Y|=3$ and $X \backslash Y = \phi)$
• D. $\forall X \in U, \forall Y \in U, (X \backslash Y = Y' \backslash X')$

Comments

@Prince07 the last statement would have been sufficient .( The only possibilities for X and Y are: {1,2,3,4,5}, {1,2,3,4,6}, {1,2,3,5,6}, {1,2,4,5,6}, {1,3,4,5,6}, {2,3,4,5,6}. So X∩Y  will NEVER be ϕ.  )

Thanks however.

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option c is true for the existential quantifier.

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we can prove option D true→ 

we know that A\B(or A-B)=A∩B’.

So X\Y=X∩Y’

Similarly Y’\X’=Y’∩(X’)’=Y’∩X.

Hence X\Y=Y’\X’. :)

Answer 1

Answer is $D$.

As $X$ and $Y$ elements of $U$, $X$ and $Y$ are subsets of $S$.

Option $A$ is wrong consider $X=\{1,2\}$ therefore $X'=\{3,4,5,6\}$, $|X|=2$ and $|X'|=4$.

Option $B$ is wrong as any two possible subsets of $S$ with 5 elements should have atleast 4 elements in common (Pigeonhole principle). Hence, $X$ intersection $Y$ cannot be null.

Option $C$ is wrong, $X$ and $Y$ can have any number of elements from $0$ to $5$. Even for the given constraint, consider $X=\{1,2\}, Y=\{3,4,5\}$ and $X\backslash Y=\{1,2\}$ which is not null.

Comments

∃X∈U,∃Y∈U,(|X|=5,|Y|=5∃X∈U,∃Y∈U,(|X|=5,|Y|=5 and X∩Y=ϕ)

this is also true as

∃X∃Y is true if we even came up with a single example

say X={{1},{2},{3},{4},{5}}

       Y={{1,2}{1,3}{1,4}{1,5}{1,6}}

|X|=5

|Y|=5

X∩Y=ϕ

here U is a powerset not a subset

∃X∈U,∃Y∈U

best answer needs correction

correct me if i am wrong

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Musa X,Y are the members of U not the subset.

X∈U means X is present in U or X is member/element of U.

X ⊂ U means every element of X must be present in U.

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Thanks for the explanation himanshu2021

Answer 2

(1) X \ Y = X - Y

(2) Y' \ X' = Y' - X' = (U - Y) - (U - X)  = X - Y  (U represents universal set for X and Y. Here it is power set.)

Equation (1) is equal to (2) So answer is (D) Part.

Answer 3

X ∕  Y=X∩(!Y)

(!Y)/(!X)=(!Y)∩(X)=X∩(!Y)

Comments

So, you assume Universal set =U = {1,2,3,4,5,6} that is why  comp{1} = {1,2,3,4,5,6} -{1} ={2,3,4,5,6}.

Plz, clarify.