Question

Consider the following relation on subsets of the set $S$ of integers between 1 and 2014. For two distinct subsets $U$ and $V$ of $S$ we say $U<V$ if the minimum element in the symmetric difference of the two sets is in $U$.
Consider the following two statements:
$S_1$ : There is a subset of $S$ that is larger than every other subset.
$S_2$ : There is a subset of $S$ that is smaller than every other subset.
Which one of the following is CORRECT?

• A. Both $S_1$ and $S_2$ are true
• B. $S_1$ is true and $S_2$ is false
• C. $S_2$ is true and $S_1$ is false
• D. Neither $S_1$ nor $S_2$ is true

Answer 1

The correct option is A Both $S_1$ and $S_2$ are true

Given the following details:
$S=\{1,2,3,4,\dots \dots 2014\}$
$U<V$ if the minimum element in symmetric difference of the two sets is in
$S_1$: There is a subset of $S$ that is larger than every other subset.
$S_2$: There is a subset of $S$ that is smaller than every other subset.
$S_1$ is true since $\varphi$ (which is subset of $S$) is larger then every other subset.
i.e. $\forall vv<\varphi$ is true since the minimum element in $v\Delta \varphi =v$, is in $v$
(Note: $v\Delta \varphi =v{\varphi }^{{}^{\prime }}+v^{{}^{\prime }}\varphi =v$)
$S_2$ is also true since $S$ (which is subset of $S$) smaller then every other subset.
i.e. $\forall vS<v$ is true since the minimum element in $S\Delta v=v^{{}^{\prime }}$, is in $S$.
(Note: $S\Delta v=S{v}^{{}^{\prime }}+S^{{}^{\prime }}v=1v^{{}^{\prime }}+0v=v^{{}^{\prime }}$)
$\therefore$ Both $S_1$ and $S_2$ are correct.