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.

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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 $\text{M}$?

Which of the following sets forms a partition of $\text{M}$?

- $\{\mathrm{A}, \mathrm{B}\}$
- $\left\{\mathrm{A}, \mathrm{B}_{,} \mathrm{A}^{\prime} \cap \mathrm{B}^{\prime}\right\}$
- $\left\{\mathrm{A} \cup \mathrm{B}, \mathrm{A}^{\prime} \cap \mathrm{B}^{\prime}\right\}$
- $\left\{\mathrm{A} \cap \mathrm{B}, \mathrm{A}-\mathrm{B}, \mathrm{B}-\mathrm{A}, \mathrm{A}^{\prime} \cap \mathrm{B}^{\prime}\right\}$

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