# Is a subset of the set $S$ of integers, is larger than every other subset or is smaller than every other subset?

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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$$S_1$$ 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!

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

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 $$S2$$.

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