Basics of Set Theory : complement of an intersection.

Question

My maths textbook says,

1) If x ∉ (A∩B)

=> x ∉ A or x ∉ B

2) If, A = {x:x is divisible by 3 and 5}

=> A' = {x:x is not divisible by 3 or x is not divisible by 5}

The italicised parts are troubling me. I am unable to visualise them or even comprehend them, especially the contextual meaning of the word "or" used in them. Please help me to understand these statements.

Answer 1

• The intersection of a set $A$ and of a set $B$ , namely the set $A\cap B$ , is defined using the logical operator " and " ( symbol : $\land$)

$A\cap B$ is the set of all objects that belong both to A and to B :

$A\cap B= \{x| x \in A \land x\in B\}$

• Consequently, in order to define the complement of the set $A\cap B $ we need to know what is the negation of an and-statement. And DeMorgan's law tells us that

"it is not the case that both sentence $P$ and sentence $Q$ are true"

is equivalent to

" either $P$ is false OR $Q$ is false".

Saying that " Peter is not both a pianist and a guitarist" means that " either Peter is not a pianist OR Peter is not a guitarist".

Note : you can check using a truth table that DeMorgan's law is actually a logical equivalence.

• If we apply this inside the set builder notation, with

$A$ = the set of all $x$ such that $3$ divides x

$B$= the set of all $x$ such that $5$ divides $x$ ,

we get :

$(A\cap B)'$

= the set of all $x$ that do NOT belong both to$ A$ and to $B$.

$ = \{x| \neg (3|x \land 5|x)\}$

$ = \{x| \neg (3|x) \lor \neg (5|x)\}$

Answer 2

We also know that $(P \implies Q) \iff (\lnot Q \implies \lnot P)$. Say that it is not true that either $x \not \in A$ or $x \not \in B$. Then $x \in A$ and $x \in B$. Then $x \in A \cap B$. Therefore $x \not \in (A \cap B) \implies (x \not \in A \lor x \not \in B)$.

Now in this particular case, say that it is not true that $x$ is divisible by both $3$ and $5$. Then $x$ is either not divisible by $3$ or not divisible by $5$ (or both).