For any non-empty set $A$ -

(a) $A \cup A' = A$
(b) $A \cup A' = A'$
(c) $A \cup A' = U$
(d) $A \cup A' = \phi$

Question

For any non-empty set $A$ -

(a) $A \cup A' = A$
(b) $A \cup A' = A'$
(c) $A \cup A' = U$
(d) $A \cup A' = \phi$

Accepted Answer

>

> [!TLDR]
> A set and its complement together cover the whole universal set, so $A \cup A' = U$.
>
> $A'$ (the complement of $A$) consists of every element of the universal set $U$ that is NOT in $A$. So $A$ contains exactly the elements $A'$ is missing, and together they contain every element of $U$ exactly once. Hence $A \cup A' = U$.
> …

Operation	What it asks	Bucket contains
Union ( $A \cup B$ )	In either?	Everything from $A$ or $B$ (or both)
Intersection ( $A \cap B$ )	In both?	Only the overlap
Difference ( $A \setminus B$ )	In $A$ but not $B$ ?	Just the part of $A$ that doesn't touch $B$
Complement ( $A^c$ )	Not in $A$ ?	Everything outside $A$ (inside $U$ )
Symmetric Difference ( $A \triangle B$ )	In exactly one?	The two crescent-shaped parts, excluding the overlap

Q.For any non-empty set $A$ -

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Q.For any non-empty set AAA -

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