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
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Answer given-
$S_1$ is true because NULL set is smaller than every other set.
$S_2$ is true because the UNIVERSAL set $\{1, 2, …, 2014\}$ is larger than every other set.
But i think its correct explanation is this-
$S_1$ is true because $∅$ is greater than any other subset of $S$.
$S_2$ is true because $\{1, 2, 3, …., 2014\}$ is a subset of $S$ that is smaller than every other subset.
Please tell which is correct and why other is wrong?
Any help is appreciated in advance!
Best Answer
Consider a smaller set. Suppose S={1,2,3,4}
Now the given 2 statements are about smallest and largest subset. So considering set S and ∅ (empty set) will be helpful.
First take U={1,2,3,4} and V={1,2} (we can take any set other than ∅ and S)
SD={3,4} (just exclude the elements which are common in the 2 sets)
Minimum element of SD is 3 which is in U and if we observe carefully minimum element will always be in U. Whatever the V is.
So acc. to the question {1,2,3,4} is smaller than any other subset of S.
Therefore, S2 is true.
Now consider U=∅ and V={1,2} (we can take any subset of S)
SD={1,2}
The symmetric difference will always be equal to V.So minimum element of SD will always exist in V when U is ∅.
So acc. to the que, ∅ is greater than any other subset of S.
Therefore, S1 is also true.
This is true even when S={1,2,3,…,2014}.
So answer is A. Both S1 and S2 are true