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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:

  • $S1$: There is a subset of $S$ that is larger than every other subset.
  • $S2$: There is a subset of $S$ that is smaller than every other subset.

Which one of the following is $\text{CORRECT}$?

  1. Both $S1$ and $S2$ are true
  2. $S1$ is true and $S2$ is false
  3. $S2$ is true and $S1$ is false
  4. Neither $S1$ nor $S2$ is true
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8 Answers

Best answer
98 votes
98 votes

Symmetric difference (SD) - suppose $A$ and $B$ are $2$ sets then symmetric difference of A and B is $(A-B)\cup(B-A) = (A\cup B)-(A\cap B).$

In question : U < V if the minimum element in the symmetric difference of the two sets is in U . Example: $\{1,2,3\} <\{2,3,4,5,6\}$ 

Symmetric difference is $\{1\} \cup \{4,5,6\}$.

Now 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 $\emptyset$ (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, according to the question $\{1,2,3,4\}$ is smaller than any other subset of $S.$ S2 is true.

Now consider 

$U=\emptyset$ 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 $\emptyset.$

So, according to the que, $\emptyset$ is greater than any other subset of $S.$ S1 is also true.

This is true even when  $S= \{1,2,3,\ldots,2014\}.$

So, answer is A. Both S1 and S2 are true

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34 votes
34 votes
S1 seems satisfied by {L} where L is largest element in S, only until we compare it to {}, where symm. diff. is {L}. Now consider {}. Any other subset of S is smaller than {} as the minimum element in their symmetric difference will be in that set. So, {}, satisfies S1, any other subset should be less than it.

S2 on the other hand, will be satisfied by S, as any other subset will be like S-{some other elements}. So symm. diff. will be {some other elems}, which will belong in S, so min. elem. will belong in S. So, that's it - (A)
11 votes
11 votes
According to the given information :

S1 is true because NULL set is smaller than every other set.

S2 is true because the UNIVERSAL set {1, 2, …, 2014} is larger than every other set.

 
Thus, both S1 and S2 are true.
5 votes
5 votes

Statement-1) is TRUE because { } > another subset because

           Let S1 denote any other subset of S other than { } 

           S1 - { } = S1 ... Now smallest element in set difference is nothing but smallest element in S1 .. So S1 < { } whatever subset you take for S1 apart from { }...

Statement-2) is TRUE because S > anyother subset of S other than S

           Let S2 be any other subset of S other than S

           S - S2 = some subset S3 of S ... Now Smallest element is nothing but element in S but not in S2 .. Since smallest element is from S ,  S < S2 whatever values you substitute for S2..

So Option A) is TRUE...

Answer:

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