In probability you combined events with "or," "and" and "not." Events are sets of outcomes, so those were set operations all along. Here they get their general names and notation, building on sets and subsets.
Idea clave
For sets inside a universe $U$: the union $A \cup B$ is everything in $A$ or $B$ (or both); the intersection $A \cap B$ is everything in both; the difference $A \setminus B$ is what is in $A$ but not $B$; the complement $A^{c} = U \setminus A$ is everything in $U$ not in $A$. De Morgan's laws hold for sets: $(A \cup B)^{c} = A^{c} \cap B^{c}$ and $(A \cap B)^{c} = A^{c} \cup B^{c}$.
Por qué funciona
Each operation is a logical connective applied to membership. $x \in A \cup B$ means $x \in A \vee x \in B$; $x \in A \cap B$ means $x \in A \wedge x \in B$; $x \in A^{c}$ means $\neg(x \in A)$. So every logical equivalence becomes a set identity. De Morgan's laws for sets are the same laws you proved with truth tables, with "or" and "and" replaced by $\cup$ and $\cap$. (Probability books often write $A'$ for the complement; it means the same as $A^{c}$.)
Worked examples
Ejemplo resuelto: Computing with a universe
Let $U = \{1, 2, \ldots, 10\}$, $A = \{2, 4, 6, 8, 10\}$ (the even numbers) and $B = \{3, 6, 9\}$ (the multiples of $3$). Find $A \cup B$, $A \cap B$, $A \setminus B$, $A^{c}$ and $(A \cup B)^{c}$.
- $A \cup B = \{2, 3, 4, 6, 8, 9, 10\}$ (each element once).
- $A \cap B = \{6\}$, and $A \setminus B = \{2, 4, 8, 10\}$.
- $A^{c} = \{1, 3, 5, 7, 9\}$, the odd numbers.
- $(A \cup B)^{c} = \{1, 5, 7\}$: in neither set. Check De Morgan: $$A^{c} \cap B^{c} = \{1, 3, 5, 7, 9\} \cap \{1, 2, 4, 5, 7, 8, 10\} = \{1, 5, 7\}.$$
Respuesta: $\{2, 3, 4, 6, 8, 9, 10\}$, $\{6\}$, $\{2, 4, 8, 10\}$, $\{1, 3, 5, 7, 9\}$, $\{1, 5, 7\}$
Ejemplo resuelto: Sizes with De Morgan
In a universe of $40$ people, $|A| = 18$, $|B| = 15$ and $|A \cap B| = 6$. How many people are in $A^{c} \cup B^{c}$?
- De Morgan: $A^{c} \cup B^{c} = (A \cap B)^{c}$.
- So the count is everyone except the $6$ people in both: $40 - 6 = 34$.
Respuesta: $34$
Error común
Complementing each piece without switching the operation. $(A \cap B)^{c}$ is not $A^{c} \cap B^{c}$. To be outside "both $A$ and $B$," you only need to miss one of them, so $(A \cap B)^{c} = A^{c} \cup B^{c}$. Check with one element: in the example above, $3$ is in $(A \cap B)^{c}$ (it is not in both), but $3$ is not in $A^{c} \cap B^{c}$ (it is in $B$).