4 4 votes Let $\text{A, B}$ be two disjoint non-empty sets. Let $\text{M}$ be the universal set and $\text{A} \cup \text{B}$ is a proper subset of $\mathrm{M}$. For any set $\mathrm{S}$, let $\mathrm{S}^{\prime}$ be the set of those elements which are in $\mathrm{M}$, but not in $\mathrm{S}$.Which of the following sets forms a partition of $\text{M}$? $\{\mathrm{A}, \mathrm{B}\}$ $\left\{\mathrm{A}, \mathrm{B}_{,} \mathrm{A}^{\prime} \cap \mathrm{B}^{\prime}\right\}$ $\left\{\mathrm{A} \cup \mathrm{B}, \mathrm{A}^{\prime} \cap \mathrm{B}^{\prime}\right\}$ $\left\{\mathrm{A} \cap \mathrm{B}, \mathrm{A}-\mathrm{B}, \mathrm{B}-\mathrm{A}, \mathrm{A}^{\prime} \cap \mathrm{B}^{\prime}\right\}$ Set Theory & Algebra goclasses-scholarship-test1 goclasses set-theory&algebra set-theory multiple-selects two-marks + – GO Classes 901 views answer comment Share Follow Print 0 reply Please log in or register to add a comment.
3 3 votes $\text{A} \cap \text{B}$ is empty set, and empty set is never a part of any partition. Partition of a set $\text{S}$ is a collection/set of non-empty disjoint subsets of $\mathrm{S}$ whose union is $\mathrm{S}$. Detailed Video Solution: https://youtu.be/tqjuxfutFHg?t=2277 PS: If the Set $S$ itself is Empty then Set $S$ has Exactly One Partition, that is $\{ \} \,\, i.e. \phi$ https://en.wikipedia.org/wiki/Partition_of_a_set#:~:text=is%20finite.-,Examples,-%5Bedit%5D GO Classes answered Aug 6, 2022 • edited Aug 8, 2022 by Deepak Poonia GO Classes comment Share Follow See all 2 Comments 2 2 Comments reply ankitgupta.1729 commented Aug 8, 2022 reply Follow flag empty set is never a part of any partition Empty set can be a partition of a set if the set is empty set itself. More precisely, empty set is a partition of an empty set. 2 2 replyShare Deepak Poonia commented Aug 8, 2022 reply Follow flag That’s right. Have put it in the answer. 0 0 replyShare Please log in or register to add a comment.
