GO Classes CS 2025 | Weekly Quiz 4 | Set Theory | Question: 8

Answer 1

Detailed Video Solution with timestamp : https://youtu.be/mOfUwNN2JP8?si=D9U1AnBcUd8-fuhj&t=2594

For Option C:
Counterexample: $S = \{1, 2, \dots, 8 \}$, here $S$ is Not a powerset of any set T.

Comments

the questions asks for some set, not saying its true for all sets right?

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The option C says . For any arbitrary set Let say S. If |S|=8 than there exist a set T such that power set of T = S.

"The statement is like for all sets there exist some T".I didn't notice that carefully while giving quiz and got this ques wrong.

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I think it says for Set S, there exists some T such that S = P(T)
Not for all but for Set S
So it should be True

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@VedantGawande, "If $S$ is a set, ....." means that $S$ can be ANY set, & that this statement is made for arbitrary set $S.$

So, Only option $A$ is true, because it is true for any set $S.$

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hey a c d must have been true if in last part it was written mod s = mod p(t)?right

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So in all the options we have to check for any set s whether the given statement is true or not.

Answer 2

Correct Answer- $A$

The reasoning for $C$ not being a solution is this-

We know that, if a set $A$ is Power set of some set $B$, then the cardinality of $A$ is given by $|A|=|P(B)|=2^{|B|}$.  (Think of it as $p\to q$)

Can you conclude that the converse of the above conditional statement ($q\to p$) is true? Answer is NO. It may be true or may not be true. It cannot be determined with certainty.

Hence, if you have a set with cardinality being equal to a power of $2$, say $2$ or $4$ or $8$, you cannot determine whether it's a power set or not.