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Official PYQ · 2014

GATE CS 2014 Set 2 Q50 — Sets, Relations and Functions

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Consider the following relation on subsets of the set SS of integers between 11 and 20142014. For two distinct subsets UU and VV of SS we say U < VU\:<\:V if the minimum element in the symmetric difference of the two sets is in UU.

Consider the following two statements:

  • S1S1: There is a subset of SS that is larger than every other subset.
  • S2S2: There is a subset of SS that is smaller than every other subset.

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

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Answer: A

To determine the truth of statements S1S1 and S2S2, let us analyze the given relation on the power set P(S)\mathcal{P}(S) of the finite set S={1,2,…,2014}S = \{1, 2, \dots, 2014\}.

For any two distinct subsets U,V⊆SU, V \subseteq S, their symmetric difference UΔV=(U∖V)∪(V∖U)U \Delta V = (U \setminus V) \cup (V \setminus U) is non-empty. Since SS is finite, UΔVU \Delta V has a unique minimum element m=min⁡(UΔV)m = \min(U \Delta V). Since mm must belong to exactly one of the two sets:

  • If m∈Um \in U, then U<VU < V.
  • If m∈Vm \in V, then V<UV < U.

This means the relation << is a total order (strictly speaking, a strict total order) on the finite set P(S)\mathcal{P}(S). Any finite totally ordered set must have a unique minimum (smallest) element and a unique maximum (largest) element.

We can explicitly identify these elements:

  • Smallest element: The set SS. For any V≠SV \neq S, the minimum element of SΔV=S∖VS \Delta V = S \setminus V is always in SS, so S<VS < V for all V≠SV \neq S. Thus, S2S2 is true.
  • Largest element: The empty set ∅\emptyset. For any V≠∅V \neq \emptyset, the minimum element of ∅ΔV=V\emptyset \Delta V = V is always in VV, so V<∅V < \emptyset for all V≠∅V \neq \emptyset. Thus, S1S1 is true.

Therefore, both statements S1S1 and S2S2 are true.

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