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

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SUMMARY

The discussion focuses on proving that if the complement of the union of two subsets A and B of a universal set is equal to the union of their complements, specifically $(A \cup B)^c = A^c \cup B^c$, then A must equal B. The proof demonstrates that if an arbitrary element x belongs to A, it must also belong to B, establishing that A is a subset of B. The symmetry of the expressions allows for a similar argument to show that B is a subset of A, thereby concluding that A equals B.

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KOO
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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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