GO Classes CS 2025 | Weekly Quiz 4 | Set Theory | Question: 10

Answer 1

A)

Let's say $\text{A} = \set{\set{2,3,4}} \text{ and B } = \set{\set{2,3}}$ then $\text{A -B} = \set{\set{2,3,4}} $ and moreover $\set{4} \not\subseteq \text{A-B}$.

$ $

$\text{Let A} = \set{4, \set{2,3,4}} \text{ and B } = \set{\set{2,3}}$ then $\text{A -B} = \set{4, \set{2,3,4}} $ and $\set{4} \subseteq \text{A-B}$.

So I think option A is $\text{NOT always true and NOT always false}$.

$ $

B) $\text{S} \subseteq \text{A} \cap\text{B} $ then $\text{S} \subseteq \text{A}$ $  \land \subseteq \text{B} $.

$ $

Assume $\text{S} \subseteq \text{A} \cap \text{B} $.

<-->$\forall \text{x}( \text{x} \in \text{S} \mapsto \text{x} \in \text{A} \cap\text{B})$

$ $

<--> $\forall \text{x}( \text{x} \in \text{S} \mapsto( \text{x} \in \text{A} \land \text{x} \in \text{B}))$

$ $

$\leftrightarrow$ $\forall \text{x}( \text{x} \notin \text{S} \lor( \text{x} \in \text{A} \land \text{x} \in \text{B}))$

$ $

$\leftrightarrow$ $\forall \text{x}( (\text{x} \notin \text{S} \lor \text{x} \in \text{A} )\land( \text{x} \notin S \lor \text{x} \in \text{B}))$

$ $

$\leftrightarrow$ $\forall \text{x}( (\text{x} \in \text{S} \mapsto \text{x} \in \text{A} )\land( \text{x} \in S \mapsto \text{x} \in \text{B}))$

$ $

$\leftrightarrow$ $\forall \text{x}( (\text{x} \in \text{S} \mapsto \text{x} \in \text{A} )) \land\forall \text{x}( \text{x} \in S \mapsto \text{x} \in \text{B})$

$ $

$\leftrightarrow$ $\text{S} \subseteq \text{A} \land \text{S} \subseteq \text{B}$

So  $\text{S} \subseteq \text{A} \cap \text{B} $ $\mapsto$ $\text{S} \subseteq \text{A} \land \text{S} \subseteq \text{B}$

So option B is $\text{TRUE.}$.

$ $

C) Assume  $\set{3,4} \subseteq \text{ A-B } $ and $ \set{1,2} \subseteq \text{B} $ and $\set{1,2,3,4} \not\subseteq A \cup B$.

$\mapsto \exists\text{x}( \text{x} \in \set{1,2,3,4} \land \text{x} \not\in A \cup B)$.

$\text{Such an $\text{x}$ can't be 1 or 2 because there are already in B so they also will be a part of}$ $A \cup B$ $\text{ and it can't be 3 or 4 because there are already in A so they also will be a part of}$ $A \cup B$.

So by proof by Contradiction

$\set{3,4} \subseteq \text{ A-B } $ and $ \set{1,2} \subseteq \text{B} $ $\mapsto$ $\set{1,2,3,4} \subseteq A \cup B$

So option C is true.

$ $

D) Similarly one can give proof for option D, it is also a True statement.

So I think option A is closest one, rest of all are True.

Comments

That's what I am also thinking, Option A is not always True and not always False it's more like a Contingency.

---

Answer is Option A. Answer key is corrected now. Refresh to recalculate your marks & rank.

Detailed Video Solution: https://youtu.be/mOfUwNN2JP8  

---

So the key point here is that we need to be careful whether the given set is arbitrary set or a well defined set and if it's an arbitrary set then we have to check for all the cases.