Prove A = B given (A∪B)ᶜ = Aᶜ∪Bᶜ

[KOO]

**Let A and B be two subsets of some universal set.
Prove that if $(A\cup B)^c$ = $A^c$ U $B^c$, then A = B.**Attempt:

Let $x\in A$. Then $x\in A\cup B$, so $x\notin(A\cup B)^c$. By hypothesis $(A\cup B)^c=A^c\cup B^c$, so $x\notin A^c\cup B^c$. In particular, then, $x\notin B^c$, and therefore $x\in B$. Since $x$ was an arbitrary element of $A$, this shows that $A\subseteq B$.

How do we show $B\subseteq A$?

 

[I like Serena]

**Let A and B be two subsets of some universal set.
Prove that if $(A\cup B)^c$ = $A^c$ U $B^c$, then A = B.**Attempt:

Let $x\in A$. Then $x\in A\cup B$, so $x\notin(A\cup B)^c$. By hypothesis $(A\cup B)^c=A^c\cup B^c$, so $x\notin A^c\cup B^c$. In particular, then, $x\notin B^c$, and therefore $x\in B$. Since $x$ was an arbitrary element of $A$, this shows that $A\subseteq B$.

How do we show $B\subseteq A$?

Start with "Let $x\in B$."
Then continue with the same argument you have - just with $A$ and $B$ swapped around...

 

[Deveno]

Or...just note that both expressions are symmetric in A and B, and union is commutative...