MHB Let A and B be two subsets of some universal set. Prove that....

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If $(A\cup B)^c = A^c \cup B^c$, then it can be proven that A equals B. The proof begins by taking an arbitrary element x from A, demonstrating that x must also belong to B, thus establishing A as a subset of B. To show B is a subset of A, a similar argument is applied by starting with an arbitrary element from B. The symmetry in the expressions for A and B confirms that both subsets contain each other. Therefore, the conclusion is A equals B.
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**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 $B\subseteq A$?
 
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KOO said:
**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...
 
Or...just note that both expressions are symmetric in A and B, and union is commutative...
 

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