Use of 'and' and 'or' in Union and intersection of sets. What would be the complement of...
$\{$x:x is a natural number divisible by 3 and 5$\}$
I checked it's solution and it kind of stumped me...
$\{$x:x is a positive integer which is not divisible by 3 or not divisible by 5$\}$
Why the word or has been used in the solution? Why not and?
 A: An element $x\in\mathbb N_+$ is in the complement of the set you mention if and only if it is not true that $x$ satisfies both the conditions that are mentioned. So $x$ will not satisfy the first or $x$ will not satisfy the second. It is possible that in that case $x$ will not satisfy both (that is not excluded by "or") but that is not necessary.
A: Let $x$ be natural by definiton. 
Let $P(x)$ be the statement "$x$ is divisible by 3".
Let $Q(x)$ be the statement "$x$ is divisible by 5".
Each statement may be either true or false, depending on $x$.
Let $A = \{x \;| \;P(x)\}$,  $B = \{x\;| \;Q(x)\}$. Such sets are called "the truth sets of $P(x)$ and $Q(x)$".  
$A$ is the set of all natural numbers divisible by 3. $B$ is a set of all natural numbers divisible by 5.
By definition we have
$$
A \cup B = \{x\;| \;P(x) \lor Q(x) \},\\
A \cap B = \{x\;| \;P(x) \land Q(x) \},\\
A^{c} = \{x\;| \; \lnot P(x)\}, \\ B^{c} = \{x\;| \; \lnot Q(x)\}.
$$
The set of all natural $x$ divisible by 3 and 5 is $A \cap B$. Its complement is $(A \cap B)^{c}=A^c \cup B^c$. 
It follows that
$$
A^c \cup B^c = \{x\;| \; \lnot P(x) \lor \lnot Q(x) \}.
$$
Note that the symbols "$\cap$" and "$\land$" look similar, and so do the symbols "$\cup$" and "$\lor$". 
P.S. For further understanding of logic and set theory (and their interconnection) I would recommend you to read the book "How to Prove It" by Daniel J. Velleman.
