Question

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

• A. $\forall XϵU(|X|=|X^{{}^{\prime }}|)$
• B. $\exists XϵU\forall YϵU(|X|=5,|Y|=5)$ and $X\cap Y=\Phi$
• C. $\forall XϵU\forall YϵU(|X|=2,|Y|=3)$ and $X$\$Y=\Phi$
• D. $\forall XϵU\forall YϵU$ ($X$ \ $Y=Y^{{}^{\prime }}$ \ $X^{{}^{\prime }})$

Answer 1

The correct option is D $\forall XϵU\forall YϵU$ ($X$ \ $Y=Y^{{}^{\prime }}$ \ $X^{{}^{\prime }})$

$S=\{1,2,3,4,5,6\}$
$U$ is the power set of $S\Rightarrow U=P\left(S\right)$
$U=\left\{\right\{\left\}\right\{1\},\{2\},\dots \{1,2\},\{1,3\},\dots \{1,2,3\},\dots \{1,2,3,4\},\dots \{1,2,3,4,5\}\dots \{1,2,3,4,5,6\left\}\right\}$
$|U|=2^6=64\text{elements}$
$TϵU\Rightarrow T^{{}^{\prime }}ϵU$
$RϵU,T$ \ $R=T-R$
(a) $\forall XϵU(|X|=|X^{{}^{\prime }}|)$ means that every subset of $S$ has same size as its complement. Clearly this is False.
(For example, $\left\{1\right\}ϵU$, and complement of $\left\{1\right\}=\{2,3,4,5,6\})$

(b)$\exists XϵU\forall YϵU(|X|=5,|Y|=5)$ and $X\cap Y=\Phi$ means that there are two 5 element subsets of $S$ which have nothing in common. This clearly False.
Since any two 5 element subsets will have atleast 4 elements in common.

(c) $\forall XϵU\forall YϵU(|X|=2,|Y|=3)$ and $X$\$Y=\Phi$ means that every 2 element subset $X$ and every 3 element subset $Y$ will have $X-Y=0$ i.e., $X\subseteq Y$.
This is clearly False as can be seen from the example $X=\{1,2\},Y=\{3,4,5\}$ where $X\text{/}\subseteq Y$

(d) $\forall XϵU\forall YϵU$ ($X$ \ $Y=Y^{{}^{\prime }}$ \ $X^{{}^{\prime }})$ means that for any two subsets $X$ and $Y$, $X$ \ $Y=Y^{{}^{\prime }}$\ $X^{{}^{\prime }}$
i.e. $X-Y=Y^{{}^{\prime }}-X^{{}^{\prime }}$.
This is clearly True since by boolean algebra
LHS $=X-Y=X{Y}^{{}^{\prime }}$.
RHS $=Y^{{}^{\prime }}-X^{{}^{\prime }}=Y^{{}^{\prime }}X$ and therefore LHS = RHS.