Trouble Proving The Principle of Duality with Regards to Sets

Question

Let's say I wanted to prove the principle of duality using a concrete example of some sets.

We will define a statement and some sets as follows:

$A∪B = \mathbb{U}$

$\mathbb{U} = \{1, \,2, \,3, \,4, \,5, \,6, \,7\}$

$A = \{1,\,2, \,3,\,4, \,5\}$

$B = \{4,\,5, \,6,\,7\}$

Now, if we apply the principle of duality, we come up with the following statement:

$A∩B = ∅$

This doesn't seem to be true, as

$A∩B = \{4, \,5\}$

Is the principle of duality only true if the sets don't intersect, or am I doing something wrong?

Answer 1

You're probably thinking of:

The complement of the union of two sets is equal to the intersection of their complements and the complement of the intersection of two sets is equal to the union of their complements. These are called De Morgan's laws.

When the second, for instance, is applied to your sets, we get: $(A\cap B )^c=A^c\cup B^c$ or $\{4,5\}^c=\{1,2,3,6,7\}=\{1,2,3,4,5\}^c\cup\{4,5,6,7\}^c $, which is true since $\{1,2,3,4,5\}^c=\{6,7\}$ and $\{4,5,6,7\}^c=\{1,2,3\}$.

Answer 2

I know this is an old question, but since you didn't get a good answer: the principle of duality only applies to equivalences for arbitrary sets. As soon as you get to concrete examples, it doesn't apply. For example, if you define C = A ∩ B, it would be very unusual for C = A ∪ B to also be true. However, for arbitrary sets A, B and C, C = A ∩ B is not a valid equivalence, so it's not valid to apply the principle of duality in the first place.

This is in contrast to something like the commutative principle: A ∪ B = B ∪ A for all sets A and B. Therefore, you can apply the duality principle to derive A ∩ B = B ∩ A.