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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:

$S_1$: 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!

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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

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The symmetric difference of the empty set (or "null set") and any other sub-set $V$ of $S$ is $V$ itself. So the minimum member of the symmetric difference is definitely in $V$. And so the empty set is smaller larger than any other sub-set of $S$.

Similarly the symmetric difference of $S$ and any other sub-set $U$ of $S$ is the complement of $U$. So the minimum member of the symmetric difference cannot be in $U$ (and we know it is in $S$). And so $S$ is larger smaller than any other sub-set of $S$.

It looks like the answer given has simply switched the statements $S1$ and $S2$.

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  • $\begingroup$ thank you for answering, but is my explanation wrong $\endgroup$ – Geeklovenerds Oct 25 '18 at 13:23
  • $\begingroup$ @Geeklovenerds Yes, $S1$ is true because $S$ is larger than every other sub-set of $S$. $S2$ is true because the empty set is smaller than every other sub-set of $S$. $\endgroup$ – gandalf61 Oct 25 '18 at 15:34
  • $\begingroup$ @ gandalf61 Can you please see the answer below and tell if it is correct or not? $\endgroup$ – Geeklovenerds Oct 26 '18 at 3:58
  • $\begingroup$ @Geeklovenerds Yes it is correct. I had my smaller/greater logic the wrong way round. I have amended my answer. $\endgroup$ – gandalf61 Oct 26 '18 at 8:56
  • $\begingroup$ @ gandalf61 just one more silly doubt- can i now say that book/given answer is wrong and mine correct? Also, thank you for replying. $\endgroup$ – Geeklovenerds Oct 26 '18 at 16:23

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